Difference between revisions of "2006 AMC 8 Problems/Problem 23"

Latest revision as of 23:24, 8 September 2025

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?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5$

Soluiton 1

This is a modular arithmetic problem.

Let the number of coins be $N$ . Then:

$N \equiv 4 \pmod{6}$

$N \equiv 3 \pmod{5}$

Solving this system, the smallest such $N$ is:

$N = 28$

Now, dividing 28 by 7:

$28 \div 7 = 4 \text{ with a remainder of } \boxed{0}$

This is the number of coins left when the box is divided equally among seven people.

Solution 2

The counting numbers that leave a remainder of $4$ when divided by $6$ are $4, 10, 16, 22, 28, 34, \cdots$ The counting numbers that leave a remainder of $3$ when divided by $5$ are $3,8,13,18,23,28,33, \cdots$ So $28$ is the smallest possible number of coins that meets both conditions. Because 28 is divisible by 7, there are $\boxed{\textbf{(A)}\ 0}$ 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 $6$ and $5$ . The smallest number that is divisible by $6$ and $5$ is $30$ , so the smallest possible number of coins in the box is $28$ and the remainder when divided by $7$ is $\boxed{\textbf{(A)}\ 0}$ .

Video Solution 1

https://www.youtube.com/watch?v=uMBev3FUoTs ~David

Video Solution 2 by WhyMath

https://youtu.be/-GteVuETb14

@@ Line 5: / Line 5: @@
 <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 </math>
-==Solution==
+==Soluiton 1==
-===Solution 1===
+This is a '''modular arithmetic''' problem.
-The counting numbers that leave a remainder of 4 when divided by 6 are
-<math>4, 10, 16, 22, 28, 34, \cdots</math> The counting numbers that leave a remainder of 3 when
+Let the number of coins be <math>N</math>. Then:
-divided by 5 are <math>3,8,13,18,23,28,33, \cdots</math> So 28 is the smallest possible number
-of coins that meets both conditions. Because <math>4\cdot 7 = 28</math>, there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left
+<cmath>
-when they are divided among seven people.
+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>
-===Solution 2===
+Now, dividing 28 by 7:
-If there were two more coins in the box, the number of coins would be divisible
+<cmath>
-by both 6 and 5. The smallest number that is divisible by 6 and 5 is <math>30</math>, so the
+\div 7 = 4 \text{ with a remainder of } \boxed{0}
-smallest possible number of coins in the box is <math>28</math> and the remainder when divided by 7 is <math>\boxed{\textbf{(A)}\ 0}</math>.
+</cmath>
-===Solution 3===
+This is the number of coins left when the box is divided equally among seven people.
-We can set up a system of modular congruencies:
+==Solution 2==
-<cmath>g\equiv 4 \pmod{6}</cmath>
+The counting numbers that leave a remainder of <math>4</math> when divided by <math>6</math> are
-<cmath>g\equiv 3 \pmod{5}</cmath>
+<math>4, 10, 16, 22, 28, 34, \cdots</math> The counting numbers that leave a remainder of <math>3</math> when
-We can use the division algorithm to say <math>g=6n+4</math> <math>\Rightarrow</math> <math>6n\equiv 4 \pmod{5}</math> <math>\Rightarrow</math> <math>n\equiv 4 \pmod{5}</math>. If we plug the division algorithm in again, we get <math>n=5q+4</math>. This means that <math>g=30q+28</math>, which means that <math>g\equiv 28 \pmod{30}</math>. From this, we can see that <math>28</math> is our smallest possible integer satisfying <math>g\equiv 28 \pmod{30}</math>. <math>28</math> <math>\div</math> <math>7=4</math>, making our remainder <math>0</math>. This means that there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left over when equally divided amongst <math>7</math> people.
+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 28 is divisible by 7, there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left
+when they are divided among seven people.
-~Champion1234
+==Solution 3==
+If there were two more coins in the box, the number of coins would be divisible
+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 <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=uMBev3FUoTs   ~David
-https://youtu.be/g1PLxYVZE_U
+==Video Solution 2 by WhyMath==
--Happytwin
+https://youtu.be/-GteVuETb14
 ==See Also==
 {{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}
 {{MAA Notice}}
+[[Category:Introductory Number Theory Problems]]

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

Page

Toolbox

Search