Difference between revisions of "2021 Fall AMC 12A Problems/Problem 4"
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<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9</math> | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9</math> | ||
| − | == Solution == | + | == Solution 1 == |
By divisibility rules, when <math>A=1,</math> the number <math>202101</math> is divisible by <math>3.</math> When <math>A=3,</math> the number <math>202103</math> is divisible by <math>11.</math> When <math>A=5,</math> the number <math>202105</math> is divisible by <math>5.</math> When <math>A=7,</math> the number <math>202107</math> is divisible by <math>3.</math> Thus, by the process of elimination we have that the answer is <math>\boxed{\textbf{(E)}\ 9}.</math> | By divisibility rules, when <math>A=1,</math> the number <math>202101</math> is divisible by <math>3.</math> When <math>A=3,</math> the number <math>202103</math> is divisible by <math>11.</math> When <math>A=5,</math> the number <math>202105</math> is divisible by <math>5.</math> When <math>A=7,</math> the number <math>202107</math> is divisible by <math>3.</math> Thus, by the process of elimination we have that the answer is <math>\boxed{\textbf{(E)}\ 9}.</math> | ||
~NH14 | ~NH14 | ||
| + | |||
| + | == Solution 2 == | ||
| + | First, modulo 2 or 5, <math>\underline{20210A} \equiv A</math>. | ||
| + | Hence, <math>A \neq 0, 2, 4, 5, 6, 8</math>. | ||
| + | |||
| + | Second modulo 3, <math>\underline{20210A} \equiv 2 + 0 + 2 + 1 + 0 + A \equiv 5 + A</math>. | ||
| + | Hence, <math>A \neq 1, 4, 7</math>. | ||
| + | |||
| + | Third, modulo 11, <math>\underline{20210A} \equiv A + 1 + 0 - 0 - 2 - 2 \equiv A - 3</math>. | ||
| + | Hence, <math>A \neq 3</math>. | ||
| + | |||
| + | Therefore, the answer is <math>\boxed{\textbf{(E) }9}</math>. | ||
| + | |||
| + | ~Steven Chen (www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
Revision as of 20:59, 25 November 2021
- The following problem is from both the 2021 Fall AMC 10A #5 and 2021 Fall AMC 12A #4, so both problems redirect to this page.
Contents
Problem
The six-digit number 
is prime for only one digit 
What is 
Solution 1
By divisibility rules, when 
the number 
is divisible by 
When 
the number 
is divisible by 
When 
the number 
is divisible by 
When 
the number 
is divisible by Thus, by the process of elimination we have that the answer is 
~NH14
Solution 2
First, modulo 2 or 5, 
.
Hence, 
.
Second modulo 3, 
.
Hence, 
.
Third, modulo 11, 
.
Hence, 
.
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
See Also
| 2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 | |
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
