Difference between revisions of "2006 AMC 8 Problems/Problem 23"
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<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 </math> | <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 </math> | ||
| − | == | + | ==Soluiton 1== |
| − | ===Solution | + | This is a '''modular arithmetic''' problem. |
| + | |||
| + | Let the number of coins be <math>N</math>. Then: | ||
| + | |||
| + | <cmath> | ||
| + | N \equiv 4 \pmod{6} | ||
| + | </cmath> | ||
| + | |||
| + | <cmath> | ||
| + | N \equiv 3 \pmod{5} | ||
| + | </cmath> | ||
| + | |||
| + | Solving this system, the smallest such <math>N</math> is: | ||
| + | |||
| + | <cmath> | ||
| + | N = 28 | ||
| + | </cmath> | ||
| + | |||
| + | Now, dividing 28 by 7: | ||
| + | |||
| + | <cmath> | ||
| + | 28 \div 7 = 4 \text{ with a remainder of } \boxed{0} | ||
| + | </cmath> | ||
| + | |||
| + | This is the number of coins left when the box is divided equally among seven people. | ||
| + | |||
| + | ==Solution 6== | ||
The counting numbers that leave a remainder of <math>4</math> when divided by <math>6</math> are | The counting numbers that leave a remainder of <math>4</math> when divided by <math>6</math> are | ||
<math>4, 10, 16, 22, 28, 34, \cdots</math> The counting numbers that leave a remainder of <math>3</math> when | <math>4, 10, 16, 22, 28, 34, \cdots</math> The counting numbers that leave a remainder of <math>3</math> when | ||
| Line 13: | Line 39: | ||
when they are divided among seven people. | when they are divided among seven people. | ||
| − | + | ==Solution 3== | |
If there were two more coins in the box, the number of coins would be divisible | If there were two more coins in the box, the number of coins would be divisible | ||
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smallest possible number of coins in the box is <math>28</math> and the remainder when divided by <math>7</math> is <math>\boxed{\textbf{(A)}\ 0}</math>. | smallest possible number of coins in the box is <math>28</math> and the remainder when divided by <math>7</math> is <math>\boxed{\textbf{(A)}\ 0}</math>. | ||
| − | ==Video Solution | + | ==Video Solution 1== |
| − | |||
| − | |||
https://www.youtube.com/watch?v=dQw4w9WgXcQ | https://www.youtube.com/watch?v=dQw4w9WgXcQ | ||
-Happytwin | -Happytwin | ||
| + | ==Video Solution 2== | ||
https://www.youtube.com/watch?v=uMBev3FUoTs ~David | https://www.youtube.com/watch?v=uMBev3FUoTs ~David | ||
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{{AMC8 box|year=2006|n=II|num-b=22|num-a=24}} | {{AMC8 box|year=2006|n=II|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 12:24, 2 September 2025
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?
Soluiton 1
This is a modular arithmetic problem.
Let the number of coins be 
. Then:
![\[N \equiv 4 \pmod{6}\]](http://latex.artofproblemsolving.com/0/9/1/091624b678ca163a9ca3ead18221e89cec6613c6.png)
![\[N \equiv 3 \pmod{5}\]](http://latex.artofproblemsolving.com/1/a/f/1af1ada00d7492743e28af4d5454ca64945e1b8c.png)
Solving this system, the smallest such is:
![\[N = 28\]](http://latex.artofproblemsolving.com/8/0/8/8082f67b0b00a28046b083f50146000c060a8b9f.png)
Now, dividing 28 by 7:
![\[28 \div 7 = 4 \text{ with a remainder of } \boxed{0}\]](http://latex.artofproblemsolving.com/6/d/d/6dd3de064d619ec2d751ee4af9260ba3e42cd295.png)
This is the number of coins left when the box is divided equally among seven people.
Solution 6
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 28 is divisible by 7, there are 
coins left
when they are divided among seven people.
Solution 3
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 .
Video Solution 1
https://www.youtube.com/watch?v=dQw4w9WgXcQ -Happytwin
Video Solution 2
https://www.youtube.com/watch?v=uMBev3FUoTs ~David
Video Solution by WhyMath
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
