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Divisors of a Number
by Total Gadha - Tuesday, 9 October 2007, 01:00 AM
  cat
cat
cat
Re: Divisors of a Number
by Software Engineer - Tuesday, 9 October 2007, 01:17 PM
  Lagey Raho TG Bhai ! ! !
Re: Divisors of a Number
by Software Engineer - Tuesday, 9 October 2007, 03:48 PM
  Rule 2 from Top:
number of even divisors of N = (b + 1)(c + 1)..  
should nt it be odd thoughtful
Re: Divisors of a Number
by SAURABH PARIHAR - Monday, 29 October 2007, 02:40 PM
  excellent article.plz post something on range and domain..
Re: Divisors of a Number
by ambar patil - Thursday, 20 March 2008, 01:10 AM
 

another killer article TG , thenks big grin 

i think in the que  how many divisors of 3636 are perfect cubes ....... i guess 20 and 30 is same i.e 1 and is counted twice .... can you throw some more light on this ??  

Re: Divisors of a Number
by Mohsin Khanji - Friday, 21 March 2008, 12:14 AM
  Hi!!!!!!!
great TG
but i aspect that u r creat more link like this soon...
smile
Re: Divisors of a Number having unit dig 5
by aimsearch aimsearch - Friday, 13 June 2008, 11:26 PM
  the no. of divisors of n as  u said  wid n=a^p b^q n so on are
(p+1)(q+1)nd so on.....den dis concept was used 2 find no. of odd factors by considering only powers of 3 ,5,7...n so on  for odd nos....so for finding d divisors wd unit digit 5...shuldnt v apply d same concept n ignoring d powrs of 2....tak into acc powers of 5 n 3 to giv result 6*5=30 instead of 24.................................plz plz plz clarify dis doubt plz


regards
Re: Divisors of a Number having unit dig 5
by Atul Singla - Saturday, 14 June 2008, 11:12 AM
 

Hi aimsearch

You are nearly right but missing one point...

Note : we need to fing the divisors with unit digit 5.

5^0 =1 so it wont help us to get 5 as the unit digit.

 

hence it should 6x4 = 24 smile

Re: Divisors of a Number having unit dig 5
by aimsearch aimsearch - Saturday, 14 June 2008, 10:35 PM
  thnx atul...so nice of u 2 rep....smile
Re: Divisors of a Number having unit dig 5
by aimsearch aimsearch - Saturday, 14 June 2008, 11:16 PM
  can sm1 plz explain d concept behind d sum of even divisors of 2^5 3^5 5^4..................plz it will b a gr8 help






regards
Re: Divisors of a Number having unit dig 5
by Chitrang Dalal - Wednesday, 18 June 2008, 02:41 PM
 

can sm1 plz explain d concept behind d sum of even divisors of 2^5 3^5 5^4

SOLUTION:

a no of the form 2n is even....

here N = 2^5 *3^5 * 5^4

for even divisors we must have 2, 2^2, 2^3, 2^4, 2^5......cannot consider 2^0 since 2^0 =1

now using the formula for sum

sum= (2 + 2^2 +2^3 +2^4 +2^5) * (3^6 -1)/(3-1) * (5^5-1)/(5-1)

     = 2(2^5 -1) * (728)/2 * (3124)/4

(for the series of 2 ....the sum of geometric series formula is used....in fact for every term i.e. 2,3,5 same formula is used)

     = 31 * 728 * 781

     = 17625608

i hope i hv helped a little to clear ur doubt

 

Re: Divisors of a Number having unit dig 5
by aimsearch aimsearch - Friday, 20 June 2008, 12:37 AM
  thnxsmile
Re: Divisors of a Number having unit dig 5
by Ramkrishna Roy - Saturday, 21 June 2008, 11:08 AM
 
 

can sm1 plz explain d concept behind d sum of even divisors of 2^5 3^5 5^4

sum= (2 + 2^2 +2^3 +2^4 +2^5) * ^5     = 2(2^5 -1) * (728)/2 * (3124)/4 ....this is what you have given the explanation......i agree that 2^0=1...so we can't consider it, then in that case we are having 5 powers of 2 i.e; 2, 2^2, 2^3, 2^4, 2^5......so by formula it has to be( 2^5-1/2-1) * (3^6 -1)/(3-1) * (5^5-1)/(5-1)......then in that way i'm not geeting the  answer....so if wud be ur great help to me if u can explain this step.

