Difference between revisions of "2024 AMC 8 Problems/Problem 1"
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==Problem 1== | ==Problem 1== | ||
− | What is the ones digit of <cmath>222, | + | What is the ones digit of: <cmath>222{,}222-22{,}222-2{,}222-222-22-2?</cmath> |
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<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== | ||
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We can rewrite the expression as <cmath>222,222-(22,222+2,222+222+22+2).</cmath> | We can rewrite the expression as <cmath>222,222-(22,222+2,222+222+22+2).</cmath> | ||
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We note that the units digit of the addition is <math>0</math> because all the units digits of the five numbers are <math>2</math> and <math>5*2=10</math>, which has a units digit of <math>0</math>. | We note that the units digit of the addition is <math>0</math> because all the units digits of the five numbers are <math>2</math> and <math>5*2=10</math>, which has a units digit of <math>0</math>. | ||
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Now, we have something with a units digit of <math>0</math> subtracted from <math>222,222</math>. The units digit of this expression is obviously <math>2</math>, and we get <math>\boxed{B}</math> as our answer. | Now, we have something with a units digit of <math>0</math> subtracted from <math>222,222</math>. The units digit of this expression is obviously <math>2</math>, and we get <math>\boxed{B}</math> as our answer. | ||
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==Solution 2== | ==Solution 2== | ||
− | 222,222-22,222 = 200,000 | + | <math>222,222-22,222 = 200,000</math> |
− | 200,000 - 2,222 = 197778 | + | <math>200,000 - 2,222 = 197778</math> |
− | 197778 - 222 = 197556 | + | <math>197778 - 222 = 197556</math> |
− | 197556 - 22 = 197534 | + | <math>197556 - 22 = 197534</math> |
− | 197534 - 2 = 1957532 | + | <math>197534 - 2 = 1957532 |
− | + | </math> | |
So our answer is <math>\boxed{\textbf{(B) } 2}</math>. | So our answer is <math>\boxed{\textbf{(B) } 2}</math>. | ||
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We only care about the unit's digits. | We only care about the unit's digits. | ||
− | 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{(B) } 2}</math>. | + | 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>. |
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+ | -unknown | ||
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+ | minor edits by Fireball9746 | ||
==Solution 4== | ==Solution 4== | ||
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)}</math> | Thus, we get the answer <math>\boxed{(B)}</math> | ||
− | ==Video Solution | + | ==Video Solution (MATH-X)== |
+ | https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (A Clever Explanation You’ll Get Instantly)== | ||
+ | https://youtu.be/5ZIFnqymdDQ?si=IbHepN2ytt7N23pl&t=53 | ||
+ | |||
+ | ~hsnacademy | ||
+ | |||
+ | ==Video Solution (Quick and Easy!)== | ||
https://youtu.be/Ol1seWX0xHY | https://youtu.be/Ol1seWX0xHY | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
− | ==Video Solution | + | ==Video Solution by Interstigation== |
− | https://youtu.be/ | + | 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== | ||
− | + | https://youtu.be/RzPadkHd3Yc | |
− | ==Video Solution by | + | ==Video Solution by WhyMath== |
− | https://youtu.be/ | + | 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}} |
Revision as of 06:13, 15 November 2024
Contents
[hide]- 1 Problem 1
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution (MATH-X)
- 7 Video Solution (A Clever Explanation You’ll Get Instantly)
- 8 Video Solution (Quick and Easy!)
- 9 Video Solution by Interstigation
- 10 Video Solution by Daily Dose of Math
- 11 Video Solution by Dr. David
- 12 Video Solution by WhyMath
- 13 See Also
Problem 1
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.
Solution 2
So our answer is .
Solution 3
We only care about the unit's digits.
Thus, ends in , after regrouping(10-2) ends in , ends in , ends in , and ends in .
-unknown
minor edits by Fireball9746
Solution 4
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
Video Solution (MATH-X)
https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130
~Math-X
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=IbHepN2ytt7N23pl&t=53
~hsnacademy
Video Solution (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.