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

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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> 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>.
  
 
~MrThinker
 
~MrThinker

Revision as of 18:32, 25 January 2024

Problem

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.

~ Dreamer1297

Solution 2

\[222,222-22,222-2,222-222-22-2\] \[= 200,000 - 2,222 - 222 - 22 - 2\] \[= 197778 - 222 - 22 - 2\] \[= 197556 - 22 - 2\] \[= 197534 - 2\] \[= 197532\] This means the ones digit is $\boxed{(B) \hspace{1 mm} 2}$ $\newline$ ~ nikhil ~ CXP

Solution 3

We only care about the unit's digits.

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

~MrThinker

Video Solution 1 (easy to digest) by Power Solve

https://www.youtube.com/watch?v=HE7JjZQ6xCk

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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