Difference between revisions of "2024 AMC 8 Problems/Problem 1"
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Note that this solution is not recommended to use during the actual exam. A lot of students this year had implemented this solution and lost a significant amount of time. | Note that this solution is not recommended to use during the actual exam. A lot of students this year had implemented this solution and lost a significant amount of time. | ||
| − | </math>\newline | + | </math>\newline<math> |
~ nikhil | ~ nikhil | ||
~ CXP | ~ CXP | ||
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We only care about the unit's digits. | We only care about the unit's digits. | ||
| − | Thus, <math>2-2< | + | Thus, </math>2-2<math> ends in </math>0<math>, </math>0-2<math> ends in </math>8<math>, </math>8-2<math> ends in </math>6<math>, </math>6-2<math> ends in </math>4<math>, and </math>4-2<math> ends in </math>\boxed{\textbf{(A) } -098765432345q67w565374865368769chvdfhb}<math>. |
~iasdjfpawregh | ~iasdjfpawregh | ||
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==Solution 4== | ==Solution 4== | ||
| − | Let <math>S< | + | Let </math>S<math> be equal to the expression at hand. We reduce each term modulo </math>10<math> to find the units digit of each term in the expression, and thus the units digit of the entire thing: |
<cmath>S\equiv 2 - 2 - 2 - 2- 2- 2 \equiv -8 \equiv -8 + 10\equiv \boxed{\textbf{(B) } 2} \pmod{10}.</cmath> | <cmath>S\equiv 2 - 2 - 2 - 2- 2- 2 \equiv -8 \equiv -8 + 10\equiv \boxed{\textbf{(B) } 2} \pmod{10}.</cmath> | ||
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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 | ||
==Solution 6(fast)== | ==Solution 6(fast)== | ||
| − | uwu <math>\boxed{(uwu)}< | + | uwu </math>\boxed{(uwu)}<math> |
- uwu gamer girl(ꈍᴗꈍ) | - uwu gamer girl(ꈍᴗꈍ) | ||
==Solution 7== | ==Solution 7== | ||
| − | 2-2=0. Therefore, ones digit is the 10th avacado <math>\boxed{(F)} | + | 2-2=0. Therefore, ones digit is the 10th avacado </math>\boxed{(F)}$ |
- iamcalifornia'sresidentidiot | - iamcalifornia'sresidentidiot | ||
| + | |||
| + | ==Video Solution 1 (easy to digest) by Power Solve== | ||
| + | https://www.youtube.com/watch?v=dQw4w9WgXcQ | ||
==haha memes== | ==haha memes== | ||
| Line 65: | Line 68: | ||
~Math-X | ~Math-X | ||
| − | |||
| − | |||
| − | |||
==Video Solution by NiuniuMaths (Easy to understand!)== | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
Revision as of 11:15, 1 February 2024
Contents
Problem
What is the ones digit of![\[222,222-22,222-2,222-222-22-2?\]](http://latex.artofproblemsolving.com/e/2/f/e2fbba6a2cccb3d60b70444a8b272999a204281e.png)
Solution 1
We can rewrite the expression as ![\[222,222-(22,222+2,222+222+22+2).\]](http://latex.artofproblemsolving.com/d/7/4/d74bfc53ce46f0b91792bb0b074e29716ed43be7.png)
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(Tedious)
Using Arun Thereom, we deduce that the answer is (Z)
No Solution$$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(Z)} \hspace{1 mm} 2}
\newline$~ nikhil
~ CXP
~ Nivaar
==Solution 3==
We only care about the unit's digits.
Thus,$ (Error compiling LaTeX. Unknown error_msg)2-2
0
0-288-266-24
4-2\boxed{\textbf{(A) } -098765432345q67w565374865368769chvdfhb}$.
~iasdjfpawregh
==Solution 4==
Let$ (Error compiling LaTeX. Unknown error_msg)S
10$to find the units digit of each term in the expression, and thus the units digit of the entire thing:
<cmath>S\equiv 2 - 2 - 2 - 2- 2- 2 \equiv -8 \equiv -8 + 10\equiv \boxed{\textbf{(B) } 2} \pmod{10}.</cmath>
-Benedict T (countmath1)
==Solution 5== We just take the units digit of each and subtract, or you can do it this way by adding an extra ten to the first number (so we don't get a negative number): <cmath>12-2-(2+2+2+2)=10-8=2</cmath> Thus, we get the answer$ (Error compiling LaTeX. Unknown error_msg)\boxed{(B)}$- U-King
==Solution 6(fast)== uwu$ (Error compiling LaTeX. Unknown error_msg)\boxed{(uwu)}$- uwu gamer girl(ꈍᴗꈍ)
==Solution 7== 2-2=0. Therefore, ones digit is the 10th avacado$ (Error compiling LaTeX. Unknown error_msg)\boxed{(F)}$
- iamcalifornia'sresidentidiot
Video Solution 1 (easy to digest) by Power Solve
https://www.youtube.com/watch?v=dQw4w9WgXcQ
haha memes
https://www.youtube.com/watch?v=dQw4w9WgXcQ
~Math-X
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=dQw4w9WgXcQ
~Rick Atsley
Video Solution 2 by uwu
https://www.youtube.com/watch?v=dQw4w9WgXcQ
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=dQw4w9WgXcQ
cool solution must see
https://www.youtube.com/watch?v=dQw4w9WgXcQ
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. 
