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Re: For positive integers x and y, x^2 = 350y. Is y divisible [#permalink]
10 May 2012, 00:48

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Expert's post

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.

awesome, thanks a lot Dabral. i am struggling with the concepts on prime factorizaiton as to when to apply to what kind of problem. appreciate your time and help..

and to Bunuel too for providing an alternate sway to solve this problem.

Re: For positive integers x and y,x^2 = 350y. Is y divisible by [#permalink]
24 Nov 2012, 21:26

1

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Sachin9 wrote:

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

Somebody please explain me why 2 is not sufficient.

From the statement we get that at least y = 2^a*7^b*z

where a,b are positive odd integers and z is a perfect square

1)x is divisible by 4. So x had at least two 2s and so x^2 has at least four 2s. So y has at least three 2s. We know y also has a 7. So y is divisible by 28. Sufficient.

2)x^2 is divisible by 28. This would be true even if y has only one 2. i.e, y = 2*7 = 14. In this case, x^2 is divisible by 28 but y is not divisible by 28. From the previous statement we can get a situation where x and y are both divisible by 28. Insufficient.

Answer is hence A.

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Re: For positive integers x and y, x^2 = 350y. Is y divisible [#permalink]
20 Feb 2014, 07:00

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