Difference between revisions of "2025 AMC 8 Problems/Problem 16"

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~Bepin999
 
~Bepin999
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= Video Solution by Pi Academy =
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https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
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==Video Solution 1 by SpreadTheMathLove==
 
==Video Solution 1 by SpreadTheMathLove==

Revision as of 12:52, 3 February 2025

Problem

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

Another Way To Compute

For $1+2+3+4+5+16+17+18+19+20$, we can add the first term and the last term, which is $21$. If we add the second term and the second-to-last term it is also $21$. There are $5$ pairs that sum to $21$, so the answer is $21 \times 5$ which equals $\boxed{\text{(C)\ 105}}$.

- leafy

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


Video Solution by Pi Academy

https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK


Video Solution 1 by SpreadTheMathLove

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

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=DtG8sG4CK4RrUlVz&t=1815 ~hsnacademy

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

See Also

2025 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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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