Difference between revisions of "2024 AMC 8 Problems/Problem 1"
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We just take the units digit of each and subtract, or you can do it this way by adding an extra ten to the first number (so we don't get a negative number): | We just take the units digit of each and subtract, or you can do it this way by adding an extra ten to the first number (so we don't get a negative number): | ||
<cmath>12-2-(2+2+2+2)=10-8=2</cmath> | <cmath>12-2-(2+2+2+2)=10-8=2</cmath> | ||
− | Thus, we get the answer <math>\boxed{(B) | + | Thus, we get the answer <math>\boxed{(B)}</math> |
- U-King | - U-King |
Revision as of 01:49, 27 January 2024
Contents
[hide]Problem
What is the ones digit of
Solution 1
We can rewrite the expression as
We note that the units digit of the addition is because all the units digits of the five numbers are and , which has a units digit of .
Now, we have something with a units digit of subtracted from . The units digit of this expression is obviously , and we get as our answer.
~ Dreamer1297
Solution 2
This means the ones digit is ~ nikhil ~ CXP
Solution 3
We only care about the unit's digits.
Thus, ends in , ends in , ends in , ends in , and ends in .
~MrThinker
Solution 4
Let be equal to the expression at hand. We reduce each term modulo to find the units digit of each term in the expression, and thus the units digit of the entire thing:
-Benedict T (countmath1)
Solution 5
We just take the units digit of each and subtract, or you can do it this way by adding an extra ten to the first number (so we don't get a negative number): Thus, we get the answer
- U-King
Video Solution 1 (easy to digest) by Power Solve
https://www.youtube.com/watch?v=HE7JjZQ6xCk
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=Ylw-kJkSpq8
~NiuniuMaths
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=L83DxusGkSY
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=IMqOfC9lZSo
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.