U need to have an approach to solve this. Lets solve this logiccally

Here is the Approach
Given that N=36k+23 Here k could be either odd or even.
we got to find the reminder when 36 multiplied by a number, is divided by 72. So in the above equation K could be of odd or even numbers. give k values starting from 0 to 4

Case 1. When k=0 (36*0)+23 /72 gives reminder 23
Case 2. when k=1 (36*1)+23 /72 gives reminder 59.How we got59 is that-36/72gives reminder36.This 36+ 23=59. )
Case 3.when k= 2 (36*2)+23 /72 gives reminder 23
Case 4.When k=3 (36*3)+23 /72 gives reminder 59
So from this it can be understood that reminder can be either 23 or 59 based on the value of k if its even or odd.
In the choice only 59 is given.
So the answer is B 59 BINGO!!

N = 36*a+23. Try a = 1, then N = 59. 59/72 = 0 with remainder 59. Answer is B.

Answer is B , but is there a generic formula or approach for this kind of problem instead of trail and error approach ?

I got B... Used the long approach
List multiples of 36
Add 53 to them and compare result to answers

So the learning from this problem can be summarized below :

If a number "N" is divided by "D" and leaves a remainder "R" , then if that same number "N" is divided by "k*D" where "k" is a positive integer , then the number of remainders that are possible are R , D+R , 2D+R , .... , (k-1)D + R.

So if the question was that what is the remainder when the number was divided by 288 , we can approach it in a mechanical way.

The possible remainders are 23 , 36+23 , 72+23 , .... , 252+23 . We can straightaway reject answer choices that are not in this list.

Sample question : A number N divided by D gives a remainder of 7 , the same number divided by 5D gives a remainder of 24. Find D [N and D are both positive integers].

Ans: The possible remainder for the 2nd case are 7 , D+7 , 2D+7 , 3D+7 , 4D+7.
So 7<> 24, case closed for 1st possibility ,
D+7 = 24 or D=17 (ok)
2D+7= 24, not ok as D cannot be fraction.
3D + 7=24, not ok as D cannot be fraction
4D+ 7 = 24, not possible as D cannot be fraction.

So possible values of D is only 17. So 17 must be present among the answer choices.

Hi guys,

what do you think about this approach ?

N = 36 A + 23
N = 72 B + X ( A, B and X are integers)

This implies that : 36 A + 23 = 72 B + X or 36 (A - 2B) = X - 23

X - 23 must be a multiple of 36. The only option is 59.

Not sure whether this is a good approach. Anyway it took me almost 2 min.

It is given that
\(n = 36p+23\)

let r be remainder when n is divided by 72 so
\(n = 72q+r\) where p and q will be integers.

from the two we have

\(36p+23 = 72q+r\)
=> \(r = 23+36p-72q\)
=> \(r = 23+36(p-2q)\)

since p and q are integers \(p-2q\) is also an integer so it can be ...-3,-2-1,0,1,2,3...

If p-2q = 0 then r =23
If p- 2q = 1 then r = 23+36 = 59 we can stop here and pick B.

You only need to know that n = 36p + 23

Now when 36p+23 / 72 what is the remainder?

Well if p = 1 then remainder is 36+23 =59, which is answer choice B

So B

Cheers
J

n= 36x + 23

If we substitute number x= 1

we get n= 59, and when n is divided by 72 we get 59 as the remainder.

We have 59 as one option, and hence we don't need to do any further calculations

B is the answer
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We can let the number be 36 + 23 = 59. When 59 is divided by 72 the remainder is 59.

Alternate Solution:

We observe that 72 is exactly twice 36. If the remainder when a number is divided by 36 is 23, then the remainder when the same number is divided by 72 can be either 23 or 23 + 36 = 59.

If this is hard to see, let’s illustrate the same phenomenon in a simple example. Let’s divide some numbers by 4 and 8 and compare the remainders:

3/4 = 0 R 3, 3/8 = 0 R 3

5/4 = 1 R 1, 5/8 = 0 R 5 (which is 1 + 4)

9/4 = 2 R 1, 9/8 = 1 R 1

14/4 = 3 R 2, 14/8 = 1 R 6 (which is 2 + 4)

The only possible remainder in the answer choices is 59.

Answer: B
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