Difference between revisions of "2021 AMC 10A Problems/Problem 4"
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==Solution 1 (Arithmetic Series)== | ==Solution 1 (Arithmetic Series)== | ||
| − | Since <cmath>\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},</cmath> we seek the sum <cmath>5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,</cmath> in which there are <math>30</math> | + | Since <cmath>\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},</cmath> we seek the sum <cmath>5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,</cmath> in which there are <math>30</math> terms. The last term is <math>5+7\cdot(30-1)=208.</math> Therefore, the requested sum is <cmath>5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\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: <cmath>\mathrm{Sum}=\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.</cmath> ~MRENTHUSIASM |
==Solution 2 (Answer Choices and Modular Arithmetic)== | ==Solution 2 (Answer Choices and Modular Arithmetic)== | ||
Revision as of 17:22, 20 May 2021
Contents
- 1 Problem
- 2 Solution 1 (Arithmetic Series)
- 3 Solution 2 (Answer Choices and Modular Arithmetic)
- 4 Solution 3
- 5 Video Solution (Simple and Quick)
- 6 Video Solution (Arithmetic Sequence but in a different way)
- 7 Video Solution (Using Arithmetic Sequence)
- 8 Video Solution 4
- 9 Video Solution by TheBeautyofMath
- 10 See Also
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 (Arithmetic Series)
Since ![\[\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},\]](http://latex.artofproblemsolving.com/3/c/c/3cc286b6c23404aef94cb7500cd7f604bb3cf900.png)
we seek the sum ![\[5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,\]](http://latex.artofproblemsolving.com/a/6/5/a65d35f502f5a93a00e576aeb6d1b99c8af2464c.png)
in which there are terms. The last term is 
Therefore, the requested sum is ![\[5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{\textbf{(D)} ~3195}.\]](http://latex.artofproblemsolving.com/4/f/5/4f5f50ede4de794fe97e1299c26b81ffa228c402.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: ![\[\mathrm{Sum}=\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.\]](http://latex.artofproblemsolving.com/c/e/6/ce65ba615dc4e73740078ca1f10a5fb4189d7a9c.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 Solution 1, taking modulo 
gives ![\[5+12+19+26+\cdots \equiv 3\cdot(0+1+2+3+4+5+6+7+8+9) = 3\cdot45\equiv5 \pmod{10}.\]](http://latex.artofproblemsolving.com/5/3/6/5369c2202d51e214dfcba30cae88782b58ddcf44.png)
The only answer choices congruent to modulo are 
and 
By a quick estimation, is too small, leaving us with 
~MRENTHUSIASM
Solution 3
The distance (in inches) traveled within each 1-second interval is:
This is an arithmetic sequence so the total distance travelled, found by summing them up is:
Or,
~BakedPotato66
Video Solution (Simple and Quick)
~ Education, the Study of Everything
Video Solution (Arithmetic Sequence but in a different way)
https://www.youtube.com/watch?v=sJa7uB-UoLc&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=4
~ North America Math Contest Go Go Go
Video Solution (Using Arithmetic Sequence)
~ pi_is_3.14
Video Solution 4
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/50CThrk3RcM?t=262
~IceMatrix
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. 
