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Some Fine Problems
by TG Team - Wednesday, 25 November 2009, 12:56 PM
 

some good quant problems for cat preparation

I was fascinated when I saw a CAT preparation website named TotalGadha for the first time. I downloaded almost every page of this MBA website on my pc and made it a regular feature of my internet browsing. Slowly I started participating in the forums, trying to answer the questions posed by intellectual community of country and started getting appreciation in the form of “Thank You” and smilies.

I started wondering why I was addicted to this site; why I was answering others’ questions on the forums. I got my answer through TG himself in one of the threads where he said “I am compulsive problem solver.” That sentence reflected me because that was true for me also. I took the message of spreading the mathematical awareness to masses at almost no cost from TG and started my journey towards CAT preparation community at my home city, Hisar.

And then came the real appreciation, may be which I was looking for- an invite from TG to join his team as he was to start his classroom Avatar “Tathagat”. I wasn’t able to join him from inception because of some commitments but now being a part of it is as challenging as rewarding.

It is challenging because you need to tackle challenges posed by largest pan India student community and rewarding because of those cute gifts in the form of smilies given by the challenge posers.

Idea is don’t run away from the challenges; because if you aren’t able to solve them no one is going to blame you. But if you solve it anyhow, fame and glory will follow. In any case you are at Profit.

Coming to the point now, here I am explaining some good questions which I have discussed in my classes at Tathagat. Some of these questions we have used in our copyCATs and some may have been discussed in TG forums also. I have catagorised the question in five categories and would like to add to this repository from time to time for the well being of online student community.

It is highly recommended to try all of the listed problems by you yourself before peeping in the solution provided.

number system problems

number system problems

number system problems

number system problems

number system problems


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Re: Some Fine Problems
by amit kheterpal - Wednesday, 25 November 2009, 03:10 PM
 

awesome article sir..smile

my take for first one:-

sum of 201 to 300 will be=>50(201+300)=25050

when each element will be multiplied it will like

101(25050)+102(25050)..............+200(25050)

=>(100+1)(25050)+(100+2)25050.........(100+100)(25050)

=>100*100(25050)+(1+2+3+4+5+6...+100)(25050)

=100*100(25050)+(5050)(25050)

=>377002500

2nd)it can be written as:-for 1 to 99

(1+2+3+4+5+6+7+8+9)(1+2+3+4+5+6+7+8+9)

for 101 to 199(101 to 110 we will not include it)

from 111 to 199 it will be like 1*(1+2+3+4+5+6+7+8+9)(1+2+3+4+5+6+7+8+9)

from 222 to 299 it will be like 2*(1+2+3+4+5+6+7+8+9)(1+2+3+4+5+6+7+8+9)

similarly continuing for others

at last we will get it like this:-

(1+1+2+3+4+5+6+7+8+9)[(1+2+3+4+5+6+7+8+9)(1+2+3+4+5+6+7+8+9)]

46*(45)^2=93150

thankyou sir..

Re: Some Fine Problems
by Rakesh Sah - Wednesday, 25 November 2009, 03:27 PM
 

"Kamal Bhai" to "Kamal Sir"..... Ek shaandaar safar Ek Jaandaar safar!!

I guess the above statement somewhat sums up your amazing journey. I am sure all the "Gadhas" of Gadhaland would be proud to have yet another gem of a teacher to provide them the necessary guidance. I am sure it must be an honour to stand with the likes of TG Sir and Dagny Mam..

Kudos to you Kamal Bhai...err...Kamal Sir smile.

TG Rocks!!

Re: Some Fine Problems
by Total Gadha - Wednesday, 25 November 2009, 03:34 PM
  Hi Rakesh,

The honour is ours also. He is an amazing person to work with. smile

Total Gadha
Re: Some Fine Problems
by TG Team - Wednesday, 25 November 2009, 04:12 PM
 

Hi Rakeshsmile

For me, there is no difference in the tag 'Bhai' or 'Sir'. As both show heart felt gratitudes of the other person only.smile

 

Re: Some Fine Problems
by Double O Seven - Wednesday, 25 November 2009, 04:15 PM
  Nice article smile
Re: Some Fine Problems
by amit kheterpal - Wednesday, 25 November 2009, 06:21 PM
 

sir, iam not getting 3rd one 0<a<b<c<d<e<n in this n can take any value there will be many values..

4)for non-negative integral solution of a+b+c+d+e<=2009

to remove the inequality we will introduce one more variable f

it will take the form of:-a+b+c+d+e+f=2009 answer will be

                                        n-1Cr-1=2009-1C6-1=2008C5

thankyou sir.

Re: Some Fine Problems
by Software Engineer - Wednesday, 25 November 2009, 06:40 PM
  Kamal Sir? Totally Shocked........... shock
Re: Some Fine Problems
by Rahul Arora - Wednesday, 25 November 2009, 06:44 PM
  Hi Sir,

First of all, a really nice article....
Kudos to TG, for finding a gem...

I would like to add my solutions for the last two solved problems, as they seem to me faster than given approaches smile

1. The quiz game question, with scores for Alice, Jane & Kathy given, and we have to find Bob's score. In this question, if we notice Alice & Jane have got 3 wrong answers each ( as their score is 70 each). And if we notice the table, they have 6 questions as different answers, so 3 each out of these 6 questions will be answered incorrectly by both of them and rest 4 questions, with same answer from both of them, are correct, i.e. question no. 1,4,6 & 9 are answered correctly by both of them for sure. Now, if we come to the Kathy's answer key, Kathy has got 4 wrong, with the help of 4 correct answers above, we can quickly check that Kathy has got question no 1,4,6 & 9 wrong, so rest 6 of her should be correct, thus, we can have the final answer key, and we can calculate Bob's score.

2. Kamal and Rajat are trying to break a 4X12 chocolate bar, and we have to find to win, Kamal should take first or second chance.
Now to break the chocolate bar completely, we have to make a total 0f 11+12*3 (11 cuts for 12 rows, and then 3 cuts for each row) cuts, i.e. 47 cuts. As they are taking turn alternately, and each of them breaks along one straight line only, thus, the person who starts the game makes the 47th cut, and wins the game and chocolates too evil.
Re: Some Fine Problems
by TG Team - Wednesday, 25 November 2009, 06:59 PM
 

Hi SEsmile

If Gadha is TOTAL here then why shoudn't shock be TOTALsmile

Re: Some Fine Problems
by BellTheCAT ... - Wednesday, 25 November 2009, 07:15 PM
  Great job Kamal....thanx a lot...
Re: Some Fine Problems
by TG Team - Wednesday, 25 November 2009, 07:15 PM
 

Thanks Rahulsmile

For the second question which you have mentioned, "11 cuts for 12 rows and then 3 cuts for each row" need not to be in that order.

But yes, total 47 cuts for getting 48 pieces of choclates separately is must. Hence the person playing first will surely be the winner.

Nice reasoning .smile

Re: Some Fine Problems
by Rahul Arora - Wednesday, 25 November 2009, 07:26 PM
  Thanks Sir!!!
In chocolates question, I too meant the same, forgot to explain it smile
Re: Some Fine Problems
by boney yeldho - Wednesday, 25 November 2009, 09:42 PM
  In question 4 regardin the missing digit in 2^29 .. modulo 9 gives remainder 1.. so sum of digits shd be of the form 9k+1..i.e 46,37,..etc.. now since only possibility is 37 (9distinct nos)
the missin no shd be 8 (0+1+2+...7+9).
I think the answer shd be 8.

