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, two 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?

Guest Jul 7, 2022

#1**+8 **

Solution by HELPMEEEEEEEEEEEEE:

The smallest number that satisfies both conditions is 28, so when divided among 7 people, there are 0 coins left.

To find the smallest number you'll need to know something called modular arithmetic which is something I cannot explain so you may have to look that up yourself.

nerdiest Jul 7, 2022

#3**0 **

*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, two 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?*

Let's try it with 22 coins.

22/6 = 3 with 4 left over

22/5 = 4 with 2 left over

Okay, so that part is good.

Now 22/7 = 3 with **1** left over.

_{.}

Guest Jul 7, 2022