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Let a be a positive integer. If n is divisible by 2^a and n is also

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Bunuel
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Let a be a positive integer. If n is divisible by 2^a and n is also [#permalink] New post  23 Jul 2015, 03:00
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Let a be a positive integer. If n is divisible by 2^a and n is also divisible by 3^(2a), then it is possible that n is NOT divisible by

A. 6
B. 3 × 2^a
C. 2 × 3^(2a)
D. 6^a
E. 6^(2a)

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Re: Let a be a positive integer. If n is divisible by 2^a and n is also [#permalink] New post  23 Jul 2015, 04:23
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Bunuel wrote:
Let a be a positive integer. If n is divisible by 2^a and n is also divisible by 3^(2a), then it is possible that n is NOT divisible by

A. 6
B. 3 × 2^a
C. 2 × 3^(2a)
D. 6^a
E. 6^(2a)

Kudos for a correct solution.


Easiest method is to assume a value of a = 1

Given n is divisible by 2^2 ---> n = 2^1*p ---> n =2p

Also, n is also divisible by 3^(2a) ---> n = 3^(2a) * q = 3^2*q = 9q

Lets look at a few numbers that are multiples of both 2 and 9 are 18,36,72....

Thus looking at the options , A-D divide the numbers 18,36,72.... while E (=6^(2) = 36 ) does not divide 18. Thus E is the correct answer.
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Re: Let a be a positive integer. If n is divisible by 2^a and n is also [#permalink] New post  23 Jul 2015, 06:12
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Bunuel wrote:
Let a be a positive integer. If n is divisible by 2^a and n is also divisible by 3^(2a), then it is possible that n is NOT divisible by

A. 6
B. 3 × 2^a
C. 2 × 3^(2a)
D. 6^a
E. 6^(2a)

Kudos for a correct solution.


Let's Find the smallest Positive Value of n

The Smallest Value of a = 1
i.e. The Smallest Value of 2^a = 2^1 = 2
i.e. The Smallest Value of 3^(2a) = 3^2 = 9

i.e. The smallest value of n Must be divisible by 2 and 9 both
i.e. Smallest value of n = 2*9 = 18

A. 6 Definitely Divisor of n
B. 3 × 2^a = 3*2^1 = 6 Definitely Divisor of n
C. 2 × 3^(2a) = 2*3^2 = 18 Definitely Divisor of n
D. 6^a = 6^1 = 6 Definitely Divisor of n
E. 6^(2a) = 6^2 = 36 NOT a Divisor of n for smallest value of n hence CORRECT OPTION

Answer: Option E
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Re: Let a be a positive integer. If n is divisible by 2^a and n is also [#permalink] New post  23 Jul 2015, 06:49
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Bunuel wrote:
Let a be a positive integer. If n is divisible by 2^a and n is also divisible by 3^(2a), then it is possible that n is NOT divisible by

A. 6
B. 3 × 2^a
C. 2 × 3^(2a)
D. 6^a
E. 6^(2a)

Kudos for a correct solution.


Since, n is divisible by 2^a and 3^(2a), it must be divisible by 6. As least value of a = 1

Only for E, 6^(2a) doesn't satisfy, if a = 1 and n=18, it is not divisible by 6^2 (i.e 36)


Hence answer is E
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Re: Let a be a positive integer. If n is divisible by 2^a and n is also [#permalink] New post  26 Jul 2015, 12:10
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Bunuel wrote:
Let a be a positive integer. If n is divisible by 2^a and n is also divisible by 3^(2a), then it is possible that n is NOT divisible by

A. 6
B. 3 × 2^a
C. 2 × 3^(2a)
D. 6^a
E. 6^(2a)

Kudos for a correct solution.


800score Official Solution:

If n is divisible by 2ᵃ and 3²ᵃ then it must be divisible by least common multiple of 2ᵃ and 3²ᵃ which equals 2ᵃ × 3²ᵃ.

Therefore the smallest possible value number n can take is 2ᵃ × 3²ᵃ which is less than answer choice (E), 6²ᵃ = 2²ᵃ × 3²ᵃ. A larger number can not be a divisor of a smaller one so 2ᵃ × 3²ᵃ is NOT divisible by 6²ᵃ. It means, that 6²ᵃ is not necessarily a divisor of n.

The right answer is choice (E).
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