Regards thoughtful
Re: Some Fine Problems
by Total Gadha - Wednesday, 25 November 2009, 09:19 PM
  SE? Where are you rey... Dagny and I were worried that you have quit preparation completely. No post, no mail, no msg...
Re: Some Fine Problems
by nidhi soni - Wednesday, 25 November 2009, 09:48 PM
 

really awesome blossom artical :D

kudos to KL

 

i coudlnt solve it cmeplte yet :D

got few prbs

prb no 1) 2^2009

Kl how did u find 2^x will give u 7 as first digit???i mean i could get it wd help of calci :O ...anyways i didnt get it :|

 

in qs 6) tht triplet question......... y did u take mod 8???

 

qs 10) preimeter 2009......why 3y?

 

 

Re: Some Fine Problems
by vivek ghiria - Wednesday, 25 November 2009, 10:14 PM
 

Hi Boney

2^29 will give reminder 5 with 9.

so sum of digits will be of the form 9K+5 = 9(4)+5 = 41

so 4 is missing. coz sum of 0+1+..9 = 45

If 9(3)+5 = 32 is not possible.

 

Re: Some Fine Problems
by bhavesh usadad - Thursday, 26 November 2009, 01:11 AM
  thanks a lot!!!!!!!!!!
really it makes gadha!!!!!!!!!!!!!!

it's like morning exercise to brain!!!!!!!!!

Re: Some Fine Problems
by Anoop Kothari - Thursday, 26 November 2009, 02:14 AM
  Ans for Swimmer prob
20 times they will cross each other
Re: Some Fine Problems
by yash gupta - Thursday, 26 November 2009, 11:47 AM
  in the problem n^2+4/n+5 , i am not able to understand the mesning of statement hcf not= to 1 means can u explain
Re: Some Fine Problems
by yash gupta - Thursday, 26 November 2009, 11:51 AM
  y we divide by 8 only in quest x^22+ y^2+ z^2=1855 can u explain the complete method , please explain my querry soon because my paper is on 28 only
Re: Some Fine Problems
by TG Team - Thursday, 26 November 2009, 12:02 PM
 

Hi Yashsmile

Question is about improper fraction in simplest form and a fraction is in its simplest form means its numerator and denominator do not share any common factor. That is HCF(numerator, denominator) = 1

If it is not so, then fraction is not in its simplest form. Isn't it simplesmile

Re: Some Fine Problems
by Ankesh Grover - Thursday, 26 November 2009, 12:38 PM
  I love the way he puts a smile after every solution...It's like " happy to have been of help"..
and trust me, you have been of tremendous help...
thanks a lot for this effort Kamal smile

Re: Some Fine Problems
by TG Team - Thursday, 26 November 2009, 01:06 PM
 

Thanks to read my mind Ankeshsmile

I feel really very happy to be of some help. smile

Hope there is no grammatical error.

 

Re: Some Fine Problems
by - SK - Thursday, 26 November 2009, 02:15 PM
 

Too good article sir!
Thanks a lotttt...

In the parabola problem, how do you take the coordinates to be exactly in Q1 and Q3 ; it can be in other ways too..??

-shiva

Re: Some Fine Problems
by anusha badhwar - Thursday, 26 November 2009, 02:23 PM
 

Sir

What do u mean by modulo 9??

Re: Some Fine Problems
by TG Team - Thursday, 26 November 2009, 02:24 PM
 

Hi Shivasmile

I have not taken the points to lie in only Ist and IIIrd quadrant.

I have just taken the coordinates of one point say B as (a, b), then coordinates of point A will be (-a, -b) because origin (0, 0) is the midpoint of line segment AB as given.

Now it depends on the values of a and b, where the points A and B will lie.

Hope I am clear.smile

Re: Some Fine Problems
by TG Team - Thursday, 26 November 2009, 02:33 PM
 

Hi Anushasmile

When I say modulo 9, that means the remainder obtained when a certain number is divided by 9.

For example: 19 mod ≡ 1 mod 9smile

And yes, why don't you see the lessons posted by SE sirsmile, he has used the term 'modulo' much more than me. smile

Re: Some Fine Problems
by amit kheterpal - Thursday, 26 November 2009, 02:59 PM
 

@TG sir/Kamal sir

could you please check my answers for the questions which i have solved..smile

thankyou..smile

 

Re: Some Fine Problems
by nidhi soni - Thursday, 26 November 2009, 03:11 PM
  KL plz expalin these too
Re: Some Fine Problems
by TG Team - Thursday, 26 November 2009, 03:47 PM
 

Amitsmile

Just waiting for some more attemptssmile

 

Re: Some Fine Problems
by TG Team - Thursday, 26 November 2009, 03:48 PM
 

Nidhismile

Where are these??

Re: Some Fine Problems
by Prateek Gupta - Thursday, 26 November 2009, 04:12 PM
  Hello Kamal Sir

First of all hearty thanks for coming up with such a wonderful session...really enjoyed solving the questions...

ONE QUESTION:

ALGEBRA:
Q1: How many sols are there to the inequality 13a+17b<221...could you please elaborate what u did after re-arranging it to give the non-zero solutions..

Q2: For the no of possible sides of the triangle with perimeter = 30, by your method, i am getting 46 but correct ans is 19...plz HELP!!!
Re: Some Fine Problems
by nidhi soni - Thursday, 26 November 2009, 04:28 PM
 

Kl i got it  fianlly

i got the eq 6x+3y+z=1005*502 :D yipeeee

my other prbs r

given tht 2^2010 is a 606-digit number......
here i coudlnt get 7 as first digit but whn i checked it on calculator i got it it was 2^49 whihc gives us first digit 7
can u plz explain this bit more sad


another is reg integral triplets which satisfy x^2+y^2+z^2=1855
why did u take mod 8 here?

Re: Some Fine Problems
by TG Team - Thursday, 26 November 2009, 06:41 PM
 

Hi Prateeksmile

For perimeter = 30, maximum side length will be 14.

So using earlier equations, I'll have a' + b' + c' = 12

which will yield 14C2 = 91 ordered pairs. Right? smile

Now this contain the triplets where

 all three are same (only one i.e. 4, 4, 4) counted once only.

exactly two are same (6 cases e.g. (0, 0, 12) ....(6, 6, 0)) counted three times each

and all three different (Let x cases e.g. (1, 2, 9)) counted six times each.

Now calculating the value of x,

6x + 18 + 1 = 91        => x = 12.

So total cases are 12 + 6 + 1 = 19smile

Where lies the problem?smile 

 

Re: Some Fine Problems
by boney yeldho - Thursday, 26 November 2009, 07:21 PM
  @vivek

You are rite.. i messed up finding the remainder.. smileThanks mate
Re: Some Fine Problems
by Prateek Gupta - Thursday, 26 November 2009, 08:09 PM
  thank u Kamal Sir for solving the problem and making me understand that even 14*3=42 can become 52 in the back of the mind..here one for u too..smile..
Re: Some Fine Problems
by Imran Basha - Thursday, 26 November 2009, 09:01 PM
  Kamal sir, Thank you for such a wonderful article
Given that 22010 is a 606-digit number whose first digit is 1, how many elements of the set S = {20, 21, 22, …, 22009} have a first digit of 4?


The answer for this i think, should be 201.
There is a pattern for first digits upto 2^10 and numbers repeat itself after every ^10
The first digits from 2^1 to 2^10 are as follows 2 4 8 1 3 6 1 2 5 1 and again from 2^11 to 2^20 the first digits are 2 4 8 1 3 6 1 2 5 1.
So There should be (2010/10 *3 +1) digits ie 604 digits in 2^2010 and since from 2^1 to 2^10 there is only one 4 therefore there are 2010/10 = 201 fours in 2^2010
By solving with your method by taking 604 digits we get 201 also. 2010-603*3 =201

Further similar patterns are seen in 3^1 to 3^21 which are again repeated from 3^22 as 3 9 2 8 2 7 2 6 1 5 1 5 1 4 1 4 1 3 1 3 1.
in 4^1 to 4^5 as 4 1 6 2 1 and so on......

