2015 AMC 12B Problem 6

Below is the professionally curated solution for Problem 6 of the 2015 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 12B solutions, or check the answer key.

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Concepts:parity basic counting

Difficulty rating: 1290

Back in 1930, Tillie had to memorize her multiplication facts from $0 \times 0$ through $12 \times 12.$ 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?

$0.21$

$0.25$

$0.46$

$0.50$

$0.75$

Solution:

The body has $13 \times 13 = 169$ entries. A product is odd exactly when both factors are odd.

There are $6$ odd numbers among $0, 1, \ldots, 12,$ giving $6 \times 6 = 36$ odd entries. The fraction is $\dfrac{36}{169} = 0.213\ldots \approx 0.21.$

Thus, the correct answer is A.

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