Difference between revisions of "1997 AJHSME Problems/Problem 11"

Latest revision as of 23:26, 4 July 2013

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

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

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

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

$11$ 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 $\boxed{A}$

1997 AJHSME (Problems • Answer Key • Resources)
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

@@ Line 11: / Line 11: @@
 <math>\boxed{\boxed{11}\times\boxed{20}}</math>
-<math>11</math> has <math>2</math> factors, and <math>20</math> has <math>\{1, 2, 4, 5, 10, 20\}</math> as factors, for a total of <math>6</math> factors.  Plugging those in:
+<math>11</math> has <math>2</math> factors, and <math>20</math> has <math>\{1, 2, 4, 5, 10, 20\}</math> as factors, for a total of <math>6</math> factors.  Plugging <math>\boxed{11} = 2</math> and <math>\boxed {20}= 6</math>:
 <math>\boxed{2 \times 6}</math>
@@ Line 24: / Line 24: @@
 * [[AJHSME Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{MAA Notice}}