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Let $x=2^6 3^4 5^2$, then how many distinct values of $|A-B|$ are possible where $A, B$ are the factors of $x$?

How to approach this problem?

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Where did this problem come from? What have you tried so far? It's easy to establish an upper bound on the number of distinct values (how many factors of $x$ are there? If $N$ is the number of factors, how many ordered pairs of numbers $\lt N$ are there?), but 'coincidences' where two distinct pairs of factors $\langle A,B\rangle$ and $\langle C,D\rangle$ have the same difference can happen, and there's no immediately obvious way of tallying the coincidences except going through all the possibilities by hand. –  Steven Stadnicki Jul 23 '12 at 16:43
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Are you sure you aren't being asked to tally how many distinct values of $|A-B|$ are possible where $AB=x$? That would be a different (and much more easily manageable by hand) problem... –  Steven Stadnicki Jul 23 '12 at 16:44
    
No, $AB$ doesn't necessarily equal to $x$. –  Hyperbola Jul 23 '12 at 17:25

1 Answer 1

There are $7\times5\times3=105$ divisors of $x$. There are at most $\binom{105}{2}+1=5461$ non-negative differences among the elements. $\binom{105}{2}$ computes the number of pairs of distinct factors and $1$ is for $0$, the difference between a factor and itself. There may be fewer than this many because some differences may be repeated.

To compute the actual number of differences, one could use a program like Mathematica. I used this code

d = Divisors[2^6 3^4 5^2]; Length[Union[Abs[Flatten[Outer[Subtract, d, d]]]]]

to compute that there are $2856$ distinct non-negative differences.

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