Difference between revisions of "2023 AMC 10B Problems/Problem 5"
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Let there be <math>n</math> numbers in the list of numbers, and let their sum be <math>S</math>. Then we have the following | Let there be <math>n</math> numbers in the list of numbers, and let their sum be <math>S</math>. Then we have the following | ||
− | < | + | <cmath>S+3n=45</cmath> |
− | < | + | <cmath>3S=45</cmath> |
− | From the second equation, <math>S=15</math>, | + | From the second equation, <math>S=15</math>. So, <math>15+3n=45</math> <math>\Rightarrow</math> <math>n=\boxed{\textbf{(A) }10}.</math> |
− | ~Mintylemon66 | + | ~Mintylemon66 (formatted atictacksh) |
==Solution 2== | ==Solution 2== | ||
Line 23: | Line 23: | ||
~vsinghminhas | ~vsinghminhas | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | If the list of numbers written on the board is <math>a_1, a_2, a_3, \ldots, a_n</math>, then we can formulate two equations: | ||
+ | |||
+ | <cmath>3n + \sum_{i=1}^{n} a_i = 45</cmath> | ||
+ | |||
+ | <cmath>3 \sum_{i=1}^{n} a_i = 45</cmath> | ||
+ | |||
+ | We can rewrite the first equation by multiplying both sides by <math>3</math>: | ||
+ | |||
+ | <math>3(3n + \sum_{i=1}^{n} a_i) = 3(45)</math> | ||
+ | |||
+ | <math>\Rightarrow 9n + 3 \sum_{i=1}^{n} a_i = 135</math> | ||
+ | |||
+ | Now, subtract the second equation from the first: | ||
+ | |||
+ | <cmath>(9n + 3 \sum_{i=1}^{n} a_i) - (3 \sum_{i=1}^{n} a_i) = 135 - 45</cmath> | ||
+ | |||
+ | <cmath>\Rightarrow 9n = 135 - 45</cmath> | ||
+ | |||
+ | <cmath>\Rightarrow 9n = 90</cmath> | ||
+ | |||
+ | <cmath>\Rightarrow n =\boxed{\textbf{(A) }10}</cmath> | ||
+ | |||
+ | ~ <math>Shalomkeshet</math> | ||
==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=SUnhwbA5_So | https://www.youtube.com/watch?v=SUnhwbA5_So | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/EuLkw8HFdk4?si=6dyj2QxkbBuNk6j7&t=951 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/-yk7ozNRrtQ | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2023|ab=B|num-b=4|num-a=6}} | {{AMC10 box|year=2023|ab=B|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 06:21, 12 December 2024
Contents
Problem
Maddy and Lara see a list of numbers written on a blackboard. Maddy adds to each number in the list and finds that the sum of her new numbers is . Lara multiplies each number in the list by and finds that the sum of her new numbers is also . How many numbers are written on the blackboard?
Solution
Let there be numbers in the list of numbers, and let their sum be . Then we have the following
From the second equation, . So,
~Mintylemon66 (formatted atictacksh)
Solution 2
Let where represents the th number written on the board. Lara's multiplied each number by , so her sum will be . This is the same as . We are given this quantity is equal to , so the original numbers add to . Maddy adds to each of the terms which yields, . This is the same as the sum of the original series plus . Setting this equal to ,
~vsinghminhas
Solution 3
If the list of numbers written on the board is , then we can formulate two equations:
We can rewrite the first equation by multiplying both sides by :
Now, subtract the second equation from the first:
~
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=SUnhwbA5_So
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/EuLkw8HFdk4?si=6dyj2QxkbBuNk6j7&t=951
~Math-X
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.