Difference between revisions of "2015 AMC 12B Problems/Problem 6"

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==Solution==
 
==Solution==
 
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There are a total of <math>(12+1) \times (12+1) = 169</math> products, and a product is odd if and only if both its factors are odd. There are <math>6</math> odd numbers between <math>0</math> and <math>12</math>, namely <math>1, 3, 5, 7, 9, 11,</math> hence the number of odd products is <math>6 \times 6 = 36</math>. Therefore the answer is <math>36/169 \doteq \boxed{\textbf{(A)} \, 0.21}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=7|num-b=5}}
 
{{AMC12 box|year=2015|ab=B|num-a=7|num-b=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:30, 4 March 2015

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

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

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

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