Difference between revisions of "2024 AMC 8 Problems/Problem 1"
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− | ==Problem | + | ==Problem== |
What is the ones digit of: <cmath>222{,}222-22{,}222-2{,}222-222-22-2?</cmath> | What is the ones digit of: <cmath>222{,}222-22{,}222-2{,}222-222-22-2?</cmath> | ||
<math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8</math> | <math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8</math> | ||
==Solution 1== | ==Solution 1== | ||
− | We can rewrite the expression as < | + | We can rewrite the expression as <math>222,222-(22,222+2,222+222+22+2)</math>. We note that the units digit of <math>22,222+2,222+222+22+2</math> is <math>0</math> because all the units digits of the five numbers are <math>2</math> and <math>5\cdot2=10</math>, which has a units digit of <math>0</math>. Now, we have something with a units digit of <math>0</math> subtracted from <math>222,222</math>, and so the units digit of this expression is <math>\boxed{\textbf{(B) } 2}</math>. |
− | |||
− | We note that the units digit of | ||
− | |||
− | Now, we have something with a units digit of <math>0</math> subtracted from <math>222,222</math> | ||
==Solution 2== | ==Solution 2== | ||
− | + | <cmath>222,222-22,222 = 200,000</cmath> | |
− | < | + | <cmath>200,000 - 2,222 = 197778</cmath> |
− | < | + | <cmath>197778 - 222 = 197556</cmath> |
− | < | + | <cmath>197556 - 22 = 197534</cmath> |
− | < | + | <cmath>197534 - 2 = 1957532</cmath> |
− | < | ||
− | </ | ||
So our answer is <math>\boxed{\textbf{(B) } 2}</math>. | So our answer is <math>\boxed{\textbf{(B) } 2}</math>. | ||
==Solution 3== | ==Solution 3== | ||
− | + | We only care about the units digits. Thus, <math>2-2</math> ends in <math>0</math>, <math>0-2</math> after regrouping(10-2) 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{(B) } 2}</math>. | |
− | We only care about the | ||
− | |||
− | Thus, <math>2-2</math> ends in <math>0</math>, <math>0-2</math> after regrouping(10-2) 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 | ||
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− | |||
− | |||
− | |||
==Solution 4== | ==Solution 4== | ||
+ | We just take the units digit of each and subtract, 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=\boxed{\textbf{(B) } 2}</cmath> | ||
− | + | == Solution 5 == | |
− | <cmath> | + | <cmath>222{,}222-22{,}222-2{,}222-222-22-2\equiv2-2-2-2-2\equiv-8\equiv\boxed{\textbf{(B) } 2}\pmod{10}</cmath> |
− | |||
− | == Video Solution | + | ==Video Solution by Central Valley Math Circle (Goes through full thought process)== |
+ | https://youtu.be/-XcShDyuZIo | ||
− | + | ~mr_mathman | |
− | + | == Video Solution 1 (Detailed Explanation) == | |
+ | https://youtu.be/jqsbMWhTYRg | ||
~ ChillGuyDoesMath :) | ~ ChillGuyDoesMath :) | ||
− | ==Video Solution (MATH-X)== | + | ==Video Solution 2 (MATH-X)== |
https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130 | https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130 | ||
− | + | ==Video Solution 3 (A Clever Explanation You’ll Get Instantly)== | |
− | |||
− | ==Video Solution (A Clever Explanation You’ll Get Instantly)== | ||
https://youtu.be/5ZIFnqymdDQ?si=IbHepN2ytt7N23pl&t=53 | https://youtu.be/5ZIFnqymdDQ?si=IbHepN2ytt7N23pl&t=53 | ||
~hsnacademy | ~hsnacademy | ||
− | ==Video Solution (Quick and Easy!)== | + | ==Video Solution 4 (Quick and Easy!)== |
https://youtu.be/Ol1seWX0xHY | https://youtu.be/Ol1seWX0xHY | ||
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==Video Solution by Daily Dose of Math== | ==Video Solution by Daily Dose of Math== | ||
− | |||
https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR | https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR | ||
Line 69: | Line 56: | ||
==Video Solution by Dr. David== | ==Video Solution by Dr. David== | ||
− | |||
https://youtu.be/RzPadkHd3Yc | https://youtu.be/RzPadkHd3Yc | ||
Latest revision as of 21:09, 7 January 2025
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Video Solution by Central Valley Math Circle (Goes through full thought process)
- 8 Video Solution 1 (Detailed Explanation)
- 9 Video Solution 2 (MATH-X)
- 10 Video Solution 3 (A Clever Explanation You’ll Get Instantly)
- 11 Video Solution 4 (Quick and Easy!)
- 12 Video Solution by Interstigation
- 13 Video Solution by Daily Dose of Math
- 14 Video Solution by Dr. David
- 15 Video Solution by WhyMath
- 16 See Also
Problem
What is the ones digit of:
Solution 1
We can rewrite the expression as . We note that the units digit of 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 , and so the units digit of this expression is .
Solution 2
So our answer is .
Solution 3
We only care about the units digits. Thus, ends in , after regrouping(10-2) ends in , ends in , ends in , and ends in .
Solution 4
We just take the units digit of each and subtract, adding an extra ten to the first number so we don't get a negative number:
Solution 5
Video Solution by Central Valley Math Circle (Goes through full thought process)
~mr_mathman
Video Solution 1 (Detailed Explanation)
~ ChillGuyDoesMath :)
Video Solution 2 (MATH-X)
https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130
Video Solution 3 (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=IbHepN2ytt7N23pl&t=53
~hsnacademy
Video Solution 4 (Quick and Easy!)
~Education, the Study of Everything
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=36
Video Solution by Daily Dose of Math
https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR
~Thesmartgreekmathdude
Video Solution by Dr. David
Video Solution by WhyMath
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.