Re: Divisors of a Number
by Total Donkey - Friday, 27 June 2008, 02:06 AM
  @Total Gadha
the product of the divisors that you have mentioned in the text  is wrong  in the last formula

It shows the product as = [ a^(a.(a+1)) .b^(b.(b+1)).c^(c.(c+1)) ] ^ (1/2)

For an example 2^2.3^2

divisors = 2^0.3^0, 2^1.3^0, 2^2.3^0, 2^0.3^1, 2^1.3^1, 2^2.3^1, 2^0.3^2, 2^1.3^2, 2^2.3^2

product = 2^9.3^9

Now as per the formula = [2^(2.3).3^(2.3)]^(1/2) = 2^3.3^3

The flaw comes when you open the brackets in the previous formula,
i.e. by mistake u have opened the braces as

(a.b.c)^(x.y.z) = a^x.b^y.c^z which is wrong

rather it shud be

(a.b.c)^(x.y.z) = a^(x.y.z).b^(x.y.z).c^(x.y.z)

Kindly enlighten if I am wrong
Re: Divisors of a Number
by Total Donkey - Friday, 27 June 2008, 02:23 AM
  @Total Gadha

I also have a doubt regarding the last solved example

It asks for the natural numbers within 1200 that are prime either to 6 or 15.

To simulate the experiment, I took 12 instead of 1200, and 3 and 6 instead of 6 and 15

So, the question consisted of numbers less than 12 prime either to 3 or 6..

Numbers prime to 3 = 1,2,4,5,7,8,10,11
Numbers prime to 6 (i.e prime both to 3 and 2) = 1,5,7,11

Now, numbers prime to either 3 or 6 = prime to one of 3 or 6 but not both
=2,4,8,10 =4 in number

Going by the stated method, we first find the numbers prime to 3
= 12[1-(1/3)] = 8
Numbers prime to 6 = 12[1-(1/2)][1-(1/3)] = 4

Numbers prime to both 3 and 6 = 12[1-(1/2)][1-(1/3)] = 4

So numbers prime to 3 and 6 both = 8 + 4 - 4 = 8

but the answer should be 4 as we counted earlier
-------------------------------------------------------------------------------------------
Going back to the solved example, I think that the problem arose when we have to find prime either to 6 or 15

so we did = N(numbers prime to 6) + N(numbers prime to 15) - N(numbers prime to both 6 and 15)

this essentially is the total numbers of numbers prime to both 6 and 15 (as we are subtracting the number of primes common to both of them)

so to find the exclusive primes, we must subtract the common primes twice (as they are counted in both primes of 6 and primes of 15)

so the correct numbers of exclusive primes should be = N(numbers prime to 6) + N(numbers prime to 15) - [ N(numbers prime to both 6 and 15) X 2 ]
-----------------------------------------------------------------------------------------------

this formula works for the example i took,

gives the answer = N(numbers prime to 3) + N(numbers prime to 6) - [ N(numbers prime to both 6 and 3) X 2 ]
= 8 + 4 - 2(4) = 4
-----------------------------------------------------------------------------------------------
Kindly clarify
Re: Divisors of a Number
by Total Gadha - Friday, 27 June 2008, 11:10 AM
  Hi Total Donkey,

Don't take 3 and 6. 3 is a divisor of 6. Prime to  means prime to 3 anyways.

Total Gadha
Re: Divisors of a Number
by Total Donkey - Friday, 27 June 2008, 04:11 PM
  @Total Gadha

Thanx for your quick reply

Btw, I didn't find it a problem of 3 being a divisor of 6.

Hence I re-simulated the experiment and would like to conclude with the reason once again:

numbers prime either to 4 or 6 in first 12 natural numbers

I solved for it. This is what I got:

Numbers prime to 4 (2.2) = 1,3,5,7,9,11
Numbers prime to 6(2.3) = 1,5,7,11
Numbers prime to both 4 and 6 = 1,5,7,11

Numbers prime to either 4 or 6 = prime to one of 4 and 6 and not both [@Total Gadha , I hope we are finding the exclusive primes, correct me if I am wrong]
= 3 and 9 = 2 in Number

As per the solved example, it is

= N(numbers prime to 4) + N(Numbers prime to 6) -  N(Numbers prime to both 6 and 4)

= 12.[1-(1/2)] + 12.[1-(1/2)].[1-(1/3)] - 12.[1-(1/2)].[1-(1/3)]
= 6 + 4 - 4 = 6 in number, which is wrong

I did another experiment as follows:

numbers prime either to 6 or 15 in first 30 natural numbers

I solved for it. This is what I got:

Numbers prime to 6(2.3) = 1,5,7,11,13,17,19,23,25,29 (10 in number)
Numbers prime to 15(3.5) = 1,2,4,7,8,11,13,14,16,17,19,22,23,26,28,29 (16 in number)
Numbers prime to both 6 and 15 = 1,7,11,13,17,19,23,29 (8 in number)

Numbers prime to either 6 or 15 = prime to one of 6 and 15 and not both
= 5,25 (from primes of 6) and 2,4,8,14,16,22,26,28 (from primes of 15)
= 10 in number

As per the solved example, it is

= N(numbers prime to 6) + N(Numbers prime to 15) -  N(Numbers prime to both 6 and 15)

= 30.[1-(1/2)].[1-(1/3)] + 30.[1-(1/3)].[1-(1/5)] - 30.[1-(1/2)].[1-(1/3)].[1-(1/5)]
= 10 + 16 - 8 =  18 in number, which is again wrong

------------------------------------------------------------------------------------------------
In both the places the answers came out to be wrong as we were going by the formula
= N(primes to a) + N(primes to b) - N(primes to both a and b), which is flawed as I understand
 
It should have been
= N(primes to a) + N(primes to b) - [2 X N(primes to both a and b)]

for the first experiment (primes either to 4 or 6 in 12), it gives
= 6 + 4 - 2(4) = 2 in number, which is correct

for the second experiment(primes to either 6 or 15 in 30), it gives
= 10 + 16 - 2(8) = 10 in number, which is again correct
------------------------------------------------------------------------------------------------
To conclude, I would like to point out that the formula is flawed as pet the Venn diagram

                                          
                 -------------------------------------------------------------------------
                 :                   Primes to 6               :           2        4             |
                 :                            |     1        7     :          8       14             |
                 :           5                |     11     13    :          16      22             |
                 :                            |      17     19    :          26      28             |
                 :            25             |      23     29   :                                   |
                 :                            |                     :                                   |
                 :                            |             Primes to 15                          |
                 -------------------------------------------------------------------------
                *One of my finest drawings :P

Now as per the formula in solved example
 = N(primes to 6) + N(primes to 15) - N(primes to both 6 and 15)
= say a + b - c

here since the primes common to 6 and 15 i.e. (1,7,11,13,17,19,23,29) have been counted in both the a and b, they should be subtracted twice to get to the region exclusive to 6 + the region exclusive to 15

so it should rather be = a + b - 2c

Phew! A long explanation smile .... Kindly enlighten .... I have tried as hard as I could to get  the point across

*Also, kindly explain if the formula of the product of divisors has been written wrongly, or was it  one of my foolish mistake

Thats all, my lord!!!
Re: Divisors of a Number
by nitin sonker - Friday, 27 June 2008, 07:53 PM
 

fine TG ur work is excellent...

u impressed me a lot kindly issue some more tips plss

Re: Divisors of a Number
by Total Gadha - Saturday, 28 June 2008, 07:26 PM
  Hi Total Donkey,

Thanks for correcting me. Have corrected the error.

