If $a$ , $b$ , and $c$ are consecutive even integers and $a < b < c$ , all of the following must be divisible by 4 EXCEPT

Answer Choices

Correct answer marked below

a+c

b+c

Correct

ac

\frac{bc}{2}

\frac{abc}{4}

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Comments

AlfonsoAug 7

Easy one. Since a,b,c are consecutive evens, a and c are both multiples of 4? Wait, not always. If a=2, b=4, c=6, then a+c=8 divisible by 4, b+c=10 not divisible by 4. But if a=4, b=6, c=8, b+c=14 not divisible by 4. So B is the exception. Done.

AliAug 16

Yeah, b and c are consecutive evens, one is 2 mod 4 and the other 0 mod 4, so their sum is 2 mod 4. Always not divisible by 4.

ArvindJun 22

I initially thought all of them would be divisible by 4 except maybe E, but then I realized b and c have different remainders mod 4. Nice question.

KejiJun 23

Same here. The 2 mod 4 sum caught me off guard.

IlyasJun 27

Yeah, the consecutive even integers alternate between 0 and 2 mod 4. Tricky but fair.

AnuragJun 5

Wait, for (D) bc/2: if b=4,c=6, then bc/2=12 divisible by 4? 12/4=3 yes. If b=6,c=8, bc/2=24 divisible by 4? 24/4=6 yes. So D is always divisible by 4. Good.

MaorJun 16

Yes, because one of b,c is divisible by 4 and the other by 2, so product divisible by 8, half gives 4.

KunalJul 9

Thanks for this. I got confused with (E) but then realized abc has two multiples of 2 and one multiple of 4? Actually three consecutive evens: one is divisible by 4, the other two are even, so product divisible by 4*2*2=16, so abc/4 divisible by 4. So E is fine. Answer is B.

What this question tests

What it tests

Your understanding of divisibility rules, least common multiples, and greatest common factors.

Common trap

Forgetting that 1 is not prime and that every integer divides 0.

If aaa, bbb, and ccc are consecutive even integers and a<b<ca < b < ca<b<c, all of the following must be divisible by 4 EXCEPT