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

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==Problem==
 
Five distinct integers from <math>1</math> to <math>10</math> are chosen, and five distinct integers from <math>11</math> to <math>20</math> are chosen. No two numbers differ by exactly <math>10</math>. What is the sum of the ten chosen numbers?
 
Five distinct integers from <math>1</math> to <math>10</math> are chosen, and five distinct integers from <math>11</math> to <math>20</math> are chosen. No two numbers differ by exactly <math>10</math>. What is the sum of the ten chosen numbers?
  
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~Soupboy0
 
~Soupboy0
  
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=Similar solution=
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==Similar solution==
  
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One efficient method is to quickly add <math>(1, 2, 3, 4, 5, 6, 7, 8, 9, and 10)</math>, which is 55. Then because you took 50 in total away from <math>(16, 17, 18, 19, 20)</math>, you add 50. 55+50=\boxed{\text{(C)\ 105}}$.
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One efficient method is to quickly add <math>(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)</math>, which is <math>55</math>. Then because you took <math>50</math> in total away from <math>(16, 17, 18, 19, 20)</math>, you add <math>50</math>. <math>55+50= \boxed{\text{(C)\ 105}}</math>.
  
 
~Bepin999
 
~Bepin999
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==Video Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=jTTcscvcQmI
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==Video Solution by Thinking Feet==
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https://youtu.be/PKMpTS6b988
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==See Also==
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{{AMC8 box|year=2025|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 19:15, 30 January 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

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 1 by SpreadTheMathLove

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

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
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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