Difference between revisions of "2021 Fall AMC 12B Problems/Problem 1"
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== Solution 3== | == Solution 3== | ||
| − | We see that the units digit must be <math>0</math>, since <math>4+3+2+1</math> is <math>0</math>. But every digit from there, will be a <math>1</math> since we have that each time afterwards, we must carry the 1. The answer choice that satisfies these conditions is <math>\boxed{\textbf{(E) }11,110}</math>.~~stjwyl | + | We see that the units digit must be <math>0</math>, since <math>4+3+2+1</math> is <math>0</math>. But every digit from there, will be a <math>1</math> since we have that each time afterwards, we must carry the 1 from the previous sum. The answer choice that satisfies these conditions is <math>\boxed{\textbf{(E) }11,110}</math>. |
| + | ~~stjwyl | ||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/p9_RH4s-kBA | https://youtu.be/p9_RH4s-kBA | ||
Revision as of 13:44, 4 December 2021
- The following problem is from both the 2021 Fall AMC 10B #1 and 2021 Fall AMC 12B #1, so both problems redirect to this page.
Contents
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
Solution 2
We have
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
Solution 3
We see that the units digit must be 
, since 
is . But every digit from there, will be a 
since we have that each time afterwards, we must carry the 1 from the previous sum. The answer choice that satisfies these conditions is .
~~stjwyl
Video Solution by Interstigation
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
