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by BTGmoderatorRO » 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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by EconomistGMATTutor » 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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by Matt@VeritasPrep » 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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by Scott@TargetTestPrep » Fri Nov 22, 2019 11:51 am
BTGmoderatorRO wrote: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
The best way to solve this problem is to try each given answer choice. However, the information given in the problem suggests that the number is divisible by 5 but not by 34. Thus, we can eliminate 74 and 680 since the former is not divisible by 5 and the latter is divisible by 34. Since the problem asks for the least possible number, let's try the smallest number from the remaining three numbers.

B) 75

75/5 = 15

75/34 = 2 R 7

Since 15 is indeed 8 more than 7, then 75 is the correct answer.

Alternate Solution:

Let n be the number. Thus, the remainder when n is divided by 34 is n/5 - 8. We have:

n = 34s + n/5 - 8

where s is some positive integer.

4n/5 = 34s - 8

Let's multiply each side by 5:

4n = 34*5*s - 40

Let's divide each side by 4:

n = (34*5*s)/4 - 10

n = (17*5*s)/2 - 10

As n is required to be an integer, the smallest value of s is s = 2. Thus, the smallest value of n is (17 * 5 * 2)/2 - 10 = 17 * 5 - 10 = 85 - 10 = 75.

Answer: B

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by Brent@GMATPrepNow » Fri Nov 22, 2019 3:56 pm
BTGmoderatorRO wrote: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
A number when divided by 5 gives a number which is 8 more than ...

gives a number??

What's the source of this question?
Brent Hanneson - Creator of GMATPrepNow.com
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