Difference between revisions of "2006 AMC 8 Problems/Problem 23"
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==Problem== | ==Problem== | ||
| − | A box contains gold coins. If the coins are equally divided among | + | 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? |
| − | six people, four coins are left over. If the coins are equally divided | + | |
| − | among | + | <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 </math> |
| − | smallest number of coins that meets these two conditions, how | ||
| − | many coins are left when equally divided among seven people? | ||
| − | (A) 0 (B) 1 (C) 2 (D) 3 (E) 5 | ||
==Solution== | ==Solution== | ||
| − | + | ===Solution 1=== | |
| − | + | The counting numbers that leave a remainder of <math>4</math> when divided by <math>6</math> are | |
| − | 4 | + | <math>4, 10, 16, 22, 28, 34, \cdots</math> The counting numbers that leave a remainder of <math>3</math> when |
| − | divided by 5 are 3 | + | divided by <math>5</math> are <math>3,8,13,18,23,28,33, \cdots</math> So <math>28</math> is the smallest possible number |
| − | of coins that meets both conditions. Because | + | of coins that meets both conditions. Because 28 is divisible by 7, there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left |
when they are divided among seven people. | 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 | If there were two more coins in the box, the number of coins would be divisible | ||
| − | by both 6 and 5. The smallest number that is divisible by 6 and 5 is 30, so the | + | by both <math>6</math> and <math>5</math>. The smallest number that is divisible by <math>6</math> and <math>5</math> is <math>30</math>, so the |
| − | smallest possible number of coins in the box is 28. | + | 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== | ||
| + | https://www.youtube.com/watch?v=dQw4w9WgXcQ | ||
| + | -Happytwin | ||
| + | |||
| + | https://www.youtube.com/watch?v=uMBev3FUoTs ~David | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/-GteVuETb14 | ||
| + | ==See Also== | ||
{{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}} | ||
Latest revision as of 17:10, 8 November 2024
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 28 is divisible by 7, 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 .
Video Solution
Video Solution
https://www.youtube.com/watch?v=dQw4w9WgXcQ -Happytwin
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
