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03 Jun 2020, 04:31
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A
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Difficulty:
95%
(hard)
Question Stats:
25% (02:52) correct
75%
(02:39)
wrong
based on 113
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For all positive integers x, \(\&x = \sqrt{x^{x+2}}\) when x is even and \(\&x = (2x^2)^{2x}\) when x is odd. What is the value of positive integer x?
(1) The highest integer power of 2 that divides &(4x) completely is 36
(2) The highest integer power of 2 that divides (&4)*x completely is 8
Are You Up For the Challenge: 700 Level Questions
(1) The highest integer power of 2 that divides &(4x) completely is 36
(2) The highest integer power of 2 that divides (&4)*x completely is 8
Are You Up For the Challenge: 700 Level Questions
03 Jun 2020, 05:11
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Nice one!
Statement 1-
4x is a even number.
\(&4x = \sqrt{(4x)^{(4x+2)}}\)
\(&4x = \sqrt{(4x)^{(2*2x+1)}}\)
\(&4x = (4x)^{(2x+1)}\)
\(&4x = 2^{(4x+2)}*x^{(2x+1)}\)
Since x is an integer, 4x+2 can never be equal to 36. Hence x must be even.
x can be 2 or 4, as for higher even values, The highest integer power of 2 that divides &(4x) exceeds 36.
At x = 4; \(2^{(4*4+2)}*4^{(2*4+1)} = 2^{36}\)
Obviously at x=2; \(2^{(4x+2)}*x^{(2x+1)} \) must be less than \(2^{36}\). We don't need to solve.
Sufficient
Statement 2-
4 is a even number.
\(&4 * x= \sqrt{(4)^{(4+2)}}* x\)
\(&4 * x= 2^6* x\)
Since highest integer power of 2 that divides &4 * x is 8, x must be equal to \(2^2*k\), where k is odd number.
x can be 4, 12, 20 and so on
Insufficient
Statement 1-
4x is a even number.
\(&4x = \sqrt{(4x)^{(4x+2)}}\)
\(&4x = \sqrt{(4x)^{(2*2x+1)}}\)
\(&4x = (4x)^{(2x+1)}\)
\(&4x = 2^{(4x+2)}*x^{(2x+1)}\)
Since x is an integer, 4x+2 can never be equal to 36. Hence x must be even.
x can be 2 or 4, as for higher even values, The highest integer power of 2 that divides &(4x) exceeds 36.
At x = 4; \(2^{(4*4+2)}*4^{(2*4+1)} = 2^{36}\)
Obviously at x=2; \(2^{(4x+2)}*x^{(2x+1)} \) must be less than \(2^{36}\). We don't need to solve.
Sufficient
Statement 2-
4 is a even number.
\(&4 * x= \sqrt{(4)^{(4+2)}}* x\)
\(&4 * x= 2^6* x\)
Since highest integer power of 2 that divides &4 * x is 8, x must be equal to \(2^2*k\), where k is odd number.
x can be 4, 12, 20 and so on
Insufficient
General Discussion
HoneyLemon
03 Jun 2020, 16:42
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Bunuel
Tough one to understand first .
IMO , A .
Please refer to the attachment .
Attachments
File comment: Rough Work for solution
for-all-positive-integers-x-x-x-x-2-when-x-is-even-and-x-2x-325790.docx [806.68 KiB]
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