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

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==Similar solution==
 
==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. <math>55+50= \boxed{\text{(C)\ 105}}</math>.
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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 50. <math>55+50= \boxed{\text{(C)\ 105}}</math>.
  
 
~Bepin999
 
~Bepin999

Revision as of 21:57, 29 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

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