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Remainder

This topic has 2 expert replies and 0 member replies

Remainder

Post Sat Sep 30, 2017 7:45 pm
A number when divided by 5 gives a number which is 8 more than the remainder obtained on dividing the same number by 34. Such a least possible number is

A. 74
B. 75
C. 175
D. 680
E. 690
Please help me out on how you solve this.
qa is b

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Post Thu Oct 12, 2017 9:28 pm
Let's call our number n. We're told that

n/5 = 8 + (remainder of n/34)

To find that remainder, we can say that n = 34*something + r, or n = 34z + r, or r = n - 34z. Once we've got that, we can replace (remainder of n/34) with r, or with n - 34z:

n/5 = 8 + n - 34z

34z = 8 + (4/5)n

170z = 40 + 4n

To solve this, we need n and z both to be integers. We want the smallest possible solutions, so let's start with z = 1. That gives n = 32.5 - not an integer!

If we try z = 2, however, we get n = 75, which is an integer, and our smallest solution.

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Post Sat Oct 07, 2017 3:29 pm
Hi Roland. This an interesting question.

To start, let x be the number we are looking for. It says that x/5 gives a number which is 8 more than the remainder... As the remainder is an integer number, then x/5 is an integer. It implies that x is divisible by 5. So, option A is out.

Now, let's try with option B.

We have that 75/5 = 15.
Now, when we divide 75 by 34, the remainder is 7.
If we add 7 + 8 we will get 15 which is the result of dividing 75 by 5.

So, B is the correct option.

I hope it helps you.

I'm available if you'd like any follow up.

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