Difference between revisions of "2016 AMC 10A Problems/Problem 14"
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<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math> | <math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math> | ||
− | == Solution == | + | ==Solution 1== |
− | + | The amount of twos in our sum ranges from <math>0</math> to <math>1008</math>, with differences of <math>3</math> because <math>2 \cdot 3 = \operatorname{lcm}(2, 3)</math>. | |
− | |||
− | The amount of twos in our sum ranges from <math>0</math> to <math>1008</math>, with differences of <math>3</math> because <math>2 \cdot 3 = lcm(2, 3)</math>. | ||
The possible amount of twos is <math>\frac{1008 - 0}{3} + 1 \Rightarrow \boxed{\textbf{(C)} 337}</math>. | The possible amount of twos is <math>\frac{1008 - 0}{3} + 1 \Rightarrow \boxed{\textbf{(C)} 337}</math>. | ||
− | + | ==Solution 2== | |
You can also see that you can rewrite the word problem into an 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 an 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 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> | + | We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have <math>\binom{2016-1}{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 | + | So, <math>2015/6\implies335\implies335+2=\boxed{\textbf{(C)}337}</math>. |
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/dHY8gjoYFXU?t=1058 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | https://youtu.be/99wDp0JcSUE | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | https://youtu.be/ZhAZ1oPe5Ds?t=2959 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==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 22:24, 17 January 2021
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 an 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, .
Video Solution
https://youtu.be/dHY8gjoYFXU?t=1058
~IceMatrix
~savannahsolver
https://youtu.be/ZhAZ1oPe5Ds?t=2959
~ pi_is_3.14
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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