Difference between revisions of "2016 AMC 10A Problems/Problem 14"
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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>. | 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>. | ||
| + | |||
| + | === Solution 3=== | ||
| + | We can utilize the stars-and-bars distribution technique to solve this problem. | ||
| + | We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have <math>(2016-1) C (2-1)</math> "total" answers. We want every third x and second y, so we divide our previous total by 6, which will result in <math>2015/6</math>. We have to round down to the nearest integer, and we have to add 2 because we did not consider the 2 solutions involving x or y being 0. | ||
| + | So, <math>2015/6</math> <math>--></math> 335 <math>--></math> 335+2 = <math>\boxed{\textbf{(C)}337}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=13|num-a=15}} | {{AMC10 box|year=2016|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:30, 15 February 2016
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
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 
.
Solution 3
We can utilize the stars-and-bars distribution technique to solve this problem.
We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have 
"total" answers. We want every third x and second y, so we divide our previous total by 6, which will result in 
. We have to round down to the nearest integer, and we have to add 2 because we did not consider the 2 solutions involving x or y being 0.
So, 
335 335+2 = .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