Total Gadha
Re: Divisors of a Number
by Total Donkey - Saturday, 28 June 2008, 09:35 PM
  @Total Gadha
my pleasure

btw, there is another minor error. I had pointed it out already.
Reiterating it.
the product of the divisors that you have mentioned in the text  is wrong  in the last formula

It shows the product as = [ a^(a.(a+1)) .b^(b.(b+1)).c^(c.(c+1)) ] ^ (1/2)

Rather, it should have been
= [a^(a.(a+1)(b+1)(c+1)) .b^(b.(a+1)(b+1)(c+1)).c^(c.(a+1)(b+1)(c+1))] ^ (1/2)

I have pointed out the reason in one of my earlier posts.

Total Donkey
hi tg
by shubham singh - Wednesday, 2 July 2008, 01:37 PM
 

please explain me again as to how to find out "how many of the first 1200 natural numbers are either prime to 6 or to 15?"

 

Re: Divisors of a Number
by daman singh - Wednesday, 2 July 2008, 10:06 PM
 

very good article.........

Can u help me with this??

 

Q A natural Number N has a total of 48 factors.Find maximum possible number of ways of writing N as a product of two co-primes.

 

thanx

Re: hi tg
by shubham singh - Thursday, 3 July 2008, 01:32 PM
  hey tg plz help me....tel me how to do this...how many numbers less than 1200 are either prime to 6 or to 15?
Re: Divisors of a Number
by manish sharma - Saturday, 6 September 2008, 09:09 PM
  as you are requires to write in maximum possible coprime, dividing 48 as so to get maximum factors. i.e. 2*2*2*2*3
number = P1^1 * P2^1 * P3^1 * P4^1 * P5^2
hence number of ways so as to write as a product of two coprime numbers is (2*2*2*2*2)/2 = 16

reply whether i am right or not..
Re: Divisors of a Number
by venkat s - Sunday, 7 September 2008, 11:19 PM
 

Hi Tg thanks for your great materials they are very useful

can u please explain this

find the highest power 0f 12 that divides 5^36-1

Re: Divisors of a Number
by Mukesh Kumar - Monday, 8 September 2008, 04:08 PM
  Boss...hats off...
Re: Divisors of a Number
by Sibsankar Dasmahapatra - Thursday, 11 September 2008, 06:18 PM
 

Hi Donkey

                     The formula for product of divisors It shows the product as = [ a^a.b^b.c^c] ^ [((a+1).(b+1).(c+1))]/2  that  does not mean = [ a^(a.(a+1)) .b^(b.(b+1)).c^(c.(c+1)) ] ^ (1/2).....that means ....

[ a^(a.(a+1).(b+1).(c+1)) * b^(b.(a+1).(b+1).(c+1)) *       c.(a+1).(b+1).(c+1)) ] ^ (1/2)...so in your example...2^2.3^2..as per formula...(2^2.9 * 3^2.9)^(1/2)...that means... 2^9.3^9...so i think formula for product for divisors is correct ..

                  Can u plz reply back....

thanks

Sibsankar


Re: Divisors of a Number
by Sibsankar Dasmahapatra - Thursday, 11 September 2008, 06:53 PM
 

Hi Manish

                           your answer is correct...

 But i can sugest u to solve the last part (i.e.--hence number of ways so as to write as a product of two coprime numbers is ) by above theory.....2^n-1..(here is n=5...).

 

thanks

Sibu

Re: Divisors of a Number having unit dig 5
by vikas sharma - Thursday, 19 February 2009, 09:57 AM
 

hi as i feel , sum of even divisors=total divisors sum-odd divisors sum

total divisors sum =17909892 from above formula

sum of odd divisors= 36-1/3-1 *55-1/5-1

so subtract it u will get ur ans.Please update me anybdy if i m wrong.

