CEMC Fermat

Algebra Difficulty 2.3 Junior Find the answer

Three real numbers $a, b,$ and $c$ have a sum of 114 and a product of 46656. If $b=ar$ and $c=ar^2$ for some real number $r$ , what is the value of $a+c$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since $b=ar, c=ar^2$ , and the product of $a, b,$ and $c$ is 46656, then $a(ar)(ar^2)=46656$ or $a^3r^3=46656$ or $(ar)^3=46656$ or $ar=\sqrt[3]{46656}=36$ . Therefore, $b=ar=36$ . Since the sum of $a, b,$ and $c$ is 114, then $a+c=114-b=114-36=78$ .

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