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

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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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New post 09 Jan 2017, 01:40
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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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New post 09 Jan 2017, 10:29
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

2)no info about a
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
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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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New post 10 Jan 2017, 18:35
==> 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.
Answer: A
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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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New post 12 Feb 2017, 21:29
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

2)no info about a
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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New post 19 Mar 2017, 18:25
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

Answer: 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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New post 20 Mar 2017, 04:42
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? &nbs [#permalink] 25 Nov 2018, 09:44
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