Difference between revisions of "2021 Fall AMC 10B Problems/Problem 1"
(→Problem) |
(→Solution 1) |
||
| Line 5: | Line 5: | ||
== Solution 1 == | == Solution 1 == | ||
| − | We see that <math>1, 2, 3,</math> and <math>4</math> each appear | + | 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 <math>10\cdot(1+10+100+1000)=\boxed{(E)11,110}</math> | we find that the sum is equal to <math>10\cdot(1+10+100+1000)=\boxed{(E)11,110}</math> | ||
~kingofpineapplz | ~kingofpineapplz | ||
Revision as of 20:32, 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 
~kingofpineapplz
