Question 238440
The key to this problem is to understand the following property of exponents: {{{(a^p)^q = a^((p*q))}}}:<ul><li>For perfect squares, q = 2: {{{(a^p)^2 = a^((p*2)) = a^((2p))}}}. This tells us that any exponent that is a multiple of 2 is a perfect square.</li><li>For perfect cube, q = 3: {{{(a^p)^3 = a^((p*3)) = a^((3p))}}}. This tells us that any exponent that is a multiple of 3 is a perfect cube.</li><li>For perfect fifth powers, q = 5: {{{(a^p)^5 = a^((p*5)) = a^((5p))}}}. This tells us that any exponent that is a multiple of 5 is a perfect fifth power.</li></ul>
So an exponent that is a prefect square, perfect cube <i>and</i> a perfect fifth power, all at the same time, will be a multiple of 2, 3 and 5. The lowest such exponent will be the Lowest Common Multiple (LCM) of these. The LCM of 2, 3 and 5 is 30.<br>
So the smallest number of cents (other than 1) would be {{{2^30 = 1073741824}}} cents which is $10737418.24.