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If a is a positive integer, then (a+2)(a+3)(a+4)(a+5) is

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If a is a positive integer, then (a+2)(a+3)(a+4)(a+5) is  [#permalink]

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New post 26 Sep 2018, 01:00
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

56% (01:08) correct 44% (01:11) wrong based on 16 sessions

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If \(a\) is a positive integer, then \((a+2)(a+3)(a+4)(a+5)\) is

A) even only when \(a\) is odd

B) odd whenever \(a\) is odd

C) divisible by 3 only when \(a\) is odd

D) divisible by 4 whenever \(a\) is even

There is no E) choice in this question I got given by my tutors.
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Re: If a is a positive integer, then (a+2)(a+3)(a+4)(a+5) is  [#permalink]

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New post 26 Sep 2018, 01:55
bettatantalo wrote:
If a is a positive integer, then (a+2)(a+3)(a+4)(a+5) is

a) even only when a is odd

b) odd whenever a is odd

c) divisible by 3 only when a is odd

d) divisible by 4 whenever a is even

There is no e) choice in this question I got given by my tutors.


one of (a+2) or (a+3) will always be odd and the other will always be even .
answer choice a and b are out
substitute a=1 (a+2) is 3 , thus answer choice c is out
substitute a=2 (a+2) is 4
substitute a=6 (a+2) is 8

therefore D
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Re: If a is a positive integer, then (a+2)(a+3)(a+4)(a+5) is  [#permalink]

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New post 26 Sep 2018, 01:56
3 consecutive numbers are always divisible by 2.
Now in the above expression there are 4 consecutive integers so we will get one more '2' when we multiply the integers.
Hence it will be divisible by 4 always.
D should be the correct choice.
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If a is a positive integer, then (a+2)(a+3)(a+4)(a+5) is  [#permalink]

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New post 26 Sep 2018, 15:05
bettatantalo wrote:
If \(a\) is a positive integer, then \((a+2)(a+3)(a+4)(a+5)\) is

A) even only when \(a\) is odd

B) odd whenever \(a\) is odd

C) divisible by 3 only when \(a\) is odd

D) divisible by 4 whenever \(a\) is even

There is no E) choice in this question I got given by my tutors.

Language in the correct answer has a trap.

Test numbers. What happens to
(a+2)(a+3)(a+4)(a+5)
-- When a is odd?
-- When a is even?

ODD: When a=1 then the first factor is (a + 2) = 3, so
when a is odd, (3 * 4 * 5 * 6) = 360

EVEN: When a=2, the first factor is 4.
When a is even, (4 * 5 * 6 * 7) = 840

Check options against test numbers
(a+2)(a+3)(a+4)(a+5) is

A) even only when a is odd: FALSE. From test cases:
The product (360, 840) is even when a is odd AND when a is even. REJECT

B) odd whenever a is odd: FALSE.
When a is odd, the product (360) is even, not odd. REJECT

C) divisible by 3 only when a is odd: FALSE
The product (360, 840) is divisible by 3 when a is odd AND when a is even. REJECT

D) divisible by 4 whenever a is even:
(1) TRUE
When a = 2 (even), the product of 840 is divisible by 4. KEEP

If concerned, check a = 2. First factor (4 + 2) = 6
(6 * 7 * 8 * 9) = 3,024. Divisible by 4. KEEP

(2) "Divisible by 4" is ALSO TRUE when a is odd
(When a = 1, product is 360, which is divisible by 4)
Possible trap, but:
What happens when a is odd does not matter.
#2 does not make #1 false. Option D is not asking about #2.*

Option D does not state that the product is divisible by 4 only when a is even.

D asks about the result when a is even.
Well, when a is even, the product of the factors is indeed divisible by 4.
Question asked, question answered.

Answer D

*In option D, we don't care about what happens in cases other than a = even.
Stay with what is asked. If or when a is even, is the product divisible by 4? Yes?
Then forget about what happens when a is odd. We are not being asked about that situation.
If D had stated "only when a is even," we would have a different scenario. We would have to account for what happens when ais odd.
-- ONLY WHEN implies: true for this situation AND NOT for other situations
Those "other situations" = a is odd
-- WHENEVER implies: true for this situation
Other situations do not matter.

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If a is a positive integer, then (a+2)(a+3)(a+4)(a+5) is &nbs [#permalink] 26 Sep 2018, 15:05
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