Difference between revisions of "2021 AMC 10A Problems/Problem 4"
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<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math> | <math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Since <cmath>\text{Distance}=\text{Speed}\times\text{Time},</cmath> we seek the sum <cmath>5(1)+12(1)+19(1)+26(1)+\cdots=5+12+19+26+\cdots,</cmath> in which there are 30 addends. The last addend is <math>5+7(30-1)=208.</math> Therefore, the requested sum is <cmath>5+12+19+26+\cdots+208=\frac{(5+208)(30)}{2}=\boxed{\textbf{(D)} ~3195}.</cmath> Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms, namely <cmath>\frac{\text{First}+\text{Last}}{2}\cdot\text{Count}.</cmath> ~MRENTHUSIASM | Since <cmath>\text{Distance}=\text{Speed}\times\text{Time},</cmath> we seek the sum <cmath>5(1)+12(1)+19(1)+26(1)+\cdots=5+12+19+26+\cdots,</cmath> in which there are 30 addends. The last addend is <math>5+7(30-1)=208.</math> Therefore, the requested sum is <cmath>5+12+19+26+\cdots+208=\frac{(5+208)(30)}{2}=\boxed{\textbf{(D)} ~3195}.</cmath> Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms, namely <cmath>\frac{\text{First}+\text{Last}}{2}\cdot\text{Count}.</cmath> ~MRENTHUSIASM | ||
Revision as of 11:59, 12 February 2021
Contents
Problem
A cart rolls down a hill, travelling 
inches the first second and accelerating so that during each successive 
-second time interval, it travels 
inches more than during the previous -second interval. The cart takes 
seconds to reach the bottom of the hill. How far, in inches, does it travel?
Solution 1
Since ![\[\text{Distance}=\text{Speed}\times\text{Time},\]](http://latex.artofproblemsolving.com/6/0/6/606216cd23e3b2f30aad3c379f14bbe7df7522e0.png)
we seek the sum ![\[5(1)+12(1)+19(1)+26(1)+\cdots=5+12+19+26+\cdots,\]](http://latex.artofproblemsolving.com/8/c/7/8c79e9f6d74d79045a0538ba9ba95d45a5f6d918.png)
in which there are 30 addends. The last addend is 
Therefore, the requested sum is ![\[5+12+19+26+\cdots+208=\frac{(5+208)(30)}{2}=\boxed{\textbf{(D)} ~3195}.\]](http://latex.artofproblemsolving.com/f/6/b/f6b8acc414c75c6fb1878ac72f3e7a796b4ddb49.png)
Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms, namely ![\[\frac{\text{First}+\text{Last}}{2}\cdot\text{Count}.\]](http://latex.artofproblemsolving.com/f/e/6/fe6482f623264044d4444569428eb2cc7d8c0d9f.png)
~MRENTHUSIASM
Solution 2 (Answer Choices and Modular Arithmetic)
From the -term sum ![\[5+12+19+26+\cdots\]](http://latex.artofproblemsolving.com/b/d/f/bdf805a7ab90ac77957f3afc6261f69f0b3a72c3.png)
in the previous solution, taking modulo 
gives ![\[5+12+19+26+\cdots \equiv 3(0+1+2+3+4+5+6+7+8+9) = 3(45)\equiv5 \pmod{10}.\]](http://latex.artofproblemsolving.com/7/6/2/76225769197f03e6a2f7780c74d49d5dc6e27519.png)
The only answer choices that are 
are 
and 
By a quick estimate, is too small, leaving us with 
~MRENTHUSIASM
Video Solution (Using Arithmetic Sequence)
~ pi_is_3.14
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