Re: Divisors of a Number having unit dig 5
by Sam Rox - Friday, 24 April 2009, 04:40 PM
  Hi sir,
         Its awesome thanks a toncool
Regards
Sam
Re: sum of reciprocals of divisors
by dibya ranjan pal - Thursday, 28 May 2009, 12:27 AM
 

All the divisors of 360, including 1 and the number itself, are summed up. The sum is 1170. What is the sum of the reciprocals of all the divisors of 360?...have no clue for this..

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Re: sum of reciprocals of divisors
by TG Team - Friday, 29 May 2009, 08:51 AM
  Hi Dibyasmile
Its 1170/360 = 3.25 smile
Re: sum of reciprocals of divisors
by dibya ranjan pal - Saturday, 30 May 2009, 11:25 AM
  Thanks kamal.......smile
Re: hi tg
by dibya ranjan pal - Saturday, 30 May 2009, 11:40 AM
 

How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?...

How to solve this???...

Re: hi tg
by dibya ranjan pal - Saturday, 30 May 2009, 11:41 AM
 

How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?...

How to solve this???...

Re: hi tg
by nishant kumar - Saturday, 30 May 2009, 04:35 PM
  a number when devide by N leaves remainder 4.When one third  of yhus number is devide by N,it leaves 29.what is the least value of such number greater than 1000??how to solve this sort of problems?
Re: hi tg
by Total Gadha - Sunday, 31 May 2009, 11:42 PM
  Let the number be T. T/3 = Nq + 29 ---> T = 3Nq + 87. 87 when divided by N leaves remainder 4 ---> N = 87 - 4 = 83. Therefore T = 87q + 4. Smallest such number greater than 1000 = 1044 + 4 = 1048
Re: hi tg
by dibya ranjan pal - Monday, 1 June 2009, 11:54 AM
 

hello tg,

i have a doubt regarding previous answer..

after getting N= 83 if we put it in T=3Nq+87...then it becomes

T=3*83*q +87

or T=249*q +87

Then the smallest no comes out to be , T=1083....am i right?

Re: hi tg
by dibya ranjan pal - Monday, 1 June 2009, 01:49 PM
 

hello tg,

i have a doubt regarding previous answer..

after getting N= 83 if we put it in T=3Nq+87...then it becomes

T=3*83*q +87

or T=249*q +87

Then the smallest no comes out to be , T=1083....am i right?

Re: hi tg
by dibya ranjan pal - Monday, 1 June 2009, 01:50 PM
 

hello tg,

i have a doubt regarding previous answer..

after getting N= 83 if we put it in T=3Nq+87...then it becomes

T=3*83*q +87

or T=249*q +87

Then the smallest no comes out to be , T=1083....am i right?

Re: hi tg
by nishant kumar - Monday, 1 June 2009, 03:32 PM
 

@Dibya

I am also getting the same answer i.e. 1083

and moreover same answer is given in the answer sheet as well.

Also 1048 is not satisfying the conditions of question and one third of this number is not an integer as well.

so I suppose 1083 is the right answer.

 

Re: hi tg
by dibya ranjan pal - Monday, 1 June 2009, 04:24 PM
 

Hi nishant,

 can u plz tell me where the answers are given???

Re: hi tg
by nidhi soni - Saturday, 6 June 2009, 07:47 PM
 

30?

m ll confused .....is this qs complete?

 

Re: hi tg
by nidhi soni - Sunday, 7 June 2009, 07:11 PM
 

heyyyyy i got my ans yes i was just missing a point :D

hey dibya see 31^2 is the highes sq which will give u number between 1-1000

so total number will b 31c2

coz 32 numbers will give u 31 numbers

31 will 30

n so on

so my final and sure ans is is 465

Re: Divisors of a Number
by cat killer - Tuesday, 9 June 2009, 10:47 PM
 

Hi,

I am not able to understand how the answer for the below question is 25x25.

