2000 CEMC Pascal Problems/Problem 2

Problem

The sum of $29 + 12 + 23$ is

$\text{ (A) }\ 32^2 \qquad\text{ (B) }\ 2^6 \qquad\text{ (C) }\ 3^4 \qquad\text{ (D) }\ 1^{64} \qquad\text{ (E) }\ 64^0$

Evaluating the sum, we get $29 + 12 + 23 = 41 + 23 = 64$ .

We now want to find what each of the answer choices are equal to until we get an answer that is equal to $64$ .

$32^2 = 1024$

$2^6 = 64$

This means that the answer is $\boxed {\textbf {(B) } 2^6}$ .

~anabel.disher