Difference between revisions of "2025 AMC 8 Problems/Problem 16"
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~Kapurnicus (someone please edit this and make it look better) | ~Kapurnicus (someone please edit this and make it look better) | ||
| − | ~Minor edit by NYCnerd | + | ~Minor edit by NYCnerd (did that) |
==Video Solution by Pi Academy== | ==Video Solution by Pi Academy== | ||
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK | https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK | ||
Revision as of 20:05, 24 February 2025
Contents
Problem
Five distinct integers from 
to 
are chosen, and five distinct integers from 
to 
are chosen. No two numbers differ by exactly . What is the sum of the ten chosen numbers?
Solution
Note that for no two numbers to differ by , every number chosen must have a different units digit. To make computations easier, we can choose 
from the first group and 
from the second group. Then the sum evaluates to 
.
~Soupboy0
~ Edited by Aoum
Another Way To Compute
For 
, we can add the first term and the last term, which is 
. If we add the second term and the second-to-last term it is also . There are 
pairs that sum to , so the answer is 
which equals 
.
- leafy
Similar solution
One efficient method is to quickly add 
, which is 
. Then because you took 
in total away from , you add . 
.
~Bepin999
Solution 4
To solve this problem, I started with the easiest/smallest case possible. In my opinion, that was
.
I then solved this equation quickly using Little Gauss's method, rearranging that into
.
I simplified this into
.
Solving this simple equation gives us the answer, which is .
~Kapurnicus (someone please edit this and make it look better)
~Minor edit by NYCnerd (did that)
Video Solution by Pi Academy
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=DtG8sG4CK4RrUlVz&t=1815 ~hsnacademy
Video Solution by Thinking Feet
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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. 
