2025 AMC 8 Problems/Problem 4: Difference between revisions
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== Problem == | |||
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 1 == | |||
We plug <math>a=100, d=-7</math> and <math>n=10</math> into the formula <math>a+d(n-1)</math> for the <math>n</math>th term of an arithmetic sequence whose first term is <math>a</math> and common difference is <math>d</math> to get <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>, <math>7</math> is subtracted from his number <math>9</math> times, so our answer is <math>100-(9 \cdot 7) = \boxed{\text{(B)\ 37}}</math> | |||
~Sigmacuber | |||
== Solution 3 == | |||
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 solution is not practical and very time-consuming. | |||
~codegirl2013, athreyay | |||
== Solution 4 == | |||
This can be thought of as an arithmetic sequence. Knowing that our first term is <math>100</math>, we have to add <math>7</math> to get to our 0th term, <math>107</math>. Our answer is then <math>107 - 10 \cdot 7 = \boxed{\text{(B)\ 37}}</math>. | |||
~Kapurnicus, NYCnerd | |||
== Video Solution 1 (Detailed Explanation) 🚀⚡📊 == | |||
https://www.youtube.com/watch?v=rf5c9ulMA2I | |||
~ ChillThingz :) | |||
== Video Solution 2 == | |||
[//www.youtube.com/jTTcscvcQmI SpreadTheMathLove] | |||
== Video Solution 3 by Daily Dose of Math == | |||
[//youtu.be/rjd0gigUsd0 ~Thesmartgreekmathdude] | |||
== Video Solution 4 == | |||
[//youtu.be/PKMpTS6b988 Thinking Feet] | |||
== Video Solution 5 == | |||
[//youtu.be/VP7g-s8akMY?si=K8Pxs_TQhlR2ntt9&t=211 ~hsnacademy] | |||
== Video Solution 6 == | |||
[//youtu.be/nwUanrEZpcQ CoolMathProblems] | |||
== Video Solution 7 == | |||
[//youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK Pi Academy] | |||
==Video Solution(Quick, fast, easy!)== | |||
https://youtu.be/fdG7EDW_7xk | |||
~MC | |||
== See Also == | |||
== | {{AMC8 box|year=2025|num-b=3|num-a=5}} | ||
{{MAA Notice}} | |||
[[Category:Introductory Algebra Problems]] | |||
Latest revision as of 10:45, 7 September 2025
Problem
Lucius is counting backward by
s. His first three numbers are
,
, and
. What is his
th number?
Solution 1
We plug
and
into the formula
for the
th term of an arithmetic sequence whose first term is
and common difference is
to get
.
~Soupboy0
Solution 2
Since we want to find the
th number Lucius says after he says
,
is subtracted from his number
times, so our answer is
~Sigmacuber
Solution 3
Using brute force and counting backward by
s, we have
.
Note that this solution is not practical and very time-consuming.
~codegirl2013, athreyay
Solution 4
This can be thought of as an arithmetic sequence. Knowing that our first term is
, we have to add
to get to our 0th term,
. Our answer is then
.
~Kapurnicus, NYCnerd
Video Solution 1 (Detailed Explanation) 🚀⚡📊
https://www.youtube.com/watch?v=rf5c9ulMA2I
~ ChillThingz :)
Video Solution 2
Video Solution 3 by Daily Dose of Math
Video Solution 4
Video Solution 5
Video Solution 6
Video Solution 7
Video Solution(Quick, fast, easy!)
~MC
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing