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> |
| − | |||
<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> | ||
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==Solution 2== | ==Solution 2== | ||
| − | <math>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> | </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 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 (A Clever Explanation You’ll Get Instantly)== | ||
| + | https://youtu.be/5ZIFnqymdDQ?si=IbHepN2ytt7N23pl&t=53 | ||
| + | |||
| + | ~hsnacademy | ||
==Video Solution 1 (Quick and Easy!)== | ==Video Solution 1 (Quick and Easy!)== | ||
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~Thesmartgreekmathdude | ~Thesmartgreekmathdude | ||
| + | |||
| + | ==Video Solution by Dr. David== | ||
| + | |||
| + | https://youtu.be/RzPadkHd3Yc | ||
==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 02:40, 25 September 2024
Contents
- 1 Problem 1
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution (A Clever Explanation You’ll Get Instantly)
- 7 Video Solution 1 (Quick and Easy!)
- 8 Video Solution (easy to understand)
- 9 Video Solution by Interstigation
- 10 Video Solution by Daily Dose of Math
- 11 Video Solution by Dr. David
- 12 See Also
Problem 1
What is the ones digit of ![\[222{,}222-22{,}222-2{,}222-222-22-2?\]](http://latex.artofproblemsolving.com/1/5/6/156c3662fe4c874a80e54887f76c302c13b9d14c.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.
−
−
Solution 2
So our answer is 
.
Solution 3
We only care about the unit's digits.
Thus, 
ends in , 
ends in 
, 
ends in 
, 
ends in 
, and 
ends in .
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):
![\[(12-2)-(2+2+2+2)=10-8=2\]](http://latex.artofproblemsolving.com/1/a/e/1ae83497079c7f1f607247a754db7a749c0380c3.png)
Thus, we get the answer 
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=IbHepN2ytt7N23pl&t=53
~hsnacademy
Video Solution 1 (Quick and Easy!)
~Education, the Study of Everything
Video Solution (easy to understand)
https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130
~Math-X
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
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. 
