2021 Fall AMC 10B Problems/Problem 1: Difference between revisions
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== Solution 1 == | == Solution 1 == | ||
We see that <math>1, 2, 3,</math> and <math>4</math> each appear in the ones, tens, hundreds, and thousands digit exactly once. Since <math>1+2+3+4=10</math>, | We see that <math>1, 2, 3,</math> and <math>4</math> each appear in the ones, tens, hundreds, and thousands digit exactly once. Since <math>1+2+3+4=10</math>, we find that the sum is equal to <cmath>10\cdot(1+10+100+1000)=\boxed{(\textbf{E})11,110}.</cmath> | ||
we find that the sum is equal to < | |||
Revision as of 20:55, 22 November 2021
Problem
What is the value of 
Solution 1
We see that 
and 
each appear in the ones, tens, hundreds, and thousands digit exactly once. Since 
, we find that the sum is equal to ![\[10\cdot(1+10+100+1000)=\boxed{(\textbf{E})11,110}.\]](http://latex.artofproblemsolving.com/1/3/7/137d953cbcd2b520d1a0ccf3a8e5dfebcdf71f8f.png)
Note: it is equally valid to manually add all 4 numbers together to get the answer.
~kingofpineapplz
