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Base system
by Total Gadha - Tuesday, 20 February 2007, 11:59 PM
 

                            image

Suppose you have a 1 000 L tank to be filled with water. The buckets that are available to you all have sizes that are powers of 3, i.e. 1, 3, 9, 27, 81, 243, and 729 L. Which buckets do you use to fill the tank in the minimum possible time?

You will certainly tell me that the first bucket you will use is of 729 L. That will leave 271 L of the tank still empty. The next few buckets you will use will 243 L, 27 L and 1 L. The use of buckets can be shown as below

image

We can say that

1 000 = 729 + 243 + 27 + 1

= 1 Ã— 3 6 + 1 Ã— 3 5 + 0 Ã— 3 4 + 1 Ã— 3 3 + 0 Ã— 3 2 + 0 Ã— 3 1 + 0 Ã— 3 0 .

The number 1 000 has been written in increasing powers of 3. Therefore, 3 is known as the ‘base’ in which we are expressing 1 000.

For example, The number 7368 can be written as 8 + 6 Ã— 10 + 3 Ã— (10) 2 + 7 Ã— (10) 3 .

The number 10 is called the 'base' in which this number was written.

Let a number abcde be written in base p, where a, b, c, d and e are single digits less than p. T he value of the number abcde = e + d Ã— p + c Ã— p 2 + b Ã— p 3 + a Ã— p 4

For example, i f the number 7368 is written in base 9,

The value of (7368) 9 = 8 + 6 Ã— 9 + 3 Ã— 9 2 + 7 Ã— 9 3 = 5408 (this value is in base 10)

There are two kinds of operations associated with conversion of bases:

1.7A conversion from any base to base ten

The number (pqrstu) b is converted to base 10 by finding the value of the number. i.e. (pqrstu) b = u + tb + sb 2 + rb 3 + qb 4 + pb 5 .

Example

38. Convert (21344) 5 to base 10.

Answer: (21344) 5 = 4 + 4 Ã— 5 + 3 Ã— 25 + 1 Ã— 125 + 2 Ã— 625 = 1474

1.7b conversion from base 10 to any base

A number written in base 10 can be converted to any base 'b' by first dividing the number by 'b', and then successively dividing the quotients by 'b'. The remainders, written in reverse order, give the equivalent number in base 'b'.

Example

  1. Write the number 25 in base 4.

                                   image

Writing the remainders in reverse order the number 25 in base 10 is the number 121 in base 4.

1.7c Addition, subtraction and multiplication in bases:

Example

40. Add the numbers (4235) 7 and (2354) 7

Answers: The numbers are written as

                                         image

The addition of 5 and 4 (at the units place) is 9, which being more than 7

would be written as 9 = 7 Ã— 1 + 2. The Quotient is 1 and written is 2.

The Remainder is placed at the units place of the answer and the Quotient

gets carried over to the ten's place. We obtain

                                          image

At the tens place: 3 + 5 + 1 (carry) = 9

Similar procedure is to be followed when multiply numbers in the same base

Example

41. Multiply (43) × (67) 8

Answer:

7 Ã— 3 = 21 = 8 Ã— 2 + 5

7 Ã— 4 + 2 = 30 = 8 Ã— 3 + 6

6 Ã— 3 = 18 = 8 Ã— 2 + 2

6 Ã— 4 + 2 = 26 = 8 Ã— 3 + 2

                                   image

For subtraction the procedure is same for any ordinary subtraction in base 10 except for the fact that whenever we need to carry to the right we carry the value equal to the base.

EXAMPLE

42. Subtract 45026 from 51231 in base 7.

Answer:

                                      image

In the units column since 1 is smaller than 6, we carry the value equal to the base from the number on the left. Since the base is 7 we carry 7. Now, 1 + 7 = 8

and 8 – 6 = 2. Hence we write 2 in the units column. We proceed the same way in the rest of the columns.

1.7D IMPORTANT RULES ABOUT BASES

Rule1. A number in base N when written in base 10 is divisible by N – 1 when the sum of the digits of the number in base N is divisible by N – 1.

