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Difference between revisions of "2024 AMC 8 Problems/Problem 1"

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==Problem 1==
 
==Problem 1==
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What is the ones digit of <cmath>222{,}222-22{,}222-2{,}222-222-22-2?</cmath>
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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==
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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==
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We only care about the unit's digits.
 
We only care about the unit's digits.
  
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Thus, <math>2-2</math> ends in <math>0</math>, <math>0-2</math> after regrouping 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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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>.
  
 
-unknown
 
-unknown
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<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>
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==Video Solution (MATH-X)==
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https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130
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 +
~Math-X
  
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
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~hsnacademy
 
~hsnacademy
  
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==Video Solution 1 (Quick and Easy!)==
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==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 (easy to understand)==
 
−
https://youtu.be/BaE00H2SHQM?si=O0O0g7qq9AbhQN9I&t=130
 
−
 
−
~Math-X
 
  
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
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https://youtu.be/RzPadkHd3Yc
 
https://youtu.be/RzPadkHd3Yc
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 +
==Video Solution by WhyMath==
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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 06:13, 15 November 2024

Problem 1

What is the ones digit of: \[222{,}222-22{,}222-2{,}222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Solution 1

We can rewrite the expression as \[222,222-(22,222+2,222+222+22+2).\]

We note that the units digit of the addition is $0$ because all the units digits of the five numbers are $2$ and $5*2=10$, which has a units digit of $0$.

Now, we have something with a units digit of $0$ subtracted from $222,222$. The units digit of this expression is obviously $2$, and we get $\boxed{B}$ as our answer.

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 $\boxed{\textbf{(B) } 2}$.

Solution 3

We only care about the unit's digits.

Thus, $2-2$ ends in $0$, $0-2$ after regrouping(10-2) ends in $8$, $8-2$ ends in $6$, $6-2$ ends in $4$, and $4-2$ ends in $\boxed{\textbf{(B) } 2}$.

-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): \[(12-2)-(2+2+2+2)=10-8=2\] Thus, we get the answer $\boxed{(B)}$

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

~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

https://youtu.be/RzPadkHd3Yc

Video Solution by WhyMath

https://youtu.be/i4mcj3jRTxM

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

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