Question 969976
Look at the prime factors of each number,
{{{180=2*2*3*3*5}}}
{{{784=2*2*2*2*7*7}}}
So the twos are already in a multiple of 3, {{{2^6}}}
The three needs 1 to make {{{3^3}}}.
The five needs 2 to make {{{5^3}}}.
The seven needs 1 to make {{{7^3}}}.
{{{m=3*5*5*7=highlight(525)}}}