EXAMPLE

43. The number 35A246772 is in base 9. This number when written in base 10 is divisible by 8. Find the value of digit A.

Answer: The number written in base 10 will be divisible by 8 when the sum of the digits in base 9 is divisible by 8.

Sum of digits = 3 + 5 + A + 2 + 4 + 6 + 7 + 7 + 2 = 36 + A. The sum will be divisible by 8 when A = 4.

Rule2. When the digits of a k-digits number written in base N are rearranged in any order to form a new k-digits number, the difference of the two numbers, when written in base 10, is divisible by N – 1.

EXAMPLE

44. A four-digit number N1 is written in base 13. A new four-digit number N2 is formed by rearranging the digits of N1 in any order. Then the difference N1 – N2 when calculated in base 10 is divisible by

(a) 9 (b) 10 (c) 12 (d) 13

Answer: c
Re: Base system
by Mitesh Chaturvedi - Friday, 6 April 2007, 12:18 PM
 

Hi TG,

   i had not gone through this lesson of yours.It is worth reading.The last two rules are very useful.

Thanks,

Mitesh 

Re: Base system
by mohammed javed - Tuesday, 15 May 2007, 06:14 PM
  Kudos!...For a useful article....The last two rules are really gud! smile
Re: Base system
by Anand Kishore - Friday, 18 May 2007, 03:58 PM
 

Thanks for the prompt explanation.

earlier i was scared by dese horrendous base probs. 

Re: Base system
by Anand Kishore - Friday, 18 May 2007, 04:00 PM
 

Thanks for the prompt explanation.

earlier i was scared by dese horrendous base probs. 

Re: Base system
by priyank daga - Tuesday, 22 May 2007, 10:31 AM
  the article was great ...can u explain multiplication in bases more clearly
Re: Base system
by shiwa k - Wednesday, 23 May 2007, 05:54 PM
 

hi .. plz clarify

41. Multiply (43) × (67) 8

Answer:

7 Ã— 3 = 21 = 8 Ã— 2 + 5

7 Ã— 4 + 2 = 30 = 8 Ã— 3 + 6

6 Ã— 3 = 18 = 8 Ã— 2 + 2

6 Ã— 4 + 2 = 26 = 8 Ã— 3 + 2

                                   image

in the last step image 365+ 322 = 3707??? 5+2= 7 ok. 6+2=0 ok. but couldnot understand 3+3 =37...

plz clarify

Re: Base system
by Total Gadha - Saturday, 26 May 2007, 07:47 AM
  Hi Shiwa,

Have corrected the mistake. Check now.

Total Gadha
Re: Base system
by gaurav kumar - Monday, 28 May 2007, 06:18 PM
 

Hi TG,

Can you please explain the multiplication in bases more clearly.

Thank you for this wonderful articlesmile

Regards,
Gaurav

Re: Base system
by Ganesh Iyer - Monday, 28 May 2007, 06:34 PM
 

hey Tg,

Good fundas...Also good way of calculating in diff bases.

Keep up the good work

 

Re: Base system
by Gaurav Malhotra - Tuesday, 29 May 2007, 02:57 PM
  Hello sir,
 Base system has always given me a tough time......this article proved to be of immense help....thnx a lot
Re: Base system
by Gaurav Malhotra - Tuesday, 29 May 2007, 03:22 PM
  Sir,
 Can u plz elaborate on the application part of the two rules.....
Re: Base system
by papai 2007 - Thursday, 7 June 2007, 01:07 PM
 

hi tg;

could u plz xplain da subtraction bit more clearly!?!

Re: Base system
by TG Team - Wednesday, 20 June 2007, 07:37 PM
 

Hi TG

Example

41. Multiply (43) × (67) 8

Answer:

7 Ã— 3 = 21 = 8 Ã— 2 + 5

7 Ã— 4 + 2 = 30 = 8 Ã— 3 + 6

6 Ã— 3 = 18 = 8 Ã— 2 + 2

6 Ã— 4 + 2 = 26 = 8 Ã— 3 + 2


                                   image

As you have multiplied (43) × (67) 8 = 3605.

My question is if 3605 in the base of 8 or not. If yes then its expansion in the base 10 must give the same result as the multiplication of 43 and 67 yield.

but this is not the case.

