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Is the product of integers M and N even?

Expert replies
by BTGmoderatorAT » Tue Dec 12, 2017 5:21 am
Is the product of integers M and N even?

(1) N can be expressed as a difference of squares of two consecutive prime numbers at least one of which is odd. M can be expressed as a product of two natural numbers P and Q, where 2P + 1= Q.
(2) N can be expressed as a difference of squares of two consecutive prime numbers which lie at a distance of 2 units. M is the sum of all the numbers from 1 to Z where (Z+1) is a multiple of 4.

What's the best way to determine whether statement 1 is sufficient? Can any experts help?
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Source: — Data Sufficiency |

by ErikaPrepScholar » Tue Dec 12, 2017 10:06 am
Let's quickly review some even/odd number rules.

even * even = even
even * odd = even
odd * odd = odd
even + even = even
even + odd = odd
odd + odd = even

For the product of integers M and N to be even, at least one of them must be even.

Statement 1

We'll deal with the sentences one at a time.

Sentence 1: With the exception of 2, all prime numbers are odd. However, we only know that ONE of our prime numbers is odd. So we could have (odd * odd) - (even * even) = odd - even = odd OR (odd * odd) - (odd * odd) = odd - odd = even. So N can be even or odd.

Sentence 2: Since 2 is even, 2P must be even, whether P is even or odd. This means 2P + 1 (which equals Q) must be odd. If Q is odd and P is either even or odd, M can be either even or odd.

Taken together, these sentences tell us that N can be even or odd and M can be even or odd. Insufficient.

Statement 2

Sentence 1: Since 2 is our only possible even prime number, and it is consecutive with only 3, which is 1 unit away, we know that both prime numbers must be odd. This gives (odd * odd) - (odd * odd) = odd - odd = even. So N is even.

At this point we don't need to look at the second sentence - whether M is even or odd, we know that M * N will be even if N is even. Sufficient.
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