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, 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 $36/169 \doteq \boxed{\textbf{(A)} \, 0.21}$.

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 $\boxed{\textbf{(A)} \, 0.21}$.

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS