Equinocs
If x2 is divisible by 216, what is the smallest possible value for positive integer x?
I came across this problem in one of my study guides. I still don't understand how to do the work even with the explanation provided.
Can someone explain how to tackle this problem?
Thanks in advance.
Dear
Equinocs,
I'm happy to respond.
First of all, if please use proper math notation. For x-squared, you can write x^2 or \(x^2\) using the math notation function. Please do not write
x2, because this will generate tremendous confusion.
If you happen to know the cubes of all the single digit numbers, that would be easier, because \(6^3 = 216\). I think all students should have those ten cubes memorized, but let's assume that you didn't know that.
A HUGE help in all kinds of questions about factors and divisible and so forth is the idea of prime factorization. See:
GMAT Math: FactorsLet's start simply by breaking up 216 into factors.
216 = 2*108 = 2*2*54
= 2*2*6*9
= 2*2*2*3*3*3
That's the prime factorization of 216. Whenever you have a particular big number in a question about find the factors or multiples etc, chances are very good that finding the prime factorization of the big number given will be helpful.
Now, another big idea about factors. When we square a number, the factors are doubled: for each single factor, we get a pair of those factors.
\(35 = 5*7\)
\(35^2 = 5*5*7*7\)
\(18 = 2*2*3\)
\(18^2 = 2*2*2*2*3*3\)
Now, let's go back and look at the prime factorization of 216. We can group some of the factors into pairs.
216 =
(2*2)*(3*3)*
2*3We have paired and unpaired factors, so to turn this into a perfect square, we would have to multiply by another 2 and another 3, so those factors are paired up.
x^2 = (2*2)*(3*3)*(2*2)*(3*3)
When we take a square-root, each pair of factors becomes a single factor.
x = 2*3*2*3 = 36
Thus, x = 36.
Does this make more sense?
Mike