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

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Note that for no two numbers to differ by <math>10</math>, every number chosen must have a different units digit. To make computations easier, we can choose <math>(1, 2, 3, 4, 5)</math> from the first group and <math>(16, 17, 18, 19, 20)</math> from the second group. Then the sum evaluates to <math>1+2+3+4+5+16+17+18+19+20 = \boxed{\text{(C)\ 105}}</math>.
 
Note that for no two numbers to differ by <math>10</math>, every number chosen must have a different units digit. To make computations easier, we can choose <math>(1, 2, 3, 4, 5)</math> from the first group and <math>(16, 17, 18, 19, 20)</math> from the second group. Then the sum evaluates to <math>1+2+3+4+5+16+17+18+19+20 = \boxed{\text{(C)\ 105}}</math>.
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~Soupboy0

Revision as of 20:48, 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