when asked how many pennies she had a collector said If I arrange them in stacks of five, none are leftover if I arrange them in stacks of six, none are leftover if I arrange them in stacks of seven one is left over. What is the least number of coins she could have
To find the least number of coins the collector could have, we need to find a number that satisfies all the given conditions: when arranging the coins in stacks of five, six, and seven.
Let's start by finding a number that is divisible by 5, leaves no remainder when divided by 6, and leaves a remainder of 1 when divided by 7.
First, let's find a number that is divisible by 5 and leaves no remainder when divided by 6. We can do this by finding the least common multiple (LCM) of 5 and 6, which is 30.
Now, we need to find a number that satisfies the third condition: leaves a remainder of 1 when divided by 7. We can do this by adding 1 to the LCM of 5 and 6 (30 + 1 = 31).
Therefore, the least number of coins the collector could have is 31.
To verify this, we can check the conditions:
- When arranging 31 coins in stacks of five, we get 6 stacks with no coins leftover.
- When arranging 31 coins in stacks of six, we also get 5 stacks with no coins leftover.
- When arranging 31 coins in stacks of seven, we get 4 stacks with one coin leftover.
So, the answer is 31 coins.
5 and 6 are factors, but 7 is not.
Explore multiples of 30.
she has 30 coins it can be divide by 5 and 6 and 7 but two will be left over
the least number of coins she could have would be 28
since it only two number away from 30