2015 AMC 12B Problem 6

Attempt Problem 6 of the 2015 AMC 12B below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

6.

Back in 1930,1930, Tillie had to memorize her multiplication facts from 0×00 \times 0 through 12×12.12 \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?
0.210.21
0.250.25
0.460.46
0.500.50
0.750.75
Answer: A
Concepts:paritybasic counting
Small hint:
A product is odd only when both factors are odd
Big hint:
Count the odd numbers from 00 to 12,12, then compare with the 13×1313 \times 13 entries
Solution:
The body has 13×13=16913 \times 13 = 169 entries. A product is odd exactly when both factors are odd.
There are 66 odd numbers among 0,1,…,12,0, 1, \ldots, 12, giving 6×6=366 \times 6 = 36 odd entries. The fraction is 36169=0.213…≈0.21.\dfrac{36}{169} = 0.213\ldots \approx 0.21.
Thus, the correct answer is A.
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