1991 AHSME Problems/Problem 14

Problem

If $x$ is the cube of a positive integer and $d$ is the number of positive integers that are divisors of $x$, then $d$ could be

(A) $200$ (B) $201$ (C) $202$ (D) $203$ (E) $204$

Solution 1: Number Sense

Solution by e_power_pi_times_i


Notice that if $x$ is expressed in the form $a^b$, then the number of positive divisors of $x^3$ is $3b+1$. Checking through all the answer choices, the only one that is in the form $3b+1$ is $\boxed{\textbf{(C) } 202}$.

Solution 2: Answer Choices

Solution by e_power_pi_times_i


Since the divisors are from $x^3$, then the answer must be something in (mod $3$). Since $200$ and $203$ are the same (mod $3$), as well as $201$ and $204$, $\boxed{\textbf{(C) } 202}$ is the only answer left.

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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