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nick1816
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Given that K^K = 2^24, where K is a positive integer, and K^2+2 Updated on: 12 Jun 2019, 03:35
Originally posted by nick1816 on 12 Jun 2019, 03:34.
Last edited by Bunuel on 12 Jun 2019, 03:35, edited 1 time in total.
Renamed the topic.
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C
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Given that \(K^K=2^{24}\), where K is a positive integer, and \(K^2+2^K\) is a multiple of \(2^a\), 'a' is also a positive integer. What is the maximum value of 'a' possible?

A. 2
B. 4
C. 6
D. 8
E. 12
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C
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Given that K^K = 2^24, where K is a positive integer, and K^2+2 12 Jun 2019, 07:53
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nick1816
Given that \(K^K=2^{24}\), where K is a positive integer, and \(K^2+2^K\) is a multiple of \(2^a\), 'a' is also a positive integer. What is the maximum value of 'a' possible?

A. 2
B. 4
C. 6
D. 8
E. 12
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K^K=2^{24}
—> k^k = (2^3)^8
—> k^k = 8^8
—> k = 8

k^2+2^k = 64 + 256 = 320

k^2+2^k is a multiple of 2^a
—> 320 is a multiple of 2^a
—> 320 = 64*5 = 2^6*5
—> Maximum value of a possible = 6

IMO Option C

Pls Hit kudos if you like the solution

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Given that K^K = 2^24, where K is a positive integer, and K^2+2 12 Jun 2019, 03:39
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K^K = 8^8,

So the expression K^2+ 2^K will become 64+ 2^8, now we can take a maximum of 2^6 out of this expression which makes a=6
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Given that K^K = 2^24, where K is a positive integer, and K^2+2 12 Jun 2019, 03:51
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First: \(K^K=2^24\); K=8

Second: \(K^2+2^K=2^6+2^8=2^6*(2^2+1)\)
So a can be max 6

IMO
Ans: C
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Given that K^K = 2^24, where K is a positive integer, and K^2+2 12 Jun 2019, 07:40
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K^K=2^24 -> K^K = (2^3)^8 = 8^8

So, K=8

K^2+2^K=8^2+2^8 = (2^6)+(2^8) =2^6∗(1+2^2) = 2^a*(factor)

So 'a' has to be 6

Option C
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Given that K^K = 2^24, where K is a positive integer, and K^2+2 15 Jul 2023, 08:09
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\(K^K=2^{24}\)

\(K^K=2^{3*8}\)

\(K^K=8^{8}\)

So K = 8.

\(K^2+2^K\) = \(8^2+2^8\) = \(2^6+2^8\) = \(2^6 (1+2^2)\)

So maximum value of a =6. Option C.
nick1816
Given that \(K^K=2^{24}\), where K is a positive integer, and \(K^2+2^K\) is a multiple of \(2^a\), 'a' is also a positive integer. What is the maximum value of 'a' possible?

A. 2
B. 4
C. 6
D. 8
E. 12
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Given that K^K = 2^24, where K is a positive integer, and K^2+2 27 Mar 2026, 05:40
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