As, (3605)8 = 3*83 + 6*82 + 0*8 + 5 =1925, but 43*67 = 2881.

Please Help......

Re: Base system
by Total Gadha - Wednesday, 20 June 2007, 11:39 PM
  Hi Kamal,

43 and 67 are also in base 8. 43 in base 8 = 35 in base 10. 67 in base 8 = 55 in base 10. 35 × 55 = 1925

Total Gadha
Re: Base system
by TG Team - Thursday, 21 June 2007, 04:53 AM
  thanks TG
for clearing the obstacle on the path to success.
Re: Base system
by Rabia Zareen - Friday, 20 July 2007, 04:16 PM
 

Hi TG,

Could u plz solve the following problem:

How many 4 digit positive integral numbers are there in base 7 if you are counting the numbers in the same base system.

Thanks,

Rabia

Re: Base system
by Total Gadha - Saturday, 21 July 2007, 01:52 AM
 

Hi Rabia,

In base 7, you can have digits from 0 to 6. The lowest 4-digit number in base 7 would be 1000 and the highest 4-digit number in base 7 would be 6666. These are 5667 numbers in all.

Total Gadha

Re: Base system
by Rabia Zareen - Saturday, 21 July 2007, 09:15 AM
 

Hi TG

The ans to the problem is 6000 in base 7.

5666 in base 7 when added to 1 in base 7 will give 6000 in base 7.

Thanks for the prompt reply.

Rabia

Re: Base system
by Rabia Zareen - Saturday, 21 July 2007, 09:30 AM
 

Hi TG,

have another problem. don't know how but once i post my doubt i am suddenly able to solve the ques myself. anyway here's the ques:

Bill & Clinton take the square of a certain no. in decimal system and express it in base 5 and 6 respectively. Then Bush comes and he takes the two representations and assuming that the expressions are in base 10 adds the nos. Which of the following cannot be the value of the unit's digit of the sum obtained?

a) 0         b) 2          c) 8            d) 6           e) 3

Thanks,

Rabia

Re: Base system
by Total Gadha - Saturday, 21 July 2007, 01:26 PM
  Ahh..you need the answer in base 7. I misread the question. 
Re: Base system
by Total Gadha - Saturday, 21 July 2007, 01:52 PM
  The unit digits of the numbers in base 5 and 6 will be the 1st remainders with 5 and 6. The squares ofnumbers give remainder 0, 1 and 4 with 5, and 0, 1, 3 and 4 with 6. Therefore unit digits of one number can be 0, 1 and 4, and unit digits of the other number can be 0, 1, 3 and 4. The unit digit of the sum can be 0, 1, 2, 3, 4, 5, 7 and 8. It cannot be 6. 
Re: Base system
by Software Engineer - Wednesday, 25 July 2007, 04:21 PM
  hi TG uncle

Exa. 40

would be written as 9 = 7 Ã— 1 + 2. The Quotient is 1 and written is 2.

its Remainder

i
love (typin) mistakes

becomputer06@gmail.com
Re: Base system
by Akon Convict - Thursday, 2 August 2007, 08:09 PM
 

Hi Tg

what if we add   (6789) base 8

                       + (2354) base 8

 

-----------------------------------
Re: Base system
by Ranvijay Singh - Thursday, 2 August 2007, 10:01 PM
  Hi Akon Convict,

How can you have digits 8 and 9 in base 8? Possibly a typosmile

Replacing 8 and 9 with largest possible digit in base 8, which is 7 and continuing with the addition:

 +1 +1 +1
   6   7   7   7
   2   3   5   4
-----------------
1 1   3   5   3

So, (6777)8 + (2354)8 = (11353)8

[Explanation: 7 + 4 = 11 = 8 * 1 + 3, so write 3 as the rightmost digit and take 1 as carry. Proceed the same way from right to left and finally when there is no pair to add the carry with
, simply write the sum of the final pair of digits first and then the carry as the leftmost digit of the total sum (Of course, if there is a carry left after the addition of the leftmost pair of digits. It's quite the same as we do standard addition in base 10, isn't it? smile)]


Vijay

Re: Base system
by ashish tyagi - Friday, 3 August 2007, 06:25 AM
  hi t g great work dude, i want to know tht i have some problem so where i can post these pls tell me , now i'm posting one problem here.............