Sir, if i am wrong let me know where i went wrong.
 
Re: Some Fine Problems
by Shashank Sharma - Friday, 27 November 2009, 12:12 AM
  2^29 will have 5 as the answer .

because if you do 2^29%9 ... you can change it as 4*(8)^3 and dividing this will give remainder as 4*(-1)^3 = -4 or 5 ... so digit is 5

please enlighten me if i am wrong
Re: Some Fine Problems
by Shashank Sharma - Friday, 27 November 2009, 12:27 AM
  In one of the question you have asked about the sum of all the digits formed by permutation of 2009... if you are taking both the zero as different

aren't you counting all the digits twice as 0029 and 0029 will come twice ..... please clear this doubt
Re: Some Fine Problems
by nidhi soni - Friday, 27 November 2009, 01:25 AM
 

phew!!!!!! now i could finished this thread :D with lots of breaks....it is really helpful to brush up some concepts

i dont know how many of following ans will b right but i know the thing tht i m working on this since yesterday noon and could compelte it now :|still if i wont get any question correct definatly me ro doongi sad :P

1) 90903250000
2) 60535
3) still trying to solve  :|
4) 2013C4???
5) S=14? :|
6)40 times?
7) no

Re: Some Fine Problems
by yash gupta - Friday, 27 November 2009, 01:48 PM
  thank a lot smile
Re: Some Fine Problems
by Vijay S - Saturday, 28 November 2009, 12:31 AM
  Hey! Great problems.. But, few of them are of JEE standard not CAT!! surprise
Re: Some Fine Problems
by TG Team - Sunday, 29 November 2009, 06:49 PM
  Leave those few Vijaysmile
Re: Some Fine Problems
by TG Team - Sunday, 29 November 2009, 06:58 PM
 

Imransmile

Your pattern of first digits from 21 to 210 is correct and it is true for the numbers in the ranges 211 to 220 and 221 to 230 also. But my dear, it doesn't continue after that.

I checked it on excel sheet. If you want I can attach it also.

I hope you can verify the pattern for higher powers also.smile

Re: Some Fine Problems
by TG Team - Sunday, 29 November 2009, 07:06 PM
 

Shashanksmile

229 = 5 mod9. That's right.

But you need to find the missing digit in the number.

It is saying that 229 is a nine digit number having all digits from 0 to 9 except one. And you need to find the missing digit.

If I make a 10 digit number using all the digits (0, 1, .., 9) then it is divisible by 9.

But our 9 digit number is giving a remainder of 5. That means the digit 4 is missing here. Because adding 4 at any place in the number it will again be divisible by 9.

Think a while and then ask me again if you are not clear.smile

Re: Some Fine Problems
by TG Team - Sunday, 29 November 2009, 07:22 PM
 

Nidhismile

Just returned after a few hectic days.

Will check your responses soon and revert you back then.

Till then there is no need to cry. smile

Re: Some Fine Problems
by Vijay S - Monday, 30 November 2009, 02:11 PM
  Hi TG/Kamal,

Wanted to propose an idea for an idea for an article regarding boundary condition checks. I usually make the mistake of marking an answer which is +/-1 the correct answer. Its really irritating that you figure out the entire logic, solve it and then find out that you would've missed counting a zero or the last number, etc., !

For Example: I would've missed out the fact that the problem set asks about whole numbers and we should consider 0 also, etc.,
Another case is when the problem set includes integers between a-b or integers from a-b, etc., Here wording is intentionally made tricky to confuse the problem solver.

Can you guys come up with some generic thumb rules while tackling such problems with answer options close to each other. (esp. counting problems) I'm sure lot of others also face this problem. smile
Re: Some Fine Problems
by TG Team - Monday, 30 November 2009, 02:36 PM
 

Vijaysmile

Best of the 'generic thumb rule' as desired by you is Common Sense and Presence of Mind.

While solving any question, these two things must be accompanied with you for sure so that there is no human error in your solution. By word - jugglery, we may make various rules regarding boundary values; like when to include or when not to include, but examiner still may prove to be a better countrpart.

So my advice is : Common Sense and Presence of Mind which everybody already possess but forget to use everytime when required. smile  

Re: Some Fine Problems
by nidhi soni - Tuesday, 1 December 2009, 09:06 PM
 

thnx kamal smile was waiting for reply

 

Re: Some Fine Problems
by TG Team - Sunday, 6 December 2009, 08:00 PM
 

Hi everybodysmile

I know all of you are busy in your exams right now.

Now I am starting to give answer to the questions, given at the end of article, one by one.smile

Here is the first one:

 


Re: Some Fine Problems
by nidhi soni - Monday, 7 December 2009, 09:54 PM
  did the same way :| but my ans is worng so calculation prb :|
Re: Some Fine Problems
by TG Team - Wednesday, 9 December 2009, 08:50 AM
 

Second nowsmile

 


Re: Some Fine Problems
by TG Team - Wednesday, 9 December 2009, 05:18 PM
 

Third: smile

 


Re: Some Fine Problems
by umesh singh - Wednesday, 9 December 2009, 06:18 PM
 

HI KAMAL

 i could not make out your way of solution of last question.pls explain it a little.

umesh

Re: Some Fine Problems
by TG Team - Wednesday, 9 December 2009, 06:51 PM
 

Hi Umeshsmile

Just take a smaller case:

If I am to find 3 natural numbers a, b and c from the set {1, 2, 3, 4, 5} such that a < b < c, then we just need to find three numbers from the given 5 in 5C3 ways and arrange them in one way for the desired arrangement of (a, b, c)

Now if the desired triplet is such that a ≤ b ≤ c, then number of ways of selecting three numbers are certainly more than earlier case because of the equality sign.

But if I find the number of ways of selecting three different numbers (a, b + 1, c + 2) from 5 + 2 = 7 ( in 7C3 ways), then again three numbers are different and b + 1 has been counted from numbers upto 6 and c + 2 has been counted from numbers upto 7.

Remember a cannot be more than 5 and b + 1 cannot be more than 6 (Why????)  

So now from all the solutions ( 7C3 ways), I just need to subtract 1 from the second number ( i.e. b + 1) and 2 from the third number (i.e. c + 2) so as to get the triplet (a, b, c) such that a ≤ b ≤ c.

So the desired number of triplets will be 7C3 .

I've tried my bit to explain it little further. But doubt may still persist as I am confident of my lowly intellect. smile

Re: Some Fine Problems
by umesh singh - Wednesday, 9 December 2009, 08:35 PM
 

hi kamal

thanks....i can't thank you enough brother. it is good of you to be humble . But it does not sound music to others to be excessively humble. i am saying half in jest,half in earnest. I also want to thank you for  your other ingenious solutions of complicated questions.

umesh

 

 

Re: Some Fine Problems
by jai prakash - Wednesday, 9 December 2009, 10:11 PM
  @kamal
if u r lowly intellect then we r non existent.
while speculation r rife with another 2009 CAT, meanwhile
can u please post some problems in QA/LR sections in main forum to keep our
heads above water for XAT test.
Plz do it.
Thanks and Regards
Re: Some Fine Problems
by TG Team - Friday, 11 December 2009, 07:14 PM
 

Thanks Umesh and JP smile for your kind words.

