Question 234483
<pre><font size = 4 color = "indigo"><b>
{{{1000 = 2^3*5^3}}} has only the prime divisors {{{2}}} and {{{5}}}.
Every divisor of {{{1000}}} is therefore of the form {{{2^p*5^q}}},
where {{{p}}} and {{{q}}} are elements of {0,1,2,3}.  Since there
are {{{4}}} choices for {{{p}}} and {{{4}}} choices for {{{q}}}, there are {{{16}}}
divisors of {{{1000}}}.  That isn't necessary to know.  But
in the product of all {{{16}}} divisors of {{{1000}}}, {{{2^0*2^1*2^2*2^3}}}
or {{{2^6}}}, and {{{5^0*5^1*5^2*5^3}}} or {{{5^6}}} occur exactly {{{4}}} times each.
Since {{{2^6}}} and {{{5^6}}} both occur {{{4}}} times each in the product
of all divisors, the product of the divisors must be
{{{(2^6)^4*(5^6)^4 = 2^24*5^24 = 10^24}}}, choice a)

Edwin</pre>