2021 Fall AMC 12A Problems/Problem 4
- 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
First, modulo or , . Hence, .
Second modulo , . Hence, .
Third, modulo , . Hence, .
Therefore, the answer is .
~NH14 ~Steven Chen (www.professorchenedu.com)
Solution 2 (Elimination)
Any number ending in is divisible by . So we can eliminate option .
If the sum of the digits of a number is divisible by , the number is divisible by . The sum of the digits of this number is . If is divisible by , the number is divisible by . Thus we can eliminate options and .
So the correct option is either or . Let's try dividing the number with some integers.
, where is . Since and are both indivisible by , this does not help us narrow the choices down.
, where is . Since , option would make divisible by . Thus, by elimination, the correct choice must be option .
~ZoBro23
Solution 3
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
divisible by .
This leaves only .
~wamofan
Video Solution (Simple and Quick)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution
~Charles3829
Video Solution by TheBeautyofMath
for AMC 10: https://youtu.be/o98vGHAUYjM?t=623
for AMC 12: https://youtu.be/jY-17W6dA3c?t=392
~IceMatrix
Video Solution
~Lucas
Video Solution
~Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.