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09 May 2012, 16:25
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
46% (02:23) correct
54%
(02:21)
wrong
based on 282
sessions
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For positive integers x and y, x^2 = 350y. Is y divisible by 28?
(1) x is divisible by 4
(2) x^2 is divisible by 28
(1) x is divisible by 4
(2) x^2 is divisible by 28
10 May 2012, 01:48
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For positive integers x and y, x^2 = 350y. Is y divisible by 28?
Given: \(x^2=2*5^2*7*y\). Notice that since \(2*5^2*7*y\) equals to some perfect square (x^2) then \(y\) must complete the odd powers of other multiples to even numbers (remember perfect square has even powers of its primes), so the least value of \(y\) is \(2*7=14\) (y is a multiple of 14). So, we need one more 2 for \(y\) in order it to be divisible by 28.
(1) x is divisible by 4 --> \(x^2\) is divisible by 4^2=16=2^4, since 350 has only one 2 then \(y\) must have the remaining 2^3, so we have that \(y\) is divisible by 2^3*7=2*28. Sufficient.
(2) x^2 is divisible by 28 --> since the least value of \(y\) is \(2*7=14\) then \(x^2=2*5^2*7*14=28*(5^2*7)\), so we already knew that \(x^2\) is divisible by 28. Not sufficient.
Answer: A.
Hope it's clear.
Given: \(x^2=2*5^2*7*y\). Notice that since \(2*5^2*7*y\) equals to some perfect square (x^2) then \(y\) must complete the odd powers of other multiples to even numbers (remember perfect square has even powers of its primes), so the least value of \(y\) is \(2*7=14\) (y is a multiple of 14). So, we need one more 2 for \(y\) in order it to be divisible by 28.
(1) x is divisible by 4 --> \(x^2\) is divisible by 4^2=16=2^4, since 350 has only one 2 then \(y\) must have the remaining 2^3, so we have that \(y\) is divisible by 2^3*7=2*28. Sufficient.
(2) x^2 is divisible by 28 --> since the least value of \(y\) is \(2*7=14\) then \(x^2=2*5^2*7*14=28*(5^2*7)\), so we already knew that \(x^2\) is divisible by 28. Not sufficient.
Answer: A.
Hope it's clear.
09 May 2012, 17:52
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Explanation attached.
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