Difference between revisions of "2021 Fall AMC 12B Problems/Problem 1"
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| + | {{duplicate|[[2021 Fall AMC 10B Problems/Problem 1|2021 Fall AMC 10B #1]] and [[2021 Fall AMC 12B Problems/Problem 1|2021 Fall AMC 12B #1]]}} | ||
| + | |||
==Problem== | ==Problem== | ||
| − | What is the value of <math>1234+2341+3412+4123 | + | What is the value of <math>1234 + 2341 + 3412 + 4123</math> |
| − | <math> | + | <math>\textbf{(A)}\: 10{,}000\qquad\textbf{(B)} \: 10{,}010\qquad\textbf{(C)} \: 10{,}110\qquad\textbf{(D)} \: 11{,}000\qquad\textbf{(E)} \: 11{,}110</math> |
== 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 find that the sum is equal to <cmath>10\cdot(1+10+100+1000)=\boxed{(\textbf{E})11,110}.</ | + | 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> |
| + | Note that it is equally valid to manually add all four numbers together to get the answer. | ||
| + | |||
| + | ~kingofpineapplz | ||
| + | |||
| + | == Solution 2 == | ||
| + | We have <cmath>1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{\textbf{(E)} \: 11{,}110}.</cmath> | ||
| + | ~Steven Chen (www.professorchenedu.com) | ||
| + | |||
| + | == 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 <math>1</math> from the previous sum. The answer choice that satisfies these conditions is <math>\boxed{\textbf{(E)} \: 11{,}110}</math>. | ||
| + | |||
| + | ~stjwyl | ||
| + | |||
| + | == Solution 4 (Brute Force)== | ||
| + | We can simply add the numbers. <math>1234 + 2341 + 3412 + 4123 = 11110 \implies \boxed{\textbf{(E)}}</math>. | ||
| + | Note: Although this would not take terribly long, it is not recommended to do this in a real contest. | ||
| − | + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | |
| + | ==Video Solution by Interstigation== | ||
| + | https://youtu.be/p9_RH4s-kBA | ||
| − | ~ | + | ==Video Solution== |
| + | https://youtu.be/I9lSM0hO39M | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/3UZHiV65WXU | ||
| + | |||
| + | ~savannahsolver | ||
| + | ==Video Solution by TheBeautyofMath== | ||
| + | For AMC 10: https://youtu.be/lC7naDZ1Eu4 | ||
| + | |||
| + | For AMC 12: https://youtu.be/yaE5aAmeesc | ||
| + | |||
| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
| + | {{AMC10 box|year=2021 Fall|ab=B|before=First Problem|num-a=2}} | ||
{{AMC12 box|year=2021 Fall|ab=B|before=First Problem|num-a=2}} | {{AMC12 box|year=2021 Fall|ab=B|before=First Problem|num-a=2}} | ||
| − | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:48, 30 January 2024
- 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/c/d/9/cd92c0d1d365e60e535571a75b7976384857893e.png)
Note that it is equally valid to manually add all four numbers together to get the answer.
~kingofpineapplz
Solution 2
We have ![\[1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{\textbf{(E)} \: 11{,}110}.\]](http://latex.artofproblemsolving.com/a/b/9/ab9f64db20cec332a367cb90f8cc5f2b812b5d36.png)
~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 from the previous sum. The answer choice that satisfies these conditions is 
.
~stjwyl
Solution 4 (Brute Force)
We can simply add the numbers. 
.
Note: Although this would not take terribly long, it is not recommended to do this in a real contest.
~ cxsmi
Video Solution by Interstigation
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
For AMC 10: https://youtu.be/lC7naDZ1Eu4
For AMC 12: https://youtu.be/yaE5aAmeesc
~IceMatrix
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. 
