If B and A are consecutive integers and A + B + X = 15, then which of

If B and A are consecutive integers and A + B + X = 15, then which of the following must be true about X?

X=15-(A+B)=15-(A+A+1)=14-2A

A. X is a multiple of 5 ... If A has units digit 2 or 7, yes, otherwise NO
X=14-2A..... X=14-2*2 or X=14-2*27...YES ... But X=14-2*2...NO

B. X is a multiple of 3 ... If A leaves a remainder of 1 when divided by 3.. yes, otherwise NO
X=14-2A..... X=14-2*4=6 or X=14-2*7=0...YES ... But X=14-2*2...NO

C. X is positive.... If A<7.. yes, otherwise NO
X=14-2A..... X=14-2*4=6 YES.... X=14-2*7=0...YES .....NO

D. X is even.Yes always
X=14-2A=2(7-A)S .....NO

E. X is odd.
Never, as C is even

D

Given: B and A are consecutive integers and A + B + X = 15

Asked: Which of the following must be true about X?

B = A+1
A+B+X=15
2A+1+X=15
2A +X = 14
X = 14-2A
X = even since 14 and 2A are even

IMO D

Archit3110

Small typing error is highlighted.

Kinshook
edited thanks

It says "which of the following," so this looks like a good problem to test cases on. The answer choices deal with evens, odds, and multiples, so we should definitely choose some even and some odd numbers. But, let's keep the numbers easy as well.

Case 1: A = 0, B = 1. Then, X = 15 - A - B = 15 - 1 = 14.

14 isn't a multiple of 5. Eliminate A.

14 isn't a multiple of 3. Eliminate B.

14 is positive. Keep C.

14 is even. Keep D.

14 isn't odd. Eliminate E.

Now, we have one answer that says "positive", and one that says "even." I'm going to try to eliminate the "positive" one by seeing if I can make X a negative number.

Case 2: When would X be a negative number? That would happen if A + B was large, so X had to be negative in order to reduce the sum back down to just 15.

So, if A = 10, B = 11, then x = 15 - 10 - 11 = -6.

X can be negative, so we can eliminate C. The correct answer is D.