Smallest Number (Division Problem)

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Smallest Number (Division Problem)

by TANMVASH » Mon Jul 29, 2013 12:32 pm
Which is the smallest number that leaves remainders 3, 4, 1, and 2 when divided successively by 4, 5, 3, and 7?
a. 579 b. 279 c. 109 d. 159 e. 187

OA is D

Can somebody please help me understand/ solve this problem.

Thanks,
Tan
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by [email protected] » Mon Jul 29, 2013 3:46 pm
Hi TANMVASH,

The concept of "remainders" is one that usually shows up 1-2 times on Test Day.

The concept is this: remainders are the "leftover pieces" when you do division (so there are no decimals and no fractions in a remainder question).

eg. 7 divided by 3 = 2 remainder 1 = 2r1

This question is NOT structured in the traditional way that the GMAT would pose the question to you (usually you would just be asked for the remainder of ONE number and you'd only be asked to perform the "remainder math" ONCE), but here is how to solve it:

This question wants us to find the smallest number that...
1) when divided by 4 gives us a remainder of 3; THEN ignore the remainder, and this new number...
2) when divided by 5 gives us a remainder of 4; THEN ignore the remainder, and this new number...
3) when divided by 3 gives us a remainder of 1; THEN ignore the remainder, and this new number...
4) when divided by 7 gives us a remainder of 2

Since the answers are numbers, we can use them against the question to find the match.

The correct answer is D. Here's why...

[spoiler]159 divided by 4 is 39r3; ignore the remainder....
39 divided by 5 is 7r4; ignore the remainder...
7 divided by 3 is 2r1; ignore the remainder...
2 divided by 7 is 0r2

Thus, the final answer is 159[/spoiler]

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by TANMVASH » Mon Jul 29, 2013 9:10 pm
Thanks Rich! I understand your point that a question like this will seldom (or perhaps NEVER) be tested on a real GMAT test, but I wanted to understand the concept and that's why I posted this question yesterday.

Thanks for the detailed explanation.

Tan

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by Matt@VeritasPrep » Tue Jul 30, 2013 8:15 am
This strikes me as a CAT question, not a GMAT one, but here's a nifty method:

A number that gives a remainder of r when divided by d is of the form dk + r, where k is an integer. (For example, 17 divided by 5 is 3 with a remainder of 2, so 17 is of the form 5*3 + 2.)

Let's start with that property.

Since we really have one number, we can set the successive division by working back to front: we start with the final division (the number is of the form 7k + 2), then plug that into the second to last division (that number is 3*(7k+2) + 1).

(If this is abstract, illustrate it with real numbers. 125 divided by 4 is 31, with a remainder of 1, so I can say 125 = 4*31 + 1. 31 divided by 6 is 5 with a remainder of 1, so 31 = 5*6 + 1. To recreate the number, I take the final equation and plug it into the earlier one: 125 = 4*(5*6+1)+1.)

Putting it all together, our equation is

Final division: 7k + 2
Second to last division: 3(7k + 2) + 1, or 21k + 7
Third to last division: 5(21k + 7) + 4, or 105k + 39
First division: 4(105k + 39) + 3, or 420k + 159

Setting k = 0 to minimize the number, we get 159. Too cool!

Rich's method (testing the answer choices) also works, and is a practical solution on test day.