2021 Fall AMC 10B Problems/Problem 1: Difference between revisions
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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 <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> | ||
Note: it is equally valid to manually add all 4 numbers together to get the answer. | |||
~kingofpineapplz | ~kingofpineapplz | ||
Revision as of 20:37, 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 
Note: it is equally valid to manually add all 4 numbers together to get the answer.
~kingofpineapplz
