2015 AMC 12B Problems/Problem 6

Problem

Back in 1930, Tillie had to memorize her multiplication facts from $0 \times 0$ to $12 \times 12$ . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?

$\textbf{(A)}\; 0.21 \qquad\textbf{(B)}\; 0.25 \qquad\textbf{(C)}\; 0.46 \qquad\textbf{(D)}\; 0.50 \qquad\textbf{(E)}\; 0.75$

Solution 1

There are a total of $(12+1) \times (12+1) = 169$ products (don't forget to include the 0's!), and a product is odd if and only if both its factors are odd. There are $6$ odd numbers between $0$ and $12$ , namely $1, 3, 5, 7, 9, 11$ hence the number of odd products is $6 \times 6 = 36$ . Therefore the answer is $\frac{36}{169} \doteq \boxed{\textbf{(A)} \, 0.21}$ .

NOTE: The exact value of $\frac{36}{169}$ is $0.21307751...$

Solution 2

Note that if we had an $11$ by $11$ multiplication table, the fraction of odd products becomes $0.25$ . If we add the $12$ by $12$ , the fraction of odd products decreases. Because $0.21$ is the only option less than $0.25$ , our answer is $\boxed{\textbf{(A)} \, 0.21}$ .

2015 AMC 12B (Problems • Answer Key • Resources)
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2015 AMC 12B Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See Also