A piece of equipment cost a certain factory Rs. 600,000. If it depreciates in value, 15% the first year, 13.5 % the next year, 12% the third year, and so on, what will be its value at the end of 10 years,
how we will solve such type questions............
Re: Base system
by Akon Convict - Friday, 3 August 2007, 09:20 AM
 

Oh thanks Ranvijay.....

u cleared my concept

Re: Base system
by Ranvijay Singh - Friday, 3 August 2007, 11:34 AM
  Hi Ashish,

There is a forum named 'CAT Quant-D.I. Forum' on TG. You may think of posting QA/DI questions there. Please find below the solution to your question:

Using formula for nth term of an AP, Tn = a + (n - 1) * d
Here a = 15, d = -1.5, n = 10
% depreciation at the end of 10th year = 15 + (10 - 1) * (-1.5) = 1.5

Using formula for Sum of n terms of an AP, Sn = n/2(a + l)
total depreciation % at the end of 10th year = 15 + 13.5 + ... + 1.5 = 10/2(15 + 1.5)
= 5 * 16.5 = 82.5

Hence, cost of the equipment after 10 years = 100 - 82.5 = 17.5 % of the original CP = 17.5 * 600000 / 100 = Rs. 105000.00


Vijay
Re: Base system
by rmmozhi prathiba - Sunday, 2 September 2007, 12:06 AM
 

hi tg.. i have a doubt how to solve the following problem:

if the numbers 2^218 and 3^109 are both evaluated converted into base 12 and written one beside the other, how many digits are there, on the whole, in the result??

 

Re: Base system
by unbreakable belief !!!!! - Wednesday, 5 September 2007, 09:33 AM
 

The unit digits of the numbers in base 5 and 6 will be the 1st remainders with 5 and 6. The squares ofnumbers give remainder 0, 1 and 4 with 5, and 0, 1, 3 and 4 with 6. Therefore unit digits of one number can be 0, 1 and 4, and unit digits of the other number can be 0, 1, 3 and 4. The unit digit of the sum can be 0, 1, 2, 3, 4, 5, 7 and 8. It cannot be 6. 

 

sir i m not getting it .please clearify it.why can"tthe square of a number can give 5 as a reminder in base 6?

Re: Base system
by sangeeth aloysius - Wednesday, 5 September 2007, 05:51 PM
  Hi TG,
         Here is what I thought about this problem. Kindly verify
The units digits of a square can be 0,1,4,5,6,9
For these nos the last digit in the two bases upon conversion are as follows
Unit Digit     Base:5   Base:6
0                  0            0
1                  1            1
4                  4            4
5                  0            5
6                  1            0
9                  4            3
Hence when these no are considered as base 10 nos and added units digit can be
0,2,8,5,1,7
Only these because it is the same number that is represented in two bases.
Hence Answer is 6,3 
Re: Base system
by ashish bhardwaj - Saturday, 15 September 2007, 11:14 PM
 

Hi prathiba ,

I think answer will be 110.I have done as follows.

2^218 => 4^109

so it has become 3^109*4^109 ==> 12^109 ...now be want it in base 12.

so if you want to write    N^n  in base N answer will be n+1 .

Need not to remember this theorm as it striked just now smile .

for e.g. 2^2 in the base 2 => 4 in the base 2 => 110 so 2+1 =3

            10^2 in the base 10 => 100 => 2+1 /

           2^4 in the base 2 it will be  10000 ...

 

Re: Base system
by rmmozhi prathiba - Sunday, 16 September 2007, 07:00 AM
 

hi ashish thank you

it was a very clear solution and the answer is right thank u..

Re: Base system
by Vinod Paramasivan - Sunday, 16 September 2007, 08:08 PM
  Hi TG/Rabia,

Can anyone explain how u got 6000 as the answer?

5666 in base 7 gives 22343.
So 5667 would be 22344.