Here is the fourth one:

 


Re: Some Fine Problems
by Yogendra Yadav - Saturday, 12 December 2009, 10:13 PM
  Under the caption COMBINATORICS,
There came a prob where one had to find out the number of triagles whose sides are integers and sum is 2009.

Really commended the way you solved.
The kind of doubt always pesters me is that when one is asked to reckon the number of such triangles, then how can one can judge whether the (x,x,y) or (x,y,x) or (y,x,x) are different cases or the same? Here u took them as same for you set up the equation 6x + 3y + z = 1005*502 ...
But considering the triangle ABC ..... (x,x,y) , (y,x,x) and (x,y,x) are the different caes. As AB=x,BC=x, CA=y is altogether different from the case where AB=x, BC=y, CA=x.

Sayonara

Re: Some Fine Problems
by Yogendra Yadav - Saturday, 12 December 2009, 10:44 PM
  Under the Caption COMBINATORICS..
In last but one pro.
I did consider the separated cases .. ie.
1) when in first and second toss, 3 coins in all were eaten
2) when in first and second toss, 4 coins in all were eaten
Then i divided each case in several other subcases and then i arrived at 189/256 as an answer.
Yah solution is a bit terse.
Could yah explain how did yah take the probabilty of any coin to be engulfed in as 3/4?

Sayonara
Re: Some Fine Problems
by Yogendra Yadav - Sunday, 13 December 2009, 09:13 AM
  Hi nidhi ... I guess are the same Nidhi smile

Yah enthu. is uttrerly outstanding. Hats off!
1) Who cares when and how many times 7 comes ... the whole thing is done when no. of time 8 or 9 comes are counted. As the first digit preceding the no. starting from 8 or 9 must have 4.

2) It is the property of any natural nos, square which was utilised here.

Sayonara
Re: Some Fine Problems
by TG Team - Sunday, 13 December 2009, 11:33 AM
 

Yogendra smile

If you see clearly, question is :

How many different triangles are there with integer sides and with the perimeter 2009?

Now if you take a triange with side lengths (3, 4, 5) you get a right angled triangle.

Now rotate with angle 900 say. You are having a different orientation of same triangle with same side lengths. Do you consider this triangle to be same as earlier or different?

Second thing: just cut the triangle from paper and turn it upside down. Again you are having different orientation ( actually mirror image) of same triangle. Do you consider this triangle to be same as earlier or different?

Actually in case of a triangle: if three sides can be joined to form a triangle then all orientations are of a same triangle only. Or in other words, three side lengths cannot form two different triangles. 

Remember this argument will not hold true for four side lengths forming a quadrilateral. Because in that case different arrangements of sides will make different quadrilaterals.

Most important point : different triangles mean - non congruent. smile

 

Re: Some Fine Problems
by Yogendra Yadav - Sunday, 13 December 2009, 12:00 PM
  Taking yah example..
(3,4,5)

1) AB=3, BC=4 and CA=5
2) AB=4, BC=5 and CA=3 .. so on
They all are congruent, but are they not different?
This is what i wanted to ask.

Sayonara
Re: Some Fine Problems
by TG Team - Sunday, 13 December 2009, 12:14 PM
 

Yogengra smile

As I have stated earlier - Different triangles mean non-congruent.

If two triangles are congruent, then you can name the vertices anyhow they are considered to be non different.

Also if you are comparing two triangles then their vertices should be named differently. I mean you cannot say triangle ABC is different from triangle BCA just by changing orientation.

Re: Some Fine Problems
by TG Team - Sunday, 13 December 2009, 12:31 PM
 

I toss 4 fair coins and eat all of the coins (they are sweet coins ) that land on heads. I toss all the uneaten coins again and then eat all the coins that land on tails. What is the probability that I have eaten at least 3 coins?

Look coins are tossed twice only. Each time the probability of a coin for not landing in heads is 1/2.

So the probability that the coin will not be eaten even after tossing twice = probability of landing tail in first toss x probability of landing tail in second toss = (1/2) x (1/2) = 1/4.

So probability of a coin to be eaten is 1 - 1/4 = 3/4. smile

I hope this is clear. 

Re: Some Fine Problems
by umesh singh - Sunday, 13 December 2009, 11:16 PM
 

Q. a girl has to climb 10 steps. she takes either 2 steps or 1 step at a time. in how many ways can she climb these 10 steps?

sorry i paste the problem here

Re: Some Fine Problems
by TG Team - Tuesday, 15 December 2009, 04:01 PM
 

Umesh smile

You can paste any problem (from anywhere PG, TG, TF, QE, AN smile)here.

Isn't the problem equivalent to finding the solutions of x + 2y = 10 and then number of arrangements of all 1's and 2's.

case - 1

(0, 5) => Number of arrangements = 5!/5! = 1

case - 2

(2, 4) => Number of arrangements = 6!/2!4! = 15

case - 3

(4, 3) => Number of arrangements = 7!/4!3! = 35

case - 4

(6, 2) => Number of arrangements = 8!/6!2! = 28

case - 5

(8, 1) => Number of arrangements = 9!/8!1! = 9

case - 6

(10, 0) => Number of arrangements = 10!/10! = 1

Total ways = 1 + 15 + 35 + 28 + 9 + 1 = 89 smile

 

Re: Some Fine Problems
by umesh singh - Tuesday, 15 December 2009, 08:04 PM
 

Sukriya bro....

What is the unit digit of [(10^20000)/((10^100) + 3)] where [] is the greatest integer function?

10^20000/10^100 + 3 = 10^19900/(1 + 3/10^100)
=> 10^19900[1 - (3/10^100) + (3/10^100)^2 + ...........
Unit place : unit place of (10^19900((3/10^100)^198 - (3/10^100)^199)
= 3^198 x100 - 3^199
Last digit of 3^199 is 7
Last digit of 3^198 x100 is 0
Last igit of 3^198 x100 - 3^199 is 3
So Last digit of [(10^20000)/((10^100) + 3)] is 3
is it right?

if yes ,is there  any better way?

umesh

Re: Some Fine Problems
by umesh singh - Tuesday, 15 December 2009, 10:36 PM
 
one more bro....
Find the positive integral solutions of equation
x1*x2*x3*x4=120
umesh
Re: Some Fine Problems
by umesh singh - Wednesday, 16 December 2009, 12:42 AM
  S = (3 + 3^2 + 3^3 + … + 3^400) – (7 + 7^2 + 7^3 + … + 7^201).
last two digits of this sum?
Re: Some Fine Problems
by TG Team - Wednesday, 16 December 2009, 10:34 AM
 

Umesh smile

As x1, x2, x3 and x4 are four positive integers, you need to just find ordered pair of solutions.

Now 120 = 23 * 3 * 5

Three similar balls (3 2's) can be distributed in (4 distinct boxes) = 6C3 = 20 ways.

One ball (1 3's) can be distributed in (4 distinct boxes) = 4C3 = 4 ways.

One ball (1 5's) can be distributed in (4 distinct boxes) = 4C3 = 4 ways.

So total number of ordered pairs = 20 * 4* 4 = 320 ways smile

Re: Some Fine Problems
by TG Team - Wednesday, 16 December 2009, 10:52 AM
 

Can you recognise that first term is greater than second one?

After that it is simple. 

S = (3 + 3^2 + 3^3 + … + 3^400) – (7 + 7^2 + 7^3 + … + 7^201)

S = (3 + 9 + 27 + 81)(1 + 34 + 38 + ... 100 terms) - (7 + 49 + 343 + 2401)(1 + 74 + 78 + ... 50 terms) - 7201

S = 120(a0) - 2800(b0) - c7 = d93 smile

Re: Some Fine Problems
by umesh singh - Wednesday, 16 December 2009, 01:03 PM
 

you did sy any thing about my solution of unit digit question?

