Difference between revisions of "2024 AMC 8 Problems/Problem 1"
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| − | ==Problem== | + | ==Problem 1== |
| − | What is the ones digit of<cmath>222, | + | What is the ones digit of <cmath>222,22 |
| + | 2-22,222-2,222-222-22-2?</cmath> | ||
| + | <math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8</math> | ||
==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== | |
| − | + | 222,222-22,222 = 200,000 | |
| + | 200,000 - 2,222 = 197778 | ||
| + | 197778 - 222 = 197556 | ||
| + | 197556 - 22 = 197534 | ||
| + | 197534 - 2 = 1957532 | ||
| − | + | So our answer is <math>\boxed{\textbf{(B) } 2}</math>. | |
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==Solution 3== | ==Solution 3== | ||
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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> 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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==Solution 4== | ==Solution 4== | ||
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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 1 (Quick and Easy!)== | |
| − | + | https://youtu.be/Ol1seWX0xHY | |
| − | ==Solution | ||
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| − | + | ~Education, the Study of Everything | |
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==Video Solution (easy to understand)== | ==Video Solution (easy to understand)== | ||
| − | https://youtu.be/BaE00H2SHQM?si= | + | https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130 |
~Math-X | ~Math-X | ||
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/ktzijuZtDas&t=36 | https://youtu.be/ktzijuZtDas&t=36 | ||
Latest revision as of 12:02, 20 April 2024
Contents
Problem 1
What is the ones digit of ![\[222,22 2-22,222-2,222-222-22-2?\]](http://latex.artofproblemsolving.com/a/0/a/a0a32e80a9a60f8adee45ccd4fa65fb37f60baa1.png)
Solution 1
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We can rewrite the expression as ![\[222,222-(22,222+2,222+222+22+2).\]](http://latex.artofproblemsolving.com/d/7/4/d74bfc53ce46f0b91792bb0b074e29716ed43be7.png)
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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 .
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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.
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Solution 2
222,222-22,222 = 200,000 200,000 - 2,222 = 197778 197778 - 222 = 197556 197556 - 22 = 197534 197534 - 2 = 1957532
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/0/4/1/041203878d9ebe56721f48cbed9e09abe44cf179.png)
Thus, we get the answer 
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
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. 