Thanks,
Vinod
Re: Base system
by naga kiran kosuru - Tuesday, 16 October 2007, 12:34 PM
 

hi TG :

This is Naga Kiran

in ur explanation for the no of four digit numbers in base 7, u did a small mistake.

ok the lowest  4 digit no in base 7 is 1000 n highest is 6666 , so, the total nos is not the difference of 6666 n 1000 but the difference between their converted nos in base 10 ie 2400-343 = 2058.......or u can even do it as :

4 digit nos in base 7 = ---- 6*7*7*7 = 2058

Re: Base system
by ishan ishu - Tuesday, 29 April 2008, 02:50 PM
 

Hi TG,

Jus wanted to clarify one thing. In the above question I used some different logic.

My logic was:

For 4 digit no.

1st place can be filled by 6 digits (1,2,3,4,5,6) as we are calculating in base 7.

similarly for 2,3 and 4th place we can use 7 digits.

Thus in all there are 6*7*7*7=2058 numbers..

and 2058 in base 7 is 6000.

But the logic you explained is by taking 1000 and 6000 as lowest and highest no and then calculating the total no's but these will also consist of no's like 1800, 1900 which doesnt exist in base 7. So is this logic correct or I am at some fault.

Please Clarify.

Thanks

Ishan

Re: Base system
by Total Gadha - Wednesday, 30 April 2008, 09:42 AM
  The lowest number is (1000)7 and the highest is (6666)7. Since you want to calculate the number of numbers (you rightly pointed out that (1800)7 will not exist) you need to subtract in base 7 itself.
Therefore the number of numbers = (6666)7 - (1000)7 + (1)7 = (6000)7 = 2058.
Re: Base system
by ishan ishu - Wednesday, 30 April 2008, 01:53 PM
 

Ok..

Thanks TG. Got it now. That was what I was thinking that how can u be wrong.wink

Regards

Ishan

Re: Base system
by ishan ishu - Friday, 16 May 2008, 12:13 AM
 

Hi TG,

TG, if you have time can you please post one lesson on Cubes too. I dont find it that easy while solving questions as I dont know the approach to solve it.

Regards

Ishan

Re: Base system
by shashi bhushan verma - Friday, 16 May 2008, 12:56 PM
 

Hi Please confirm the answer.

The solution as per me is : the unit digit of the square number in base 10 can be 0,1,4,5,6,9. The remainder for the same in the base 5 is 0,1,4,0,1,4 and in base 6 is 0,1,4,5,0,3. adding in sequence. the possible value of the unit digit can be 0,2,8,5,1,7. So the values that can not be unit's digit is 6,3.

Please let me know if my solution is correct or not.

Re: Base system
by sunil garg - Tuesday, 20 May 2008, 04:54 PM
 

zero could'nt come in unit place

am i right????????

............do or die for CAT08

SKGARG

Re: Base system
by Rohan Agarwal - Thursday, 29 May 2008, 06:59 PM
  Hello TG and Ashish....
this is 9 moths old prob ...but i read it today...and didn't get the solution...can you please help me out......
Re: Base system
by ATOM ANT - Sunday, 1 June 2008, 08:50 AM
 

How many 4 digit positive integral numbers are there in base 7 if you are counting the numbers in the same base system.

what is the logic between 6*7*7*7 = 2058

is it not 7*7*7*7 as repetition is allowed...

Kindly explain.

Re: Base system
by ATOM ANT - Sunday, 1 June 2008, 08:53 PM
 

if the numbers 2^218 and 3^109 are both evaluated converted into base 12 and written one beside the other

They are not multiplied, but written along each other... how can we get 12^109.. Kindly clarify

thanks TG

Re: Base system
by ATOM ANT - Saturday, 7 June 2008, 07:36 PM
  Pls..Someone clarify my doubt...sad
Re: Base system
by Total Gadha - Saturday, 7 June 2008, 11:06 PM
  Hi Atom,

You cant get. Work with small powers and check.

Total Gadha
Re: Base system
by vijay kumar - Tuesday, 10 June 2008, 11:24 AM
 

Hi all,

Can somebody explain me how TG arrived at rule2 in the above lesson.

Any help would be greatly appreciated.

 

regards,

Vijay.

