Competition problem — Number theory

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Number theory Difficulty 2.5 Junior Find the answer

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?

Pick one

Solution

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

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.