How many divisors are perfect cubes in 36^36?

I am getting answer as 24x24...and the article has 25x25 as ans..not able to understand which one I missed to count.

Regards

SK

 

Re: hi tg
by sapna jain - Wednesday, 17 June 2009, 12:25 AM
 

hey nidhi

cud u pls elaborate ur solution....

diff of squares of two numbers.......??

Re: hi tg
by sapna jain - Wednesday, 17 June 2009, 01:00 AM
 

i think the ans shud b 499 (no. of odd no.s  between 1 and 1000 excepting 1)

reason...the diff of squares of two consecutive no.s increase in the airthmetic progression denoted by odd no.s 2^2 - 1^2 = 3  3^2 - 2^2 =5...and so on...till we get 999 = 500^2 - 499^2

So the no.s wud be 3, 5, 7, 9, 11, .....999

 

Re: hi tg
by sapna jain - Thursday, 18 June 2009, 06:03 PM
 

Sorry! In the above solution...we wud also add all multiples of 4 beginning at 8 (diff of 2 consecutive odd or 2 consecutive even no.) = 249

Hence total numbers between 1 and 1000 that can be written as diff of squares of two positive integers is 499 + 249 = 748

Re:How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?...
by yogesh bansal - Sunday, 21 June 2009, 11:19 PM
 

smilehi dibya...

soln.2 ur prblm..

every odd no. nd a multiple of 4 can be expressed as difference of square of 2 non negative integers.

as 4n = 2n+2n = (n+1)2-(n-1)2

2n+1 =n2+2n+1-n2 =(n+1)2 - n2

so, ans is 750.

smile

Re: hi tg
by yogesh bansal - Sunday, 21 June 2009, 11:26 PM
 

hey nishant kumar

m getting ans as 1083..nd i hv verified ...m correct..

Re:How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?...
by sapna jain - Thursday, 25 June 2009, 01:16 AM
  you will not be able to xpress 1 & 4 as a difference...so 748
Re:How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?...
by saitu gupta - Friday, 26 June 2009, 11:34 AM
  Dear Sapna

the question asks for non negative integers which also includes 0(zero as non negative)

therefore u can express 1 and 4 as difference of (1^2-0^2)and (2^2-0^2)respectively..thus the answer is 750 and not 748

hope u got that..for further clarification ..mail me

abiding_saitu@yahoo.in

regards
saitu
Re:How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?...
by yogesh bansal - Friday, 26 June 2009, 09:05 PM
 

1 = 1^2 - 0^2

4 = 2^2 - 0^2

Re:How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?...
by sapna jain - Sunday, 28 June 2009, 11:24 PM
 

thanks for the clarification friends...i misread the question as natural nos.

Sapna

Re: Divisors of a Number
by Rohit Ghosh - Monday, 29 June 2009, 01:40 AM
  Hi , can anybody help me in explaining , why in the last problem 2*320 is done ?
Re: Divisors of a Number
by akhil mal - Thursday, 19 November 2009, 04:03 PM
  in the last Q is'nt it should be (prime no to 6+ prime no to 15 - prime no to both 6 and 15.)
Re: Divisors of a Number
by amar goswami - Thursday, 19 November 2009, 08:38 PM
  Nope Akhil, we have to completely neglect the numbers which are prime to both 6 and 15. We had counted them twice hence subtracted twice.smile
Re: Divisors of a Number
by akhil mal - Thursday, 19 November 2009, 09:42 PM
 

@ amar goswami

we have to count numbers which r prime to 6 or 15. now we count no which r prime to 6 then for 15 then for which r prime to both 6 and 15 so we have to subtract them once only .

or if i understand it wrongly then please explain it again 

 

Re: Divisors of a Number
by amar goswami - Thursday, 19 November 2009, 10:20 PM
  okay, i got what u meant..  Tu mera hi bhai hai.. smile i also missed 'either' when i attempted it first time tongueout.
well, read the question carefully.. question says numbers which are prime EITHER to 6 or 15.(ek time pe ek hi, dono nahi, if u include both, either wali condition violate ho jayegi). Hence we are excluding the numbers which are prime to both of them...
Thanks to TG for putting such wonderful questions here.. these are the mistake we tend to make in CAT also..

Re: Divisors of a Number
by Pallav Jain - Friday, 20 November 2009, 04:33 AM
  Hello Guys !!

Can anyone tell me the how to solve this question ?

If K = 231 × 319, then how many positive divisors of K2 are less then K but do not divide K?

1228
1028
639
589
1024

Cheers !!!
pallav
Re: Divisors of a Number
by akhil mal - Friday, 20 November 2009, 06:21 PM
 

@amar goswami

smilenow we have to learn english first before doing quant questions

thx a lot to u and to tg sir for such a nice Question

Re: Divisors of a Number
by ankit dikshit - Friday, 27 November 2009, 05:27 PM
  great work toal donkey.. blessed to have students like u in TG fraternity..
thanks to TG 2
Re: Divisors of a Number
by tapas mani shyam - Sunday, 29 November 2009, 01:06 AM
  Hi Pallav

Is the answer 639??
Re: Divisors of a Number
by Suraj Saluja - Sunday, 24 October 2010, 10:30 PM
 
Hey guys

Please help me in solving this -


N is a (n + 1) digit positive integer which is in the form anan - 1an - 2 . . . a2a1a0, where ai (i = 0, 1, 2, . . ., n) are digits and a
n width=150, thus N = an 10n + an-1 10n - 1 + . . . + a1 10 + a0, where 0 width=13 ai width=13 9 and anwidth=150. We define F (N) = (a+ 1) (an - 1 + 1) . . . (a+ 1) (a+ 1) For Example F(3407) = (3 + 1) (4 + 1) (0 + 1) (7 + 1) = 160.
Identify the number of two digit numbers such that F(N) = N + 1.

a. 9
b. 1
c. 6
d. 5
Re: Divisors of a Number
by Nikhil Sinha - Friday, 29 October 2010, 02:11 PM
  Hi Suraj,

Just saw your question....I think I solved it, but you won't believe, am not sure of the proper logic..

Anyways, here it is:

We require nos such as F(N)= N+1 (N has to be a 2 digit no)

First of such no is 19, as F(19) =(1+1)(9+1)=20= 19 +1
Next is the next two digit no ending in a 9, i.e 29 as F(29)= (2+1)(9+1)= 30= 29+ 1

Now, going by the trend, we can see that all of the below:

39, 49, 59...99 will be showing similar property..

So, basically we have 9 such 2 digit nos.


Now I need to take a closer look at my solution, but still thought to post it by that time... smile

Nikhil
Re: Divisors of a Number
by Mithil Shah - Tuesday, 2 November 2010, 07:32 PM
  hey,nice article ..

If we can solve below q with approach given above , pl help ..

Q> How many odd numbers between 150 and 350 are divisible neither by 9 nor by 11?

Re: Divisors of a Number
by R V - Tuesday, 9 November 2010, 10:55 PM
  i am not sure about my approach. kindly correct me incase i go wrong.

odd numbers between 150 and 350 are divisible neither by 9 nor by 11 = total odd numbers between 150 and 200 -(odd multiples of 9) - (odd multiples of 11) + (odd multiples of 11 and 9)

total odd numbers between 150 and 350 is = (350-150)/2 = 100.

odd multiples of 9 = [total multiples of 9 in the range ]/ 2 (since every alternate multiples is a even number for 9 (as well as 11)). (= 11)

where [] stands for smallest integer value.

