Question 1195200
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Find the least integer by which 8100 must be multiplied to obtain a perfect cube
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<pre>
8100 = 81*100 = 9^2*10^2 = 2^2*3^4*5^2.


To make a perfect cube, we should have indexes at prime factors multiple of 3.


From here, it is clear that the least such integer number to multiply is  2*3^2*5 = 9*10 = 90.


Then the product is  8100*90 = 729000.


<U>CHECK</U>.  {{{root(3,729000)}}} = 90.


<U>ANSWER</U>.  The least integer by which 8100 must be multiplied to obtain a perfect cube is 90.
</pre>

Solved and explained.