Re: Base system
by ATOM ANT - Friday, 13 June 2008, 10:20 PM
  Thanks TG. I got it.
 I will do permutations and combinations to understand my first doubt.If I still do not get that i ll ask you.
Re: Base system
by srinivasa reddy - Saturday, 14 June 2008, 04:16 PM
  12^n1 = 4^109   n1=109 (log4)12

12^n2=3^109      n2=109 (log3)12

no of digits n1+n2= 109 (log12)12=109

individually how to determine the digits??is it 107
Re: Base system
by Total Donkey - Thursday, 26 June 2008, 05:05 PM
  @vijay

The basic funda is sonething i can explain to you

for simplicity let us take a number having 4 digits (abcd)base n

now, m showing the result for a few ramdom permutation (rearrangement), i hope you can generalize it:

lets take dbac

abcd = a.n^3 + b.n^2 + c.n^1 + d.n^0

dbac = d.n^3 + b.n^2 + a.n^1 + c.n^0

now abcd - dbac = a(n^3 - n^1) + b(n^2 - n^2) + c(n^1 - n^0) + d(n^0 - n^1)

= (n-1) [ a(n^2 + n^1 + 1) + c - d]

so we can get (n-1) as the common factor

If you might be wondering that it worked in case of my permutation.. let me explain
without loss of generality ... we can take one of the term of
abcd - (permutation of abcd) =  a[n^x - n^y]

= a.n^y.[n^(x-y) - 1]

and (n-1) will surely be a factor of [n^(x-y) - 1] (unless x-y=0 which makes the whole term Zero)

Hence, for each of the term we can take (n-1) safely as the common factor.

So, the number obtained after subtracting would be divisible by (n-1) and its factors ..

@Total Gadha

the number in base n is not necessarily divisible by (n-1) [when the sum of digits is divisible by (n-1)] in base 10 only ...
it is divisible by (n-1) in any base .... isn't it?
Kindly clarify



Re: Base system
by Ankur B - Thursday, 7 August 2008, 01:43 PM
  I think we will have to add 2 to it....Ans should be 109 + 2 = 111
Re: Base system
by Partha D - Thursday, 7 August 2008, 04:56 PM
 

12^n1 = 4^109   n1=109 (log4)12

12^n2=3^109      n2=109 (log3)12

n1+n2= 109 (log12)12=109

The answer is 109+1=110

Let's take the numbers 2^7=128 and 5^7=78125 in base=10. If we place these numbers side by side, total no. of digits will be 8.

Now,

log(2^7)10 + log (5^7)10

=7* log(2*5)10

=7

So, we have to add 1 to get the no. of digits.

But I could not derive any general formula for it.

Re: Base system
by Prateek Agarwal - Thursday, 7 August 2008, 08:45 PM
 

How many 4 digit positive integral numbers are there in base 7 if you are counting the numbers in the same base system.

what is the logic between 6*7*7*7 = 2058

is it not 7*7*7*7 as repetition is allowed...

Kindly explain.

Hi Atom,

The first digit cant be zero so we cant take 7 for first place.

( possible digits for thousands place are 1,2,3,4,5,6)if wwe take 0 here it will be a three digit no

Re: Base system
by ravi mr. - Monday, 6 October 2008, 04:25 PM
  the problem requires one to write the digit side by side and need not multiply it as u have done........so think for some other solution.........

by rsn
Re: Base system
by ravi mr. - Monday, 6 October 2008, 04:59 PM
  the problem requires one to write the digit side by side and need not multiply it as u have done........so think for some other solution.........

by rsn
Re: Base system
by Neelesh Sethi - Sunday, 26 October 2008, 12:10 AM
  heres a prob if binary nos from 100 to 1000000 are written find the no of 1s in it
Re: Base system
by ATOM ANT - Sunday, 26 October 2008, 06:26 PM
  Hi Neelesh,

The answer is 124.
In a n-digit number the first number is always 1.Each digit-place can be filled in 2 ways... so the no:of ways for a n-git no = 2n-1 ways...
smile
Re: Base system
by Neelesh Sethi - Monday, 27 October 2008, 12:53 PM
  hey thnx for trying but the ans is 189.I have the solution but cannot understand it.


