2024 AMC 8 Problems/Problem 1

Revision as of 09:19, 12 March 2024 by Mc ade (talk | contribs) (Solution 2)

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 (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 $/boxed{B]$ (Error compiling LaTeX. Unknown error_msg)

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

See Also

2024 AMC 8 (ProblemsAnswer KeyResources)
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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