Difference between revisions of "2024 AMC 8 Problems/Problem 1"
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Using Arun Thereom, we deduce that the answer is (Z) | Using Arun Thereom, we deduce that the answer is (Z) | ||
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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> |
~ nikhil | ~ nikhil | ||
~ CXP | ~ CXP | ||
| Line 30: | Line 28: | ||
We only care about the unit's digits. | We only care about the unit's digits. | ||
| − | Thus, < | + | 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 | ||
| Line 36: | Line 34: | ||
==Solution 4== | ==Solution 4== | ||
| − | Let < | + | 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 < | + | Thus, we get the answer <math>\boxed{(B)}</math> |
- U-King | - U-King | ||
==Solution 6(fast)== | ==Solution 6(fast)== | ||
| − | 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 < | + | 2-2=0. Therefore, ones digit is the 10th avacado <math>\boxed{(F)}</math> |
- iamcalifornia'sresidentidiot | - iamcalifornia'sresidentidiot | ||
Revision as of 22:02, 3 February 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2(Tedious)
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6(fast)
- 8 Solution 7
- 9 Video Solution 1 (easy to digest) by Power Solve
- 10 Video Solution (easy to understand)
- 11 Video Solution by NiuniuMaths (Easy to understand!)
- 12 Video Solution 2 by uwu
- 13 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 14 cool solution must see
- 15 See Also
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)
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.
~ nikhil
~ CXP
~ Nivaar
Solution 3
We only care about the unit's digits.
Thus, 
ends in , 
ends in 
, 
ends in 
, 
ends in 
, and 
ends in 
.
~iasdjfpawregh
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:
![\[S\equiv 2 - 2 - 2 - 2- 2- 2 \equiv -8 \equiv -8 + 10\equiv \boxed{\textbf{(B) } 2} \pmod{10}.\]](http://latex.artofproblemsolving.com/3/9/d/39dd43bbf71a3102bc3e6addcd0fa9bd9a724945.png)
-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):
![\[12-2-(2+2+2+2)=10-8=2\]](http://latex.artofproblemsolving.com/0/4/1/041203878d9ebe56721f48cbed9e09abe44cf179.png)
Thus, we get the answer 
- U-King
Solution 6(fast)
uwu 
- uwu gamer girl(ꈍᴗꈍ)
Solution 7
2-2=0. Therefore, ones digit is the 10th avacado 
- iamcalifornia'sresidentidiot
Video Solution 1 (easy to digest) by Power Solve
https://www.youtube.com/watch?v=dQw4w9WgXcQ
Video Solution (easy to understand)
https://youtu.be/BaE00H2SHQM?si=_8lhp8-dzNxZ-eUQ
~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. 
