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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 59% (01:51) correct 41% (01:44) wrong based on 121 sessions

### HideShow timer Statistics If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?

1) a=even
2) b=odd

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Re: If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?  [#permalink]

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1
MathRevolution wrote:
If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?

1) a=even
2) b=odd

1) if a=even then
(ab+2)(ab+3)(ab+4)= consecutive numbers where (ab+2)=even number
thus for any 3 consecutive number where first one is even
thier product is always divisible by 2*2*3 =12
suff

if a is odd then (ab+2)(ab+3)(ab+4)= 3 consecutive number where first one is odd
then it may not be divisible by 12
let ab=1*1
(ab+2)(ab+3)(ab+4)= 3*4*5...not div by 12
but if a is even then expression is divisible by 12 (as shown in (1)

Not suff

Ans A
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7597
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?  [#permalink]

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==> If you modify the original condition and the question, since (ab+2)(ab+3)(ab+4) is the product of 3 consecutive integers, so it is always divisible by 3. Then, since (ab+2)(ab+3)(ab+4) is divisible by 3, in order for it to be divisible by 12, it needs to be divisible by 4, so you must get ab+2=even. For con 1), if a=even, you get ab=even, then you get ab+2=even and ab+4=even, so it is always divisible by 4, and hence it is yes.

The answer is A. It is a CMT 4(A) question.
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If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?  [#permalink]

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rohit8865 wrote:
MathRevolution wrote:
If a and b are integers, is $$(ab+2)(ab+3)(ab+4)$$ divisible by 12?

1) a=even
2) b=odd

1) if a=even then
$$(ab+2)(ab+3)(ab+4)$$= consecutive numbers where (ab+2)=even number
thus for any 3 consecutive number where first one is even
thier product is always divisible by 2*2*3 =12
suff

if a is odd then $$(ab+2)(ab+3)(ab+4)= 3$$ consecutive number where first one is odd
then it may not be divisible by 12
let $$ab=1*1$$
$$(ab+2)(ab+3)(ab+4)=$$ $$3*4*5$$...not div by $$12$$
but if a is even then expression is divisible by 12 (as shown in (1)

Not suff

Ans A

Dear rohit8865 & MathRevolution, Under statement (2), why the product of $$(3)(4)(5) = 60$$ is not divisible by $$12$$?

$$a*b=3*3=9$$, $$(odd*odd)$$

$$(ab+2)(ab+3)(ab+4)=(9+2)(9+3)(9+4)=1716$$, which is divisible by $$12$$.

$$a*b=2*3=6$$. $$(even*odd)$$

$$(ab+2)(ab+3)(ab+4)=(6+2)(6+3)(6+4)=720$$, which is divisible by $$12$$.
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Re: If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?  [#permalink]

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If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?

Let analyze question before solving. Consider ab=x

If x=0........2*3*4 is divisible by 12
If x=2........4*5*6 is divisible by 12
If x=4........6*7*8 is divisible by 12

If x is even, then it is divisible by 12

If x=1.......3*4*5 is divisible by 12

If x=3.......5*6*7 is NOT divisible by 12

1) a=even

If a is even, then ab must be even. Examples above proves the case.

Sufficient

2) b=odd

If b is odd , then no info about a. If a is even like above, so it is divisible by 12

if a is odd, it might be or not be divisible by 12 as illustrated by odd examples.

Insufficient

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Re: If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?  [#permalink]

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MathRevolution wrote:
If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?

1) a=even
2) b=odd

product of 3 consecutive integers are given. therefore it could be even, odd, even or odd, even, odd. but it will definitely have a multiple of three

St 1: a is even. therefore the sequence will be even , odd, even. two even and a multiple of 3 .Definitely divisible by 12. ANSWER
St 2: b is odd. totally depends if a is odd or even. INSUFFICIENT

Option A
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Re: If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?  [#permalink]

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_________________ Re: If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?   [#permalink] 25 Nov 2018, 10:44
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