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If 375y=x^2 and z and y are positive integers....

Expert replies
by alextheodosi » Sun Mar 03, 2013 5:23 pm
Hi,

Not sure if I am just being dim on this one....taken from Veritas arithmetic problem set

If 375y=x^2 and x andy are positive integers, then which of the following must be an integer?

1) y/15
2) y/30
3)(y^2)/25

a)1 only
b) 3 only
c) 1 and 2
d)1 and 3
e all

There are worked answers which equate x to equal 75. I can follow the worked answers from this point but find it convenient that they land on x=75 and I simply to do not understand why it does so.

Please help!

Al
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Source: — Problem Solving |

by Anurag@Gurome » Sun Mar 03, 2013 6:40 pm
alextheodosi wrote:If 375y=x^2 and x and y are positive integers, then which of the following must be an integer?

1) y/15
2) y/30
3)(y^2)/25
375 = 3*125 = 3*5*(5²)
Hence, to make 375y a perfect square y must contain at least one 3 and one 5, i.e. y must be a multiple of 15.

Hence, only 1 and 2 must be an integer.

The correct answer is C.
alextheodosi wrote:There are worked answers which equate x to equal 75. I can follow the worked answers from this point but find it convenient that they land on x=75 and I simply to do not understand why it does so.
It is not necessary to find out the value of x. However, to clear your doubts, as I've explained above, y must be a multiple of 15. Hence, minimum possible value of y is 15. Hence, minimum possible value of x² = 375*15 ---> minimum possible value of x = √(375*15) = √(25*15*15) = 5*15 = 75
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