Question

10. A box contains gold coins. If the coins are equally divided among six friends, four coins are left over. If the coins are equally divided among five friends, 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 friends?

Answer 1

0 coins are left because total no of coins are 28 in the box

Answer 2

Step-by-step explanation:

Given  A box contains gold coins. If the coins are equally divided among six friends, four coins are left over. If the coins are equally divided among five friends, 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 friends?

• So the gold coins are equally divided among 6 friends and 4 coins are left over. Also if the coins are equally divided among 5 friends 3 coins are left over.
• So the numbers 4,10,16,22,28,34,40…. gives a remainder 4 when divided by 6.
• Similarly the numbers 3,8,13,18,23,28,33,38…..leaves a remainder 3 when divided by 5.
• So in this case the smallest possible coin that meets both conditions is 28 since 4 x 7 = 28
• Therefore there will be 0 coins left when they are divided among seven people.

Reference link will be

https://brainly.in/question/12413519