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If a and b are odd integers such that a - b > 7,

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by NandishSS » Sat Nov 19, 2016 3:19 pm
If a and b are odd integers such that a - b > 7, what is the smallest possible positive difference between a and an even number less than b?

A) 7

B) 8

C) 9

D) 10

E) 11

OA:C
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by melguy » Sat Nov 19, 2016 6:42 pm
We are starting from 1 as we want the lowest possible value. But the option is red is not possible as we need to have an even number less than B.

The next option after that is A = 11 B = 3
So 11 - 2(an even integer less than B) = 9

Answer is C
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by DavidG@VeritasPrep » Sun Nov 20, 2016 6:12 am
NandishSS wrote:If a and b are odd integers such that a - b > 7, what is the smallest possible positive difference between a and an even number less than b?

A) 7

B) 8

C) 9

D) 10

E) 11

OA:C
You can also a little logic and the answer choices. If a is ODD and we subtract an EVEN, we'll get an ODD result. (ODD - EVEN = ODD.) So we know the answer is odd. Eliminate B and D. We know that a - b > 7, so if we were to subtract a number smaller than b, the difference would have to be even bigger than 7. Eliminate A. So we just want the smallest odd remaining, which would be C, 9.
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by Matt@VeritasPrep » Fri Nov 25, 2016 3:09 pm
Let's say that c is the nearest even number to b that is less than b.

Then we know c = b - 1.

We want a - c.

Since a - b > 7, the smallest possible value of (a - b) = 8.

Since c = (b - 1), the smallest possible value of a - c = (a - (b - 1)) = (a - b) + 1 = 9.
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