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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{(A) } 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>.
  
 
~iasdjfpawregh
 
~iasdjfpawregh
−
~vockey(removed the grief)
+
~vockey
  
 
==Solution 4==
 
==Solution 4==

Revision as of 21:03, 3 February 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.

i am smart

~ Dreamer1297

Solution 2(Tedious)

Using Arun Thereom, we deduce that the answer is (Z)

Note that this solution is not recommended to use during the actual exam. A lot of students this year had implemented this solution and lost a significant amount of time. $\newline$ ~ nikhil ~ CXP ~ Nivaar

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

~iasdjfpawregh ~vockey

Solution 4

Let $S$ be equal to the expression at hand. We reduce each term modulo $10$ to find the units digit of each term in the expression, and thus the units digit of the entire thing:

\[S\equiv 2 - 2 - 2 - 2- 2- 2 \equiv -8 \equiv -8 + 10\equiv \boxed{\textbf{(B) } 2} \pmod{10}.\]

-Benedict T (countmath1)


Solution 5

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 $\boxed{(B)}$

- U-King

Solution 6(fast)

uwu $\boxed{(uwu)}$

- uwu gamer girl(ꈍᴗꈍ)

Solution 7

2-2=0. Therefore, ones digit is the 10th avacado $\boxed{(F)}$

- iamcalifornia'sresidentidiot

Video Solution 1 (easy to digest) by Power Solve

https://www.youtube.com/watch?v=dQw4w9WgXcQ

Video Solution (easy to understand)

https://youtu.be/BaE00H2SHQM?si=_8lhp8-dzNxZ-eUQ

~Math-X

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=dQw4w9WgXcQ

~Rick Atsley

Video Solution 2 by uwu

https://www.youtube.com/watch?v=dQw4w9WgXcQ

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=dQw4w9WgXcQ

cool solution must see

https://www.youtube.com/watch?v=dQw4w9WgXcQ

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. AMC logo.png

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