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2024 AMC 8 Problems/Problem 1: Difference between revisions

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==Solution 2(Tedious)==
==Solution 2(Tedious)==


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


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.
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.

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 (B)

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

These problems are copyrighted © by the Mathematical Association of America.

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