1997 AJHSME Problems/Problem 11

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Problem

Let $\boxed{N}$ mean the number of whole number divisors of $N$. For example, $\boxed{3}=2$ because 3 has two divisors, 1 and 3. Find the value of

\[\boxed{\boxed{11}\times\boxed{20}}.\]

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 24$

Solution

$\boxed{\boxed{11}\times\boxed{20}}$

$11$ has $2$ factors, and $20$ has $\{1, 2, 4, 5, 10, 20\}$ as factors, for a total of $6$ factors. Plugging $\boxed{11} = 2$ and $\boxed {20}= 6$:

$\boxed{2 \times 6}$

$\boxed{12}$

$12$ has factors of $\{1, 2, 3, 4, 6, 12\}$, so $\boxed{12} = 6$, and the answer is $\boxed{A}$

See also

1997 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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
All AJHSME/AMC 8 Problems and Solutions

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