Difference between revisions of "2024 AMC 8 Problems/Problem 1"
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− | ==Problem | + | ==Problem== |
− | What is the | + | What is the last 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) } 8\qquad\textbf{(E) } 10</math> | |
− | <math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } | ||
==Solution 1== | ==Solution 1== | ||
− | + | 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 can rewrite the expression as < | ||
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− | We note that the units digit of | ||
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− | Now, we have something with a units digit of <math>0</math> subtracted from <math>222,222</math> | ||
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==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 = 197532</cmath> | ||
+ | So our answer is <math>\boxed{\textbf{(B) } 2}</math>. | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==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>222{,}222-22{,}222-2{,}222-222-22-2\equiv2-2-2-2-2\equiv-8\equiv\boxed{\textbf{(B) } 2}\pmod{10}</cmath> | ||
+ | |||
− | + | == Solution 6== | |
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− | + | We can ignore the other digits and just do <math>22-2-2-2-2-2</math>. Because you are subtracting five <math>2s</math> and <math>2\cdot5 = 10</math>, you subtract <math>10</math> from <math>22</math>. This gives us 12, so the last digit is <math>\boxed{\textbf{(B) } 2}</math>. | |
− | ==Solution | + | == Video Solution 1 (Detailed Explanation) 🚀⚡📊 == |
+ | Youtube Link ⬇️ | ||
− | + | https://youtu.be/jqsbMWhTYRg | |
− | + | ~ ChillGuyDoesMath :) | |
− | == | + | == Video by MathTalks_Now == |
− | + | https://www.youtube.com/watch?v=crn37TRMLv4 | |
− | |||
− | |||
− | + | -rc1219 | |
− | |||
− | + | ==Video Solution by Central Valley Math Circle (Goes through full thought process)== | |
+ | https://youtu.be/-XcShDyuZIo | ||
− | ==Video Solution ( | + | ==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)== | |
+ | https://youtu.be/5ZIFnqymdDQ?si=IbHepN2ytt7N23pl&t=53 | ||
− | ==Video Solution | + | ==Video Solution 4 (Quick and Easy)== |
+ | https://youtu.be/Ol1seWX0xHY | ||
+ | |||
+ | ==Video Solution 5 Interstigation== | ||
https://youtu.be/ktzijuZtDas&t=36 | https://youtu.be/ktzijuZtDas&t=36 | ||
− | ==Video Solution | + | ==Video Solution 6 Daily Dose of Math== |
+ | https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR | ||
− | https://youtu.be/ | + | ==Video Solution 7 Dr. David== |
+ | https://youtu.be/RzPadkHd3Yc | ||
− | + | ==Video Solution 8 WhyMath== | |
+ | https://youtu.be/i4mcj3jRTxM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2024|before=First Problem|num-a=2}} | {{AMC8 box|year=2024|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 08:46, 16 February 2025
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6
- 8 Video Solution 1 (Detailed Explanation) 🚀⚡📊
- 9 Video by MathTalks_Now
- 10 Video Solution by Central Valley Math Circle (Goes through full thought process)
- 11 Video Solution 2 (MATH-X)
- 12 Video Solution 3 (A Clever Explanation You’ll Get Instantly)
- 13 Video Solution 4 (Quick and Easy)
- 14 Video Solution 5 Interstigation
- 15 Video Solution 6 Daily Dose of Math
- 16 Video Solution 7 Dr. David
- 17 Video Solution 8 WhyMath
- 18 See Also
Problem
What is the last 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
Solution 6
We can ignore the other digits and just do . Because you are subtracting five
and
, you subtract
from
. This gives us 12, so the last digit is
.
Video Solution 1 (Detailed Explanation) 🚀⚡📊
Youtube Link ⬇️
~ ChillGuyDoesMath :)
Video by MathTalks_Now
https://www.youtube.com/watch?v=crn37TRMLv4
-rc1219
Video Solution by Central Valley Math Circle (Goes through full thought process)
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
Video Solution 4 (Quick and Easy)
Video Solution 5 Interstigation
https://youtu.be/ktzijuZtDas&t=36
Video Solution 6 Daily Dose of Math
https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR
Video Solution 7 Dr. David
Video Solution 8 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.