Difference between revisions of "2024 AMC 8 Problems/Problem 1"
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~ Dreamer1297 | ~ Dreamer1297 | ||
− | ==Solution 2== | + | ==Solution 2 (boring)== |
− | + | 222,222-22,222 = 200,000 | |
+ | 200,000 - 2,222 = 197778 | ||
+ | 197778 - 222 = 197556 | ||
+ | 197556 - 22 = 197534 | ||
+ | 197534 - 2 = 1957532 | ||
+ | |||
+ | So our answer is <math>/boxed{B]</math> | ||
==Solution 3== | ==Solution 3== |
Revision as of 08:19, 12 March 2024
Contents
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.
i am smart
~ Dreamer1297
Solution 2 (boring)
222,222-22,222 = 200,000 200,000 - 2,222 = 197778 197778 - 222 = 197556 197556 - 22 = 197534 197534 - 2 = 1957532
So our answer is $/boxed{B]$ (Error compiling LaTeX. Unknown error_msg)
Solution 3
We only care about the unit's digits.
Thus, ends in , ends in , ends in , ends in , and ends in .
~iasdjfpawregh ~hockey
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
- FU-King nice
Video Solution 1 (Quick and Easy!)
~Education, the Study of Everything
Video Solution (easy to understand)
https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130
~Math-X
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=36
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.