Difference between revisions of "2021 AMC 10A Problems/Problem 11"
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==Solution 3 (Educated Guess)== | ==Solution 3 (Educated Guess)== | ||
− | Note that choices <math>\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}</math> are congruent to <math>0,1,0,1,2</math> modulo <math>3,</math> respectively. Since only one of | + | Note that choices <math>\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}</math> are congruent to <math>0,1,0,1,2</math> modulo <math>3,</math> respectively. Since only one of these choices is correct, we pick <math>\boxed{\textbf{(E)} ~8}</math> due to its uniqueness. |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 10:43, 7 May 2021
Contents
Problem
For which of the following integers is the base- number not divisible by ?
Solution 1
We have which is divisible by unless The only choice congruent to modulo is
~MRENTHUSIASM
Solution 2 (Easy)
Vertically subtracting we see that the ones place becomes 0, and so does the place. Then, we perform a carry (make sure the carry is in base !). Let . Then, we have our final number as
Now, when expanding, we see that this number is simply .
Now, notice that the final number will only be congruent to if either , or if (because note that would become , and would become as well, and therefore the final expression would become . Therefore, must be . Among the answers, only 8 is , and therefore our answer is
- icecreamrolls8
Solution 3 (Educated Guess)
Note that choices are congruent to modulo respectively. Since only one of these choices is correct, we pick due to its uniqueness.
~MRENTHUSIASM
Video Solution (Simple and Quick)
~ Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=XBfRVYx64dA&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=10
~North America Math Contest Go Go Go
Video Solution 3
~savannahsolver
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.