Total number of 1’s on first position = 61.

total number of 1s after 1st position =128

Since total number of 1’s = 128 + 61 = 189


Re: Base system
by ATOM ANT - Tuesday, 28 October 2008, 08:04 AM
  Hi Neelesh.. Its indeed 189... I did not read the question carefully...

For any n-digit number find the no:of arrangements and in turn the no:of ones...

for example from 1000 to 1111
[Fix the first digit as 1.It is not going to change]

1000 - 1*1 =1
1100 - 3!/2! ways of writing the nos with first digit as 1
So no:of Xones= 3!/2! * 2 = 6
1110 - 3!/2! * 3 = 9
1111 - 1*4 = 4
So total no:of ones in a 4-digit no = 20

Likewise do for other cases...
Re: Base system
by Neelesh Sethi - Tuesday, 28 October 2008, 01:21 PM
  hey thnx got it now
Re: Base system
by ravin verma - Wednesday, 5 November 2008, 12:09 AM
 

cool  ]

okee not to much effectivvvvveeee

 

Re: Base system
by aashish biala - Wednesday, 12 August 2009, 11:32 AM
 

hi TG,

thanks a lot. before this i did not know how to add in bases. thankyou.

cheers

aashish biala

Re: Base system
by vikas tiwari - Wednesday, 6 October 2010, 10:57 AM
  Hello Sir,
new to totalgadha, one doubt regarding d expression

(7368)base 9= 8+6A-9+3A-81+7A-729.

what is A over here.....havin difficulty in understanding how to convert to otherbase and what shud be the value of A?

Please explain.....
Re: Base system
by Manas Padhi - Wednesday, 13 October 2010, 01:21 PM
 

Hi TG Sir,

Your rule 1 says a number in base n is divisible by n-1 in base 10 if sum of the digits is divisible by n-1.

Whereas SE Sir under Rule 1 of Division and Bases section has explained lucidly that a number in base n is divisible by n-1 in base n if sum of the digits is divisible by n-1.

Please help us with the correct rule.

Regards,

Manas

Re: Base system
by nitish mayawala - Monday, 8 November 2010, 12:22 PM
 

Q:The product of two numbers ‘231’ and ‘ABA’ is ‘BA4AA’ in a certain base system (where base is less than 10), where A and B are distinct digits. What is the base of that system?

Re: Base system
by nitish mayawala - Monday, 8 November 2010, 12:31 PM
 

Hey shashi can u plz tel me what about 4 and 9 coz we r getting sum as 0,1,2,5,7,8 ..here 3,6,4,9 are missing??

 

Re: Base system
by amit kheterpal - Monday, 8 November 2010, 04:10 PM
 

Converting into decimal base system

Let a be the base

(231)a * (ABA)a=(BA4AA)a

Final equation Iam getting is

a^3(2A-B)+a^2(A+B)+2a(A+B+2)+ 2A+B=0

Not able to find value of a

Please explain.

 

Re: Base system
by Nikhil Sinha - Friday, 12 November 2010, 01:33 AM
  Hey Amit,

This one is a tough problem, often discussed in various forums.. (Hope such probs dont come my way :D)

P.S.: U got the start for this one though.. smile

Here is the soln to the problem:

(231)n * (ABA)n=(BA4AA)n

(2n+1)(n+1)(An^2 + Bn + A) = BA4AA
BA4AA is divisible by n+1, hence (A+4+B) - (A+A) = 0 or (n+1)*k,where k is an integer.

OR (another way of saying the same thing)

231 = 11*21 (in any base system)
so we know that BA4AA has 11 as a factor
applying divisibility rule of 11 we get B+4-A= 0 or n+1(here n is the base of the system)

so B+4-A= n+1 ---------(1)

now take the product
we get 231*ABA= 2An^4+(2B+3A)n^3+(3A+3B)n^2+(3A+B)n+A
Replace all Bs with n+A-3 (from (1)) then compare it with BA4AA

so we get 2A+2= B comparing power of n^4 and
4A-3=A
so A=1 and B =4
therefore, n =B-A+3 =6


Nikhil cool