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.

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1991 AHSME Problems/Problem 14

Contents

Problem

Solution 1: Number Sense

Solution 2: Answer Choices

See also