Question 1133242
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We are looking for the smallest positive integer that is both 2 times the square of some positive integer and 5 times the 5th power of another positive integer:<br>
{{{2x^2 = 5y^5}}}<br>
There is a single known factor of 2 on the left and a single known factor of 5 on the right.  That means both x and y must both contain factors of only 2 and/or 5.  So<br>
let {{{x^2 = (2^a)(5^b)}}}, where a and b are even; and
and {{{y^5 = (2^c)(5^d)}}}, where c and d are multiples of 5<br>
Then<br>
{{{2*(2^a)(5^b) = 5(2^c)(5^d)}}}<br>
{{{(2^(a+1))(5^b) = (5^(d+1))(2^c)}}}<br>
From that we know<br>
{{{c = a+1}}} and {{{b = d+1}}}<br>
For the smallest positive integer for which all those are true, we need to have c = d = 5; that makes a=4 and b=6.<br>
And we finally have<br>
{{{x^2 = (2^4)(5^6)}}}  -->  {{{2x^2 = (2^5)(5^6)}}}
{{{y^5 = (2^5)(5^5)}}}  -->  {{{5y^2 = (2^5)(5^6)}}}<br>
So the smallest positive integer that is both 2 times the square of one positive integer and 5 times the 5th power of another positive integer is<br>
{{{(2^5)(5^6) = 500000}}}<br>
{{{500000 = 2(250000) = 2(500^2) = 2x^2}}}
{{{500000 = 5(100000) = 5(10^5) = 5y^5}}}