Difference between revisions of "2025 AMC 8 Problems/Problem 4"
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Lucius is counting backward by <math>7</math>s. His first three numbers are <math>100</math>, <math>93</math>, and <math>86</math>. What is his <math>10</math>th number? | Lucius is counting backward by <math>7</math>s. His first three numbers are <math>100</math>, <math>93</math>, and <math>86</math>. What is his <math>10</math>th number? | ||
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 47</math> | <math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 47</math> | ||
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==Solution== | ==Solution== | ||
Revision as of 19:16, 30 January 2025
Contents
Problem
Lucius is counting backward by 
s. His first three numbers are 
, 
, and 
. What is his 
th number?
Solution
By the formula for the 
th term of an arithmetic sequence, we get that the answer is 
where 
and 
which is 
.
~Soupboy0
Solution 2 (Brute Force)
You could brute force(not really recommended) the sequence, and we get: ![\[100, 93, 86, 79, 72, 65, 58, 51, 44, 37\]](http://latex.artofproblemsolving.com/1/3/d/13d3a72a35f5d071e574e4871c41c69a0733a166.png)
Therefore, our answer is 
~athreyay
Vide Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
Video Solution by Thinking Feet
See Also
| 2025 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