BTW, extremely thanks for these wonderful solutions

um

Re: Some Fine Problems
by umesh singh - Wednesday, 16 December 2009, 01:07 PM
 

S = (3 + 3^2 + 3^3 + … + 3^400) – (7 + 7^2 + 7^3 + … + 7^201)

S = (3 + 9 + 27 + 81)(1 + 34 + 38 + ... 100 terms) - (7 + 49 + 343 + 2401)(1 + 74 + 78 + ... 50 terms) - 7201

S = 120(a0) - 2800(b0) - c7 = d93

pls elaborate the line ....

umesh

Re: Some Fine Problems
by TG Team - Wednesday, 16 December 2009, 06:43 PM
 

Umesh smile

I have just paired 4 terms together for 3's and 7's

So from 31 to 3400 we'll have 400/4 = 100 pairs each of whose sum is ending in zero. Similarly from 71 to 7200 we'll have 200/4 = 50 pairs each of whose sum is ending in 00.

Hope this helps. smile 

Re: Some Fine Problems
by umesh singh - Wednesday, 16 December 2009, 08:22 PM
 

hi kamal bro....

i did it like that

Y= 3(3^400-1)/2 -7(7^201-1)/6

Phi(100)=40

 Y'=Y/100= 3(3^40k-1)/2 -7(7x7^40k1-1)/6

    = 00-07=93

 Is it right ?

though getting answer but i am not sure about this way,for i read some where that do not sum the series while calculating last digits...

pls let me know about your views

Re: Some Fine Problems
by umesh singh - Thursday, 17 December 2009, 12:42 PM
 

ONe more...........

In how many ways can 8 persons sit at a round dining table in such a manner that all will not have the same neighbours in any 2 arrangements?

Re: Some Fine Problems
by TG Team - Friday, 18 December 2009, 07:05 PM
 

Umesh smile

As I understand, you are asking the number of arrangements of 8 people around a round table in such a way that no people have same neighbours in any two arrangements.

That means for person 1 - I can choose different neighbours from 7 persons (2 to 8) in such a way that in no two arrangements he has same neighbours.

So I cannot have more than 3 arrangements.

For example - If I make pairs as (2 - 1 - 8), (5 - 1 - 6), (3 - 1 - 4), then there cannot be the fourth arrangement for person 1 with different neighbours from earlier three cases. 

Now question left is that we need to ensure different neighbours for other persons also in each of the three cases. To my goodness, it is possible as shown below:

 


Re: Some Fine Problems
by umesh singh - Friday, 18 December 2009, 09:39 PM
 

pls kamal

look into my last question solution......

give your opinion?

since i am a bit sceptical about that?

 

Re: Some Fine Problems
by umesh singh - Friday, 18 December 2009, 09:44 PM
 

thanks brother!!!!!!

i knew ....you would ....

umesh

Re: Some Fine Problems
by umesh singh - Monday, 21 December 2009, 06:20 AM
 

hi problem solver

one more

If a,b,c are natural numbers , then how many number of unordered (a,b,c) satisfy
(1+1/a)(1+1/b)(1+1/c) = 2 ?

Re: Some Fine Problems
by umesh singh - Monday, 21 December 2009, 12:39 PM
 

ek aur sar dard(headache)---

If r, s, t are prime numbers and p, q are positive integers such that LCM of p, q is r^2*t^4*s^2, then the number of ordered pair (p, q) is

Re: Some Fine Problems
by TG Team - Monday, 21 December 2009, 04:00 PM
 

Hi Umesh smile

I don't know whether you have answer to previous question or not. But if you have it, then verify:

Required number of ordered pairs = 5 x 9 x 5 = 225 smile

I'll explain it later if you want me to. smile

Re: Some Fine Problems
by umesh singh - Monday, 21 December 2009, 05:05 PM
 

Hats off.....brother

right brother...plsssssssssssssssssssssssss explain.

hey .... you are 'PANACEA' for me.

umesh

 

 

Re: Some Fine Problems
by TG Team - Monday, 21 December 2009, 05:30 PM
 

If r, s, t are prime numbers and p, q are positive integers such that LCM of p, q is r2t4s2, then the number of ordered pair (p, q) is :

First of all a basic but most important point - Atleast one of p and q must have r2, t4, and s2

Powers of r (as r0, r1 or r2 ) can be distributed to p and q in 32 i.e. 9 ways.

But according to basic point mentioned above I need to subtract the ways in which only r0, r1 have been distributed to p and q, i.e. 22 = 4 ways.

So finaly powers of r can be distributed in = 9 - 4 = 5 ways.

Similarly powers of t can be distributed in = 25 - 16 = 9 ways.

and powers of s can be distributed in = 9 - 4 = 5 ways.

So total number of ordered pairs (p, q) = 5 x 9 x 5 = 225 smile

Can you tell me who is hsemu ?

Re: Some Fine Problems
by umesh singh - Monday, 21 December 2009, 06:15 PM
 

hi kamal,

 it is me brother.how come you know the person.BTW , i am overwhelmed with your logical skill.thanks again

umesh

Re: Some Fine Problems
by TG Team - Monday, 21 December 2009, 07:46 PM
 

Hi Umesh smile

Thanks for your encouraging words.

I asked you about hsemu|umesh because I found someone is posting our questions (from TG and TGRMO) on other sites which we found to be offensive.

You know it very well that it takes a lot of intellectual blood and sweat to create a good question or solve an olympiad level problem. Most of the material on TG and TGRMO has been created by our constant hard work. And I don't think that you are lacking quality discussion in TG forums.

Anyhow, you should not post any of our question on some other website as it is violation of copyright laws. And still if you want to discuss some questions on other websites then do mention the source of question by giving a link to us.

I hope this should not hamper your mathematical extravaganza. smile

Re: Some Fine Problems
by umesh singh - Monday, 21 December 2009, 10:19 PM
 

hi Kamal

        i did not have the idea brother. i will never do it again.

umesh

Re: Some Fine Problems
by umesh singh - Tuesday, 22 December 2009, 12:36 AM
 
Let N be the number of consecutive zeros at the right end of the decimal representation of the product 1!.2!.3!.4!.5!.....99!.100!. Find the remainder when N is divided by 1000?
 
i worked out.and got answer....which is rightbut took inordinate  time ,because i did not noticed sequence disturbance at 25,50 and 75.
Total = (5 + 10 + 15 + 20 + 25.................+ 110) - 25 - 55 - 85 + 24 = 1124 ........{24 zeroes for 100!}
 
 give some kamal's KAMAL.........

 
Re: Some Fine Problems
by TG Team - Wednesday, 23 December 2009, 11:53 AM
 

Hi Umeshsmile

To find the number of consecutive zeroes at the end of the decimal representation of the given number is same as finding highest power of 5 contained in this number.

Given number can be written as :

1!.2!.3!....100! = (1)(1.2)(1.2.3)...(1.2.3...100) = 1100299398...1001

Now N = sum of powers of multiples of 5 + sum of powers of multiples of 52

=> N = (96 + 91 + ... + 1) + (76 + 51 + 26 + 1) = 970 + 154 = 1124.

