## Tuesday, February 7, 2012

### this week's MGRE Math Beast Challenge problem

This one doesn't look too hard:

x, y, and z are three consecutive multiples of 3 such that x < y < z.

Quantity A
The remainder when the sum of x + 1, y – 2, and z + 3 is divided by 9

Quantity B
2

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

Go to it! My answer will appear in the comments. If you want to work on the problem yourself, be sure not to peek.

_

#### 1 comment:

Kevin Kim said...

This one seems pretty clear-cut. I say the answer is (C): the quantities are equal. Why?

First, let's rename x, y, and z, which we know are three consecutive multiples of 3 (examples: {3, 6, 9} or {6, 9, 12}, etc.).

Let x = 3q. A multiple of 3 is always 3 times something.

Let y = 3(q + 1) = 3q + 3.

Let z = 3(q + 2) = 3q + 6.

If that's the case, then

x + 1 = 3q + 1

y - 2 = (3q + 3) - 2 = 3q + 1

z + 3 = (3q + 6) + 3 = 3q + 9

Add those together, and you get

9q + 11

Divide that by 9, and you get

q + 1, remainder 2.

So that's why I'm picking (C).

You might ask yourself: why assign the value of 3q to x? The three consecutive multiples are all at a distance of 3 from each other on a number line, so why not write the values this way:

x = q

y = q + 3

z + q + 6

?

Wouldn't that be simpler?

Yes, it would be, but then you'd have no guarantee that the numbers were, in fact, multiples of 3. For example, you could have sets like the following:

{2, 5, 8}

{5, 8, 11}

{4, 7, 10}

--all the above sets have numbers that are at a distance of 3 from each other, but as you see, not one of those sets contains an actual multiple of 3. That's why you have to set x up as 3q.

And speaking of q:

QED.