2025 AMC 8 Problems/Problem 16: Difference between revisions
| Line 11: | Line 11: | ||
==Similar solution== | ==Similar solution== | ||
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}}$ | 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}}$ | ||
~Bepin999 | ~Bepin999 | ||
Revision as of 21:54, 29 January 2025
Five distinct integers from 
to 
are chosen, and five distinct integers from 
to 
are chosen. No two numbers differ by exactly . What is the sum of the ten chosen numbers?
Solution
Note that for no two numbers to differ by , every number chosen must have a different units digit. To make computations easier, we can choose 
from the first group and 
from the second group. Then the sum evaluates to 
.
~Soupboy0
Similar solution
One efficient method is to quickly add 
, which is 55. Then because you took 50 in total away from , you add 50. 55+50= \boxed{\text{(C)\ 105}}$
~Bepin999