Hence N mod1000 = 124 smile

Re: Some Fine Problems
by umesh singh - Wednesday, 23 December 2009, 03:35 PM
 

hi kamal....                                                                                                    ya .... you messured up to my expectation...BTW, yesterday i was in my office party in powai(mumbai). i met with a bunch of youngester doing their graduation... i asked him ..have you heard of site totalgadha.com? Most of them were not aware of this site ,barring two or three. i told them about your ,or rather,our site .Accidently, i ran into few of them while going office.one of them told me...Bhaiya ,the site is really quite good .....but how come you know volume about the site.....i told them ...in the same fashion as you came to know.

umesh

Re: Some Fine Problems
by TG Team - Wednesday, 23 December 2009, 05:05 PM
 

Hi Umesh smile

It's good to see you promoting our (I am including you too) efforts to the mass for their own benefit.

That's really a act of benevolence. smile

 

 

Re: Some Fine Problems
by umesh singh - Thursday, 24 December 2009, 10:00 PM
 

hi bro...

kaise karenge?

Finding last non-zero digit of 8293!

Re: Some Fine Problems
by umesh singh - Friday, 25 December 2009, 11:10 AM
 

I want to find out number of negative terms in (x-y -z) ^ (2n+1)

 

Total number of terms=(2n+1+3-1)C2= (2n+3) C2= [(2n+3) (2n+2)]/2

Let y +z=m

Number of positive terms or negative terms in (x-m) ^ (2n+1) is

= (2n+2)/2= (n+1)

Positive terms=ax^ (2n+1) +bx^ (2n-1) m^2+cx^ (2n-3) m^4+…. +p x^1 m^2n

Number of positive terms= 1+3+5+7+9+-----+ (2n+1) =n^2

Number of negative terms= (2n+3) (2n+2)/2-n^2= (2n+3) (n+1)-n^2= (2n^2+5n +3-n^2) = (n^2+5n+3)

Number of negative terms=Ax^ (2n) m^1+Bx^ (2n-2) m^3+Cx^ (2n-4) m^5 +---Pm^ (2n+1) =2+4 +6+----+ (2n+2) = (n+1) (n+2)

i know something is amiss in calculating number of negative terms? do not know where ?

help bro...

Re: Some Fine Problems
by umesh singh - Friday, 25 December 2009, 07:52 PM
 

hi kamal ....

aaj ke problem jo ...eaten away hours is

70770777077770.................up to 119 digits when divided by 440 leaves a remainder of (a)107 (b)77 (c)370 (d)70 ?

umesh

Re: Some Fine Problems
by umesh singh - Friday, 25 December 2009, 08:40 PM
 

i have not heard you since today morning .......probably enjoying ....'3 idiot,

 

Re: Some Fine Problems
by TG Team - Saturday, 26 December 2009, 02:11 PM
 

Hi Umesh smile

Was busy in household affairs. Anyhow back to work now. smile

As we know  440 = 5 × 8 × 11

So we need to check the individual remainders with 5, 8 and 11 only.

It's clearly evident that last three digits of the number are 770, so options (a) and (b) are negated automatically.

Now the given number = 2 mod8 and from options (c) and (d) only option (c) satisfies. Hence must be the answer.

But if I want to verify this remainder, I need to calculate the remainder with 11 also. So have a look at the proceedings below:

 


Re: Some Fine Problems
by umesh singh - Saturday, 26 December 2009, 02:32 PM
 

thanks kamal ....

i had problem in verfying for 11. when you have time ,just see two other problem as well.

regards

umesh

 

 

Re: Some Fine Problems
by umesh singh - Saturday, 26 December 2009, 07:14 PM
 

hi kamal ....

i came across a problem...

x1 +x2+x3=13

x1<=x2<=x3

number of ways of distribution under the condition....waise maine kar liya hain aur answer 21 hain .....par my approach is not as sweet and short as your approach ....can we drive any formula ?

 

Re: Some Fine Problems
by TG Team - Sunday, 27 December 2009, 11:51 AM
 

Hi Umesh smile

I used to do this type of question by a bit lengthy way. But your large number (8293!) has made me think alternatively.

1

×

2

×

3

×

4

×

5(1)

×

6

×

7

×

8

×

9

×

5(2)

11

×

12

×

13

×

14

×

5(3)

×

16

×

17

×

18

×

19

×

5(4)

21

×

22

×

23

×

24

×

5(5)

×

26

×

27

×

28

×

29

×

5(6)

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

8281

×

8282

×

8283

×

8284

×

5(1657)

×

8286

×

8287

×

8288

×

8289

×

5(1658)

From a block of every 10 consecutive numbers starting from 1 (as above), we have two numbers which are multiple of 5 and when multiplied by two 2’s they will generate two zeroes.

Now leaving the multiples of 5 intact, I remove the two 2’s from the remaining eight numbers from same block of 10 numbers.

Taking a general block, I have the product as

(10x + 1)(10x + 2)(10x + 3)(10x + 4)(10x + 5)(10x + 6)(10x + 7)(10x + 8)(10x + 9)(10x + 10)

Or (10x + 1)(10x + 2)(10x + 3)(10x + 4)(10x + 6)(10x + 7)(10x + 8)(10x + 9) leaving two multiples of 5 aside.

Or 100x2(4a) + 10x(4b) + 1×2×3×4×6×7×8×9

Or 100ax2 + 10bx + 18144 taking two 2’s aside.

So finally I am having unit digit of a block of 10 consecutive numbers as 4 (leaving multiples of 5 and taking out two 2's)

Now coming to the original question:

Last non-zero digit of 8293! = Last non-zero digit of 8290!×1×2×3 = 4829×6×Last non-zero digit of 1658! = 4×2×4165× Last non-zero digit of 330! = 2×433×Last non-zero digit of 66! = 8×6×46×last non-zero digit of 12! = 8×2×4×2 = 8.

So the last non-zero digit of 8293! is 8.

You can verify it here also.

 http://www.wolframalpha.com/input/?i=8293%21+mod10^2071&t=ietb01

Re: Some Fine Problems
by TG Team - Sunday, 27 December 2009, 05:20 PM
 

Hi Umesh smile

(x - y - z)2n + 1 = (2n + 1)!/a!b!c! (x)a(-y)b(-z)c

where a + b + c = 2n + 1

Now any term of the expression will be negative only when b + c is negative or a is positive. Values for a are varying from 0 to 2n and simultaneously number of solutions for b + c are varing from 2n + 1 to 1.

On an average, b + c = (2 + 2n + 2)/2 = n + 2 and there are n + 1 different values for a.

So in all, there are (n + 1)(n + 2) negative terms in the expansion smile  

Re: Some Fine Problems
by jai prakash - Monday, 28 December 2009, 08:42 AM
  @ Kamal.
for TGites, u r Rancho of  3 Idiots.
Re: Some Fine Problems
by TG Team - Monday, 28 December 2009, 12:31 PM
 

JP smile

It's nice to see that this page hasn't been limited to two persons only.

BTW who is this Rancho? I've not seen the movie.

One personal info: Aamir Khan and Albert Einstein ( and obviously so many others) share their date of birth (14th March) with me.

In mathematical world, this day (3/14) is celebrated as Pi - Day and we are also going to hold some contest this year in RMO section. smile  

Re: Some Fine Problems
by umesh singh - Monday, 28 December 2009, 08:41 PM
 

On an average, b + c = (2 + 2n + 2)/2 = n + 2 and there are n + 1 different values for a.

So in all, there are (n + 1)(n + 2) negative terms in the expansion

these lines are beyound me brother?

Re: Some Fine Problems
by umesh singh - Tuesday, 29 December 2009, 09:29 AM
 

hi kamal......

