Difference between revisions of "2025 AMC 8 Problems/Problem 4"
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| − | ==Problem== | + | == Problem == |
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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> | ||
| − | ==Solution== | + | == Solution 1 == |
| − | By the formula for the <math>n</math>th term of an arithmetic sequence, we get | + | By the formula for the <math>n</math>th term of an arithmetic sequence, we get the answer <math>a+d(n-1)</math> where <math>a=100, d=-7</math> and <math>n=10</math> which is equal to <math>100-7(10-1) = \boxed{\text{(B)\ 37}}</math>. |
~Soupboy0 | ~Soupboy0 | ||
| − | ==Solution 2== | + | == Solution 2 == |
| − | + | Since we want to find the <math>9</math>th number Lucius says after he says <math>100</math>, our answer is <math>100-(9 \cdot 7) = \boxed{\text{(B)\ 37}}</math> | |
~Sigmacuber | ~Sigmacuber | ||
| + | ==Solution 3 (Not the most practical)== | ||
| − | + | Using [[brute force]] and counting backward by <math>7</math>s, we have <math>100, 93, 86, 79, 72, 65, 58, 51, 44, \boxed{\text{(B)\ 37}}</math>. | |
| − | + | Note that this is not the most practical solution, and it is very time-consuming. | |
| − | |||
| − | + | ~codegirl2013 | |
| − | + | == Video Solution 1 by SpreadTheMathLove == | |
| − | |||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| − | ==Video Solution by Daily Dose of Math== | + | == Video Solution 2 by Daily Dose of Math == |
https://youtu.be/rjd0gigUsd0 | https://youtu.be/rjd0gigUsd0 | ||
| Line 35: | Line 35: | ||
~Thesmartgreekmathdude | ~Thesmartgreekmathdude | ||
| − | ==Video Solution by Thinking Feet== | + | == Video Solution 3 by Thinking Feet == |
| + | |||
https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
| − | ==See Also== | + | == See Also == |
| + | |||
{{AMC8 box|year=2025|num-b=3|num-a=5}} | {{AMC8 box|year=2025|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 14:52, 31 January 2025
Contents
Problem
Lucius is counting backward by 
s. His first three numbers are 
, 
, and 
. What is his 
th number?
Solution 1
By the formula for the 
th term of an arithmetic sequence, we get the answer 
where 
and 
which is equal to 
.
~Soupboy0
Solution 2
Since we want to find the 
th number Lucius says after he says , our answer is 
~Sigmacuber
Solution 3 (Not the most practical)
Using brute force and counting backward by s, we have 
.
Note that this is not the most practical solution, and it is very time-consuming.
~codegirl2013
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution 2 by Daily Dose of Math
~Thesmartgreekmathdude
Video Solution 3 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. 
