2024 AMC 8 Problems/Problem 1

==Problem==uad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

==Problem==uad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8 $==Solution 1== + ==Problem==uad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$ −

− We can rewrite the expression as $222,222-(22,222+2,222+222+22+2).$

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− 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 $0$ .

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− 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.

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− i am smart

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− ~ Dreamer12

Solution 2 (boring)

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

~iasdjfpawregh ~hockey

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 $\boxed{(B)}$

- FU-King nice

Video Solution 1 (Quick and Easy!)

https://youtu.be/Ol1seWX0xHY

~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

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AJHSME/AMC 8 Problems and Solutions

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