odd multiples of 11 = total multiple of 11 in the range /2.(=9)

odd multiples of 11 and 9 = total multiples of 99 in the range/2.
( =1)

so according to me , answer should come out to be:
100 - 11 - 9 + 1 = 81.
Re: Divisors of a Number
by R V - Tuesday, 9 November 2010, 11:00 PM
  @suraj, my answer is also 9 and here is the solution:

let the 2 digit number be ab.
hence ab can also be written as 10a + b;

now F(ab) = (a+1)(b+1)= ab+a+b+1

for F(ab) = ab +1;

we have:

ab + a + b + 1 = 10a + b +1;
hence, we have

ab = 9a;

b = 9;

and a can be any value including 1-9. hence 9 different values

Re: Divisors of a Number
by Kamal Joshi - Thursday, 1 September 2011, 01:14 AM
  @Pallav,

Is this a valid question? All divisors of K^2 less than K will be = (divisors of K) - 1 (Subtracting one coz K is not less than K). and every divisor of k will ofcourse divide K.

Am I missing any point here?
TG sir, pls enlighten.

Regards,
Kamal



Re: Divisors of a Number
by In lonely planet i live - Sunday, 11 September 2011, 02:05 PM
  Sir,

i wanted to ask here a question. sorry if it is repeated:

Is 2222^7777 + 7777^2222 divisible by 99?


for me its divisible:

as 99 = 9 * 11 ( co-prime). so first test with 11, its obvious that it will be divisible by 11. Then comes 9: 2222^7777 will have reminder 8 and 7777^2222 will have reminder 1. Hence 8 + 1 = 9 agin divisible by 9. Hence total no is divisible by 99.



Please let me know the correctness of the answer. Thanks in advance.
Re: Divisors of a Number
by TG Team - Sunday, 11 September 2011, 04:59 PM
 

That's correct In lonely planet i leave smile

Kamal Lohia

Re: Divisors of a Number
by In lonely planet i live - Sunday, 11 September 2011, 05:44 PM
  Thanks a lot sir!!!
Re: Divisors of a Number
by AJAYSANTHOSH BALMANI - Tuesday, 20 September 2011, 08:32 PM
  How many of the first 1200 natural numbers are either prime to 6 or 15?
I think the answer is

400+640-320=400+320=720(coz either prime to 6 or 15)

And not 400(320 subtracted twice)

If above one is not the right answer please explain.
Re: Divisors of a Number
by rimmi rimmi - Sunday, 30 September 2012, 09:57 PM
  thanks a lot sir smile

I am stuck with this question sad can anyone please solve this smile


123456789101112131415161718192021222324252627.....
upto4041424344 div by 45? what will be the quotient and remainder?
Re: Divisors of a Number
by neha shugani - Tuesday, 3 December 2013, 10:30 PM
  thankuu TG sir for such a gr8 article.. m new to this site nd m loving it..

PLz help me wid this ques.

In case of some particular dates, when written in the format 'mm/dd/yy', it is found that yy = (mm) * (dd)
for e : 22nd april 1988 is written as 04/22/88, where 88 = 4 *22

Between the year 1950 and 2000 what was the maximum possible number of days in a single year, in which this could hav happened?

Re: Divisors of a Number
by neha shugani - Tuesday, 3 December 2013, 10:52 PM
  589 is the ans?
Re: Divisors of a Number
by TG team - Saturday, 7 December 2013, 03:54 PM
  Hi nehasmile
To find the maximum number of days in a single year we have to look out for an year with maximum number of prime factors.
So, a few years between 50 and 100 with more than 3 prime factors are : 54,60,72,80,81,88,90 and 96.
Since the number of months can be up to 12 and number of days can be up to 31,so on a careful look we find that the only the options 60 and 90 have the maximum days.
CASE 1: 1960
2/30/60
3/20/60
4/15/60
5/12/60
6/10/60
10/6/60
12/5/60

CASE 2:1990
2/45/90(not possible)
3/30/90
5/18/90
6/15/90
9/10/90
10/9/90
Clearly,1960 is the answer.
Hope i am clearsmile