Difference between revisions of "2015 AMC 8 Problems/Problem 20"
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</math> | </math> | ||
+ | ==Solution== | ||
+ | So let there be <math>x</math> pairs of <math>1</math> dollar socks, <math>y</math> pairs of <math>3</math> dollar socks, <math>z</math> pairs of <math>4</math> dollar socks. | ||
+ | |||
+ | We have <math>x+y+z=12</math>, <math>x+3y+4z=24</math>, and <math>x,y,z \ge 1</math>. | ||
+ | |||
+ | Now we subtract to find <math>2y+3z=12</math>, and <math>y,z \ge 1</math>. | ||
+ | It follows that <math>y</math> is a multiple of <math>3</math> and <math>2y</math> is a multiple of <math>6</math>, so since <math>0<2y<12</math>, we must have <math>2y=6</math>. | ||
+ | |||
+ | Therefore, <math>y=3</math>, and it follows that <math>z=2</math>. Now <math>x=12-y-z=12-3-2=\boxed{\textbf{(D)}~7}</math>, as desired. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=19|num-a=21}} | {{AMC8 box|year=2015|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:41, 25 November 2015
Ralph went to the store and bought 12 pairs of socks for a total of $24. Some of the socks he bought cost $1 a pair, some of the socks he bought cost $3 a pair, and some of the socks he bought cost $4 a pair. If he bought at least one pair of each type, how many pairs of $1 socks did Ralph buy?
Solution
So let there be pairs of dollar socks, pairs of dollar socks, pairs of dollar socks.
We have , , and .
Now we subtract to find , and . It follows that is a multiple of and is a multiple of , so since , we must have .
Therefore, , and it follows that . Now , as desired.
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AJHSME/AMC 8 Problems and Solutions |
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