remainder problem

[scoutkb]

If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is devided by t?

A-2
B-4
C-8
D-20
E-45

Can anyone help? I thought i understood what was being tested when i talked to a MGMAT tutor...but apparently i didnt.

[ayang]

This is a tricky problem, and requires one to understand each of the givens.

s and t are each positive integers - that is, they are whole numbers.

s/t = 64.12. That is, s divided by t results in a remainder of .12t left over.

The question then asks which could be the remainder of s/t. That is, which number could represent the .12t?

In order to find this, it is necessary to remember that t must be an integer. Therefore, the correct answer will enable t to be a whole number.

Or, remainder = .12t

Divide both sides by .12 yields

remainder/.12 = t

Of the answer choices:

A-2. 2/.12 does not yield an integer.
B-4 4/.12 does not yield an integer
C-8 8/.12 does not yield an integer.
D-20 20/.12 does not yield an integer
E-45 45/.12 yields an integer. E is the correct answer.

Another way of looking at it that makes the calculations easier is to simplify .12 into a fraction:

12/100
6/50
3/25

Thus, to divide by 3/25 is to multiply by 25/3. The correct answer must be divisible by 3 in order to make t an integer.

I hope that this explanation is helpful! - Andrew

[r.avengraus]

I've struggled with this question and read many explanations regarding the answer. Yours is by far the clearest and most easy to understand. Thank you very much.

I believe my biggest problem with this problem is not understanding what the .12 represented. I initially thought that the fractional part of any division operation was the remainder, so when the question asked "what could be the remainder" I was clueless.

With your explanation, I now see the .12 represents a fraction of the divisor (or t). And multiplying the two brings me back to the "left-over" portion of the division operation (aka, the remainder) had the two numbers gone smoothly into one another.

Thanks once again!

[Ben Ku]

I'm glad that helped.

[vikramsumer]

[color=#0000FF]Gosh.... This reply makes me think that OG should now be written by MGMAT guys.

Cracking reply, I checked over 3 sources (including OG) and this by far the best answer.


I am sure I will never get such questions incorrect. You guys are AWESOME !![/color]

[jnelson0612]

Thanks vikram! :-)

[meenal121212]

Very good explanation ...:) I like the way it has been explained . I have gone through the OG book but did not understand the question . SO, searched the question through Manhattan forum .. Finally , I am able to do the problem ... Thank you Manhattan staff :)

[tim]

Glad to hear it!