2006 AMC 8 Problems/Problem 23
Contents
Problem
A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?
Solution
Solution 1
The counting numbers that leave a remainder of when divided by are The counting numbers that leave a remainder of when divided by are So is the smallest possible number of coins that meets both conditions. Because , there are coins left when they are divided among seven people.
Solution 2
If there were two more coins in the box, the number of coins would be divisible by both and . The smallest number that is divisible by and is , so the smallest possible number of coins in the box is and the remainder when divided by is .
Solution 3
We can set up a system of modular congruencies: We can use the division algorithm to say . If we plug the division algorithm in again, we get . This means that , which means that . From this, we can see that is our smallest possible integer satisfying . , making our remainder . This means that there are coins left over when equally divided amongst people.
~Champion1234
Video Solution
https://youtu.be/g1PLxYVZE_U -Happytwin
https://www.youtube.com/watch?v=uMBev3FUoTs
See Also
2006 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.