As for this question-

x1 +x2+x3=13 wherex1<=x2<=x3

number of ways for distribution-

bro in the case of three,it is sigma(6)=21,what i did ,i cheked maximum possible value of x2 under the given condition

but it is possible for 3 variable ,but what if four variable?

umesh

Re: Some Fine Problems
by TG Team - Wednesday, 30 December 2009, 12:06 PM
 

Umesh smile

For all n + 1 different even values for 'a', number of solutions for b + c are linearly varying from 2 to 2n + 2. That means on an average, I can assume (only for calculation purpose) b + c has n + 2 number of solutions for each of n + 1 different values of 'a', giving me desired (n + 1)(n + 2) negative terms in all.

Hope this is clear now.smile 

Re: Some Fine Problems
by TG Team - Wednesday, 30 December 2009, 12:13 PM
 

Umesh smile

Can you elaborate that ∑6 part?

 

Re: Some Fine Problems
by umesh singh - Wednesday, 30 December 2009, 06:22 PM
 

hi kamal.......

x1 +x2+x3=13 where x1<=x2<=x3

(0,0,13),(0,1,12)........(0,6,7)=7 ways

(1,1,11),(1,2,10).........(1,6,6)=6 ways

(2,2,9),(2,3,8),(2,4,7).....(2,5,6)=4 ways

(3,3,7),(3,4,6).....(3,5,5)= 3 ways

(4,4,5) =1 ways

1+3+4+6+7=21 ways=sigma(6)

just tried to formulate ....

Re: Some Fine Problems
by TG Team - Wednesday, 30 December 2009, 07:16 PM
 

Umesh smile

Just sharing some thoughts over the problem.

Number of non - negative integral solutions of

x1 + x2 + x3 = 13 where x1 ≤ x2 ≤ x3

is same as number of non negative integral solutions of

x + 2y + 3z = 13

or number of positive integral solutions of

x + 2y + 3z = 16

which is given by the expression, {162/2(1)(2)(3)} = {256/12} = 21.

Here {x} denotes the nearest integer to x.

I've given this formula earlier in some thread earlier also. But I took time to respond because I don't have proof to it. And also what will happen if number of variables is more than 3.

So please excuse me for that. Give me some time to ponder over this. smile

Re: Some Fine Problems
by umesh singh - Wednesday, 30 December 2009, 09:41 PM
 

hi kamal....

you invariably do something ingeiously in course of solving question.....that leave my mouth agape......do you love to give me complex( in jest)?

is baar ye line hain..

which is given by the expression, {162/2(1)(2)(3)} = {256/12} = 21.

umesh

Re: Some Fine Problems
by umesh singh - Friday, 1 January 2010, 12:39 AM
 

hi kamal

wishing you very^(100000000000000000000000000000000000000000000000000

000000000000000000000000000000000000000) happy new year

umesh

 

Re: Some Fine Problems
by umesh singh - Friday, 1 January 2010, 01:38 AM
 

Each term in the expansion of (x-y-z)^(2n+1) will  be in the form kx^a(-y)^b(-z)^c=k(-1)^(b+c)x(X^a)x(Y^b)x(Z^c) such that a+b+c=2n+1 for some nonnegative integer, and a,b,c are non-negative.

Note  that this term will be negative if (b+c) =2p+1 for some non-negative integer p.All the possible value of p range from 0 to n,so there are (n+1) possible value of (b+c).                          

            The number of nonnegative integer pairs (b,c) that satisfy b+c=m for some integer m is (m+1)C1=(m+1).The value of ‘a ‘follows from each distinct pair(b,c) so we simply count the ordered pair (b,c) to find the number of ordered triplet (a,b,c) such that the term is negative.

Thus -

Sigma [(2p+1) +1], where p varies from 0 to n = 2[sigma (p) +sigma (1)] = (n+1) (n+2).

 

 

Each term in the expansion of (x-y-z)^(2n) will  be in the form kx^a(-y)^b(-z)^c=k(-1)^(b+c)x(X^a)x(Y^b)x(Z^c) such that a+b+c=2n for some nonnegative integer, and a,b,c are non-negative.

                                                          Note  that this term will be negative if (b+c) =2p+1 for some non-negative integer p.All the possible value of p range from 0 to (n-1),so there are  n possible value of (b+c).                          

                                                                         The number of nonnegative integer pairs (b,c) that satisfy b+c=m for some integer m is (m+1)C1=(m+1).The value of ‘a ‘follows from each distinct pair(b,c) so we simply count the ordered pair (b,c) to find the number of ordered triplet (a,b,c) such that the term is negative.

Thus -

Sigma [(2p+1) +1], where p varies from 0 to n-1 = 2[n (n-1)/2 + (n)] =2n (n+1)/2=n (n+1)

 

 

if you find that everything is ok ,then give your concurrence  

Re: Some Fine Problems
by umesh singh - Friday, 1 January 2010, 11:09 PM
 

Do, Re, Mi, Fa, So, La, Ti, want to have their pictures taken with at least one of the other six. However, Do and Ti as well as Mi and Fa do not go along well with each other,and because of this, they do not want to be seen in any picture with the other. How many pictures are possible?

Possible groups(7,0),(2,5),(3,4)(3,2,2)
=7C7+7C2+7C4 +(7C3X4C2x2C2)=7+21+35+35x6=7x(1+3+5+5x6)=7x(39)=2 73

Possibilities to be subtracted-
1.(7,0)=1
2.(2,5)= 1(when both in the group of two) +5C3(when both are in group of 5)
=1 +10=11
3.(3,4)=5C1(when both are in group of 3) +5C2(when both are in group of 4)=15
4.(3,2,2)=5C1(when both are in group of 3) +1(when both in group of 2)=6
total=1+11+15+6=33
required cases=273-33=240

pls see this solution......if i missed something pls let me know

Re: Some Fine Problems
by TG Team - Saturday, 2 January 2010, 07:59 PM
 

Umesh smile

For simplicity, I am reframing the question as there are seven persons A, B, C, D, E, F and G who want to have picture with atleast one of the other six. However A and B as well as C and D do not go along well with each other and hence cannot be photographed together. How many pictures are possible?

Here is the beauty solution:

From A and B , selection can be made in 3 ways (A or B or none of them)

From C and D, selection can be made in 3 ways ( C or D or none of them)

For E, F, G selection can be made in 23 ways ( 2 ways for each; selected or not selected)

Now out of total 3223 = 72 cases I need to subtract the cases when only one person is selected ( i.e. 7 ways; any one of the seven) and when no person is selected ( i.e. 1 way only).

So, in all, total pictures possible = 72 - 7 - 1 = 64. smile

Re: Some Fine Problems
by TG Team - Monday, 18 January 2010, 08:32 PM
 

Answer to Fifth question now:

Though it is quite late still quest for knowledge never ends. smile

 


Re: Some Fine Problems
by umesh singh - Monday, 18 January 2010, 10:43 PM
 

yaar  ...

see the question and help

Let m and n be positive integers such that1=<m<n. In their decimal representations, the last three digits of1978^m are equal, respectively, to the last three digits of1978^n. Find m and n such that n+m has its least value.

Re: Some Fine Problems
by TG Team - Tuesday, 19 January 2010, 11:47 AM
 

Umesh smile

This question belongs to IMO 1978.

1978 = 2 × 23 × 43

As 1978m and 1978n are having same last three digits, so their difference must be divisible by 1000 = 23 × 53.

Given that m, n are positive integers such that 1 ≤ m < n

So 1978n – 1978m = 1978m(1978n – m – 1) ≡ 0 mod1000

(1978n – m – 1) is an odd number so do not contain any powers of 2 so smallest value of m = 3 so that 1978m has a factor of 23.

Now 1978n – 3 – 1 ≡ 0 mod125 as 19783 do not have any factor of 5.

=> 1978n – 3 ≡ 1 mod125.

Using Euler’s theorem we know that 1978100 ≡ 1 mod125 ,though it may not be the smallest power but smallest power must be a factor of 100 for sure.

