Odd even divisbility

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faraz_jeddah: Master | Next Rank: 500 Posts; Posts: 358; Joined: Thu Apr 18, 2013 9:46 amFollowed by: 7 members; Location: Jeddah, Saudi Arabia; Thanked: 42 times; GMAT Score:730

by faraz_jeddah » Mon May 27, 2013 1:45 am

If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer?

a - a/4
b - a/12
c - a/24
d - a/27
e - a/72

OA is E

What is the best strategy here?

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Source: Beat The GMAT — Problem Solving | Mon May 27, 2013 1:45 am

srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members; Location: Seattle, WA; Thanked: 86 times

by srcc25anu » Mon May 27, 2013 12:31 pm

12^3 = 2^6 * 3^3
Odd No = a^2 / (2^6 * 3^3)
if a^2 = 2^6 * 3^3 then a^2 / 12^3 = 1 (odd) but a in that case will be 2^3 * (3root3) and a will NOT be an integer.
If a^2 = 2^6 * 3^4 then a^2 / 12^3 = 3 (odd) then a = 2^3 * 3^2 = 72(a = Integer)
From ans choices a/72 = 1 (odd)
Ans E

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apoorva.rattan: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Wed Oct 07, 2009 3:55 pm

by apoorva.rattan » Mon May 27, 2013 5:28 pm

What happens when a^2 / (2^6 * 3^3)=5 or 7 ?

Wouldn't that make C the best answer choice ?

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srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members; Location: Seattle, WA; Thanked: 86 times

by srcc25anu » Mon May 27, 2013 5:54 pm

To answer your question: what happens when a^2 / (2^6 * 3^3)=5 ?
We must recognize that a has to be some multiple of even powers of any integer.
to make a^2 / (2^6 * 3^3)=5, the numerator should be 12^3 * 5
But note that we Cannot write 12^3 * 5 in the form of a^2 where a is an integer.
if a^2 = 12^3 * 5 then a = root(12^3 * 5) = 24(root15). BUT a is NOT an INTEGER.
So 5 cannot be the answer of a^2/12^3 where a is an INTEGER
Same for 7
Hope this is clear

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apoorva.rattan: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Wed Oct 07, 2009 3:55 pm

by apoorva.rattan » Mon May 27, 2013 6:13 pm

Ahh.. Ok

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