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Multiple of 5 and 125. x=a^2*b?

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by nk_81 » Fri Feb 11, 2011 10:20 pm
X is a multiple of 5, X=a^2 *b, a and b are integers. Which of the following must be the multiple of 125?

1) a.b
2) a.b^2
3) a^2.b^2
4) a^3.b^3
5) 2.a.b
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by nk_81 » Fri Feb 11, 2011 10:21 pm
Can someone show me the approach to this problem?
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by Anurag@Gurome » Fri Feb 11, 2011 10:39 pm
If X is a multiple of 5, then either a^2 or b or both are multiples of 5.
Let us solve by example.
Now, 80 = 4^2 * 5. So let a = 4, b = 5, X = 80.
So, 80 is a multiple of 5.
Now ab = 20 is not a multiple of 125.
a *b^2 = 4 * 5^2 = 100 is not a multiple of 125.
a^2 * b^2 = 4^2 * 5^2 = 400 is not a multiple of 125.
a^3 * b^3 = 4^3 * 5^3 = 64 * 125, which is a multiple of 125.
2ab is 2*4*5 = 40 which is not a multiple of 125.

The correct answer is hence 4).
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by banibhusan » Sat Feb 12, 2011 11:56 am
I took a different approach. Since X is a multiple of 5, so we can conclude that either a or b or both (since a & b are integers, a can't take the value of 5^1/2 in which case a^2 would be a multiple of 5) is a multiple of 5. In case of a^2, assuming that 5 appears in the list of factors of 'a' at least once we can can make sure that its divisible by 125, i.e. 5^3, by multiplying it by simply another 'a'. So it becomes a^3 and its certain to be divisible by 125.

Similarly in case of b, assuming that 5 appears in the list of factors of 'b' at least once, we can multiply it by b^2 to increase the no of '5' factors to 3.

So a^3*b^3 is the answer. I hope I am clear. :)
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