Difference between revisions of "2015 AMC 12B Problems/Problem 6"
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Back in 1930, Tillie had to memorize her multiplication facts from <math>0 \times 0</math> to <math>12 \times 12</math>. 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? | Back in 1930, Tillie had to memorize her multiplication facts from <math>0 \times 0</math> to <math>12 \times 12</math>. 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? | ||
| − | <math>\textbf{(A)}\; | + | <math>\textbf{(A)}\; 0.21 \qquad\textbf{(B)}\; 0.25 \qquad\textbf{(C)}\; 0.46 \qquad\textbf{(D)}\; 0.50 \qquad\textbf{(E)}\; 0.75</math> |
| − | ==Solution== | + | ==Solution 1== |
| + | There are a total of <math>(12+1) \times (12+1) = 169</math> products (don't forget to include the 0's!), 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>\frac{36}{169} \doteq \boxed{\textbf{(A)} \, 0.21}</math>. | ||
| + | NOTE: The exact value of <math>\frac{36}{169}</math> is <math>0.21307751...</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | Note that if we had an <math>11</math> by <math>11</math> multiplication table, the fraction of odd products becomes <math>0.25</math>. If we add the <math>12</math> by <math>12</math>, the fraction of odd products decreases. Because <math>0.21</math> is the only option less than <math>0.25</math>, our answer is <math>\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}} | ||
Latest revision as of 08:31, 31 October 2024
Contents
Problem
Back in 1930, Tillie had to memorize her multiplication facts from 
to 
. 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?
Solution 1
There are a total of 
products (don't forget to include the 0's!), and a product is odd if and only if both its factors are odd. There are 
odd numbers between 
and 
, namely 
hence the number of odd products is 
. Therefore the answer is 
.
NOTE: The exact value of 
is 
Solution 2
Note that if we had an 
by multiplication table, the fraction of odd products becomes 
. If we add the by , the fraction of odd products decreases. Because 
is the only option less than , our answer is 
.
See Also
| 2015 AMC 12B (Problems • Answer Key • Resources) | |
| 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. 
