GMAT Number Theory

Gennadiy · **Joined:** Sun May 30, 2010 2:23 am **Posts:** 480

questioner wrote:

Please explain this some again in detail. I wasnt able to understand the explanation, probably with numbers substituted. Thanks.

If there is any specific step in any reasoning above, let me know, and I'll go over it again.

There can be NO shortcuts for statement (2) by considering some specific values, because this statement gives us enough information to proof that n²ª is a multiple of mª for ANY positive integers.

However, you can try a simpler case, let's say for a = 1. The explanation is the same, just put 1 instead of a, whenever you meet it. The question itself will become:

If m and n are positive integers, is n² a multiple of m?
(1) n is a multiple of m/2
(2) n is a multiple of 2m

questioner · **Joined:** Sun May 30, 2010 3:15 am **Posts:** 407

Let n = 2, m = 5. 2m = 10 is a divisor of n or 2, but n²ª (or 4ª) is not a multiple of mª (or 5ª).

Gennadiy · **Joined:** Sun May 30, 2010 2:23 am **Posts:** 480

Quote:

2m = 10 is a divisor of n or 2.

2m = 10 is a multiple of n = 2, while n = 2 is a divisor of 2m = 10.

questioner · **Joined:** Sun May 30, 2010 3:15 am **Posts:** 407

In your explanation of why answer A is insufficient, you give 2 examples of why the answer is no, which is an answer. You might inlcude an example of when the answer is yes sometimes and no other times, making A insufficient.

Gennadiy · **Joined:** Sun May 30, 2010 2:23 am **Posts:** 480

questioner wrote:

You might inlcude an example of when the answer is yes sometimes and no other times, making A insufficient.

That's exactly what we do.

(1) n is a multiple of m/2

Ex. 1:
m = 6, n = 3.
Then 3 is a multiple of 6/2 = 3. BUT 3²ª = 9ª is clearly NOT a multiplier of 6ª. -> NO.

Ex. 2:
m = 6, n = 6.
Then 6 is a multiple of 6/2 = 3. AND 6²ª is clearly a multiplier of 6ª. -> YES.

dave · **Joined:** Mon Nov 26, 2012 5:39 pm **Posts:** 11

Can you expand on the second point? I understand why if 2m is a divisor of n, then m is a divisor of n – but how do we get from there to understanding that n²ª is a multiple of m²ª?

Gennadiy · **Joined:** Sun May 30, 2010 2:23 am **Posts:** 480

Quote:

…, then m is a divisor of n – but how do we get from there to understanding that n²ª is a multiple of m²ª?

We know that n is a multiple of m (in other words n is divisible by m).

So n² is a multiple of m², because
n² = n × n
m² = m × m
AND n²/m² = (n/m) × (n/m) is an integer, because each factor is an integer.

n³ is a multiple of m³
…
n²ª is a multiple of m²ª

radz1806 · **Joined:** Tue Aug 21, 2012 3:07 am **Posts:** 1

A way to answer this question would be

statement 1)
first statement needs to be an integer as n is a multiple of m/2

so if you write it as n/(m/2) = 2(n/m) and this needs to be an integer for it to be an integer we have two options:

thus n/m can be 1/2 -> not an integer NO
or n/m can be 1,2,3,4,.......---> integers yes

thus Not sufficient

statement 2)
it states that n/2m is an integer

now separate it as (1/2) * (n/m) and this is an integer

thus for this to be an integer naturally (n/m) has to be an integer which cancels out the 2 in the denominator------> yes only

Thus statement 2 is sufficient as if n is multiple of m then n^2 will also be a multiple of m

Gennadiy · **Joined:** Sun May 30, 2010 2:23 am **Posts:** 480

Quote:

statement 1)
first statement needs to be an integer as n is a multiple of m/2

First statement is not a number. However you can say that n, m/2 and n / (m/2) must be integers, according to definitions of divisibility and a multiple.

Quote:

so if you write it as n/(m/2) = 2(n/m) and this needs to be an integer for it to be an integer we have two options:

thus n/m can be 1/2 -> not an integer NO
or n/m can be 1,2,3,4,.......---> integers yes

thus Not sufficient

You are making the conclusion of whether n²ª is a multiple of mª or not, based on n/m . But it is NOT correct.

n/m can be a fraction, while n²ª will be a multiple of mª . Take, for example, n = 2 and m = 4.
n/m = 1/2 – a fraction
n²ª = 4ª is divisible by mª = 4ª

Quote:

statement 2)
it states that n/2m is an integer

now separate it as (1/2) * (n/m) and this is an integer

thus for this to be an integer naturally (n/m) has to be an integer which cancels out the 2 in the denominator------> yes only

Thus statement 2 is sufficient as if n is multiple of m then n^2 will also be a multiple of m

This reasoning is correct. For formality, just add the last step.
n² is a multiple of m, so
(n²)ª is a multiple of (m)ª .

fedana · **Joined:** Mon Nov 26, 2012 4:59 pm **Posts:** 8

On the GMAT, would the number 7 be considered a multiple of the number 3.5? I am wondering, whether when picking numbers for this problem, I could have picked n = 7 and m = 7 (since n is multiple of m/2, and 7 is (or not?) a multiple of 3.5)

GMAT Number Theory

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