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2025 AMC 8 Problems/Problem 16: Difference between revisions

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==Vide Solution 1 by SpreadTheMathLove==
==Vide Solution 1 by SpreadTheMathLove==
https://www.youtube.com/watch?v=jTTcscvcQmI
https://www.youtube.com/watch?v=jTTcscvcQmI
==Video Solution by Thinking Feet==
https://youtu.be/PKMpTS6b988

Revision as of 18:59, 30 January 2025

Five distinct integers from $1$ to $10$ are chosen, and five distinct integers from $11$ to $20$ are chosen. No two numbers differ by exactly $10$. What is the sum of the ten chosen numbers?

$\textbf{(A)}\ 95 \qquad \textbf{(B)}\ 100 \qquad \textbf{(C)}\ 105 \qquad \textbf{(D)}\ 110 \qquad \textbf{(E)}\ 115$

Solution

Note that for no two numbers to differ by $10$, every number chosen must have a different units digit. To make computations easier, we can choose $(1, 2, 3, 4, 5)$ from the first group and $(16, 17, 18, 19, 20)$ from the second group. Then the sum evaluates to $1+2+3+4+5+16+17+18+19+20 = \boxed{\text{(C)\ 105}}$.

~Soupboy0

Similar solution

One efficient method is to quickly add $(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)$, which is $55$. Then because you took $50$ in total away from $(16, 17, 18, 19, 20)$, you add $50$. $55+50= \boxed{\text{(C)\ 105}}$.

~Bepin999

Vide Solution 1 by SpreadTheMathLove

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

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

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