Checking all the factors of 100, it can be easily observed that smallest value of n – 3 = 100

=> n = 103.

So m + n = 3 + 103 = 106. smile

Re: Some Fine Problems
by umesh singh - Friday, 22 January 2010, 08:05 PM
 

hi kamal .......

just point up my mistake ...

|x|+|y|+|z|=15  

 CASE I
If none of them is zero.
Total positive solutions 14C2=91
Now as x, y, z each of them may take positive or neg. value
Total solutions =91*8=728
CASE II
If one of them is zero
Total solutions 14*4*3C2=168
CASE III
If two of them are zeros
Total solutions 6
A total of 728+168+6=902 

But if i approach in other way-                                                                                                                                                                                                                                                       for There are 8 ways for representing the nos. with + or -
so total = 17C2*8 = 1088
If two of them are zeros, zeros with + and- are also factored in,  

hence ways to be excluded = (3C1x2^3)/2=6

if one of them is zero then terms to be excluded is=(2^3x3C2x14C1)/2=168

total terms to be excluded=174

hence total values = 1088 - 174 =914  but answer is 902

Ye aur 12 koun se minus karne hain boss.... pls help

with regards

umesh                

Re: Some Fine Problems
by barry white - Saturday, 18 September 2010, 07:35 AM
  Amazing article.... ur too good..

Btw I did my part in clicking on that share link

Cheers
Re: Some Fine Problems
by siddharth jain - Wednesday, 22 September 2010, 02:21 PM
  i was working on question 3...i constructed a set of prime number (67,19,2,83,5) whose sum is 176 ... while the min sum should not be less than 207 as u have shown... is i m rgt...tell me if i m commiting any mistake,,,
 siddharth
Re: Some Fine Problems
by TG Team - Thursday, 23 September 2010, 01:22 PM
 

Hi Siddharth smile

You have missed the digit 4. smile

Re: Some Fine Problems
by rajan bh - Thursday, 7 October 2010, 01:22 AM
 

Hi Kamal,

Pls help me in 5 th solution:

n>334

so least value should be 335

as 334*6 = 2004 (so even in 334 throws prob of sum of 2009 will be 0)

Secondly, can u pls elaborate on

(7-a1)+ (7-a2)+ ...(7-an) = S ??  I dint get this

Regards,

Rajan

summation of reciprocal
by nuzhat jahan - Saturday, 16 October 2010, 01:25 PM
 

hi please suggest and explain the logic of the following series

1+1/2+ 1/3 +1/4 + ........+ 1/50

Re: Some Fine Problems
by rakesh kumar - Thursday, 14 April 2011, 12:09 AM
  sir
we cannot get the 2009 as sum when 334 dice are rolled, the
maximum value is 2004
then probability is zero
i didnt get the 7-a1...........
moreover it cant be less than 335
Re: Some Fine Problems
by TG Team - Friday, 15 April 2011, 11:01 AM
 
Hi Rajan and Rakesh smile

Yes, you are right. Here n > 334 so least value of n will be 335.

Now see, in a regular hexagonal dice sum of numbers on opposite faces is always 7 i.e. (1, 6), (2, 5), (3, 4). So probability of obtaining x is same as that of 7 - x.
In case of single dice, it is not clearly evident as probability of obtaining any number from 1 to 6 is same i.e. 1/6.

In case of two die, if I see that probability to obtain a sum of 2 is 1/36 because there is only one case possible (1 + 1) out of 36 ordered outcomes or if I am looking for unordered outcomes then this probability is 1/21. Now this probability is same as that of obtaining a sum of 12 because again only one case is possible i.e. (6 + 6). Notice the key point that 1 + 6 = 7 and 2(7) - 2 = 12.

Let's check out one more sum. To obtain a sum of 4, there are three ordered favorable outcomes; 1 + 3, 2 + 2, 3 + 1. Similarly to obtain a sum of 2(7) - 4 = 10 also there are corresponding three ordered favorable outcome; 6 + 4, 5 + 5, 4 + 6. Notice the key point here again that 1 + 6 = 7, 3 + 4 = 7, 2 + 5 = 7.
I hope it is clear now. smile

I am attaching the corrected version of the answer below.

Kamal Lohia
   


Re: summation of reciprocal
by TG Team - Friday, 15 April 2011, 11:23 AM
 
Hi Nuzhat smile

In this series, we can find the range of the sum as follows:

1 + 1/2 + 1/3 + 1/4 + ..... + 1/50 < 1 + (1/2 + 1/2) + (1/4 + 1/4 + 1/4 + 1/4) + (1/8 + 1/8 + ....8 terms) + (1/16 + 1/16 + ...16 terms) + (1/32 + 1/32 + ....19terms) = 1 + 1 + 1 + 1 + 1 + 19/32 = 5 + 19/32 = 179/32 ~ 5.6

Also 1 + 1/2 + 1/3 + 1/4 + .... + 1/50 > 1 + (1/3 + 1/3) + (1/9 + 1/9 + ..6 terms) + (1/27 + 1/27 + ...18 terms) + (1/81 + 1/81 + ...23 terms) = 1 + 2/3 + 2/3 + 2/3 + 23/81 = 3 + 23/81 = 266/81 ~ 3.3

So finally we can say for given expression that
3.3 < (1 + 1/2 + 1/3 + ....1/50) < 5.6

If I try to find this expression exactly, it comes out to be 4.5 approximately. So our above range is quite close to the actual value. But there is no method to find the actual value other than calculating it. smile

Kamal Lohia

 
Re: Some Fine Problems
by Neha Arora - Wednesday, 24 August 2011, 10:29 AM
  great site...great questions..thankss they are really helpful smile smile smile
Re: Some Fine Problems
by Neha Arora - Wednesday, 24 August 2011, 10:36 AM
  Can anyone pls explain the solution : 13a + 17b <221

How 13(a'+1) + 17(b'+1) =221 and (17-1)(13-1)-1 came into picture?
Re: Some Fine Problems
by Neha Arora - Wednesday, 24 August 2011, 12:48 PM
  hi Kamal

Can u pls elaborate on {162/2(1)(2)(3)} = {256/12} = 21.

How to solve these type of problems?

x + 2y + 3z = 16 or x + 2y + 3z <= 16

Thanks smile
Re: Some Fine Problems
by Rahul Patil - Friday, 2 September 2011, 07:38 AM
  Hii Kamal Sir, Can U please explain what does m=n (mod x) means???
How do we use it in number system problems???
Re: Some Fine Problems
by priyanka bansal - Friday, 9 September 2011, 05:58 PM
  kindly solve dis question plzzzzzzz sad

a 12 digit number is divisible by 72 consist of 4 and 6 only. find the no. of 6(largest possible number)
Re: Some Fine Problems
by TG Team - Friday, 9 September 2011, 06:37 PM
  666 444 444 444
Re: Some Fine Problems
by Kamal Joshi - Thursday, 13 October 2011, 09:05 PM
  Sir, 666 444 444 444 is not divisible by 8, it s divisible by 9, though. Don't you think the answer should be 664 444 444 464? because we are looking for a no. divisible by 72?
Re: Some Fine Problems
by TG Team - Friday, 14 October 2011, 12:35 PM
 

Hi Kamal smile

You are right. Number should be 664 444 444 464. But question was asking about the number of 6's in the number that should be 3 only. smile

Kamal Lohia

Re: Some Fine Problems
by Pritam Mishra - Monday, 17 August 2015, 10:50 PM
  i get addicted to totalgadha ....ur post just mesmerised me....last three digits of 2988^678 ..what will be its easy technique?