Difference between revisions of "2015 AMC 12B Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | + | 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 01:30, 4 March 2015
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
There are a total of 
products, 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 
.
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. 
