Difference between revisions of "2016 AMC 10A Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
− | You can also see that you can rewrite the word problem into a equation <math>2x</math> + <math>3y</math> = <math>2016</math>. Therefore the question is just how many multiples of 3 subtracted from 2016 will be an even number. We can see | + | You can also see that you can rewrite the word problem into a equation <math>2x</math> + <math>3y</math> = <math>2016</math>. Therefore the question is just how many multiples of <math>3</math> subtracted from 2016 will be an even number. We can see that <math>x = 1008</math>, <math>y = 0</math> all the way to <math>x = 0</math>, and <math>y = 672</math> works, with <math>y</math> being incremented by <math>2</math>'s.Therefore, between <math>0</math> and <math>672</math>, the number of multiples of <math>2</math> is <math>\boxed{\textbf{(C)}337}</math>. |
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+ | ==See Also== | ||
+ | {{AMC10 box|year=2016|ab=A|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Revision as of 21:52, 3 February 2016
Contents
Problem
How many ways are there to write as the sum of twos and threes, ignoring order? (For example, and are two such ways.)
Solution 1
The amount of twos in our sum ranges from to , with differences of because .
The possible amount of twos is .
Solution 2
You can also see that you can rewrite the word problem into a equation + = . Therefore the question is just how many multiples of subtracted from 2016 will be an even number. We can see that , all the way to , and works, with being incremented by 's.Therefore, between and , the number of multiples of is .
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 10 Problems and Solutions |
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