MATH SOLVE

3 months ago

Q:
# paula and ricardo are serving cupcakes at a school party. if they arrange the cupcakes in groups of 2,3,4,5 or 6, they always have exactly one cupcake leftover. what is the smallest number of cupcakes they could have?

Accepted Solution

A:

Find the smallest number that is divisible by 2, 3, 4, 5, 6 and add 1.

We need the least common multiple of 2, 3, 4, 5, 6.

2 = 2

3 = 3

4 = 2^2

5 = 5

6 = 2 * 3

LCM = product of common and not common prime factors with larger exponent.

LCM = 2^2 * 3 * 5 = 4 * 3 * 5 = 60

To always have a remainder of 1, you need of add 1 to 60.

The number is 61.

Check:

61/2 = 30 remainder 1

61/3 = 20 remainder 1

61/4 = 15 remainder 1

61/5 = 12 remainder 1

61/6 = 10 remainder 1

We need the least common multiple of 2, 3, 4, 5, 6.

2 = 2

3 = 3

4 = 2^2

5 = 5

6 = 2 * 3

LCM = product of common and not common prime factors with larger exponent.

LCM = 2^2 * 3 * 5 = 4 * 3 * 5 = 60

To always have a remainder of 1, you need of add 1 to 60.

The number is 61.

Check:

61/2 = 30 remainder 1

61/3 = 20 remainder 1

61/4 = 15 remainder 1

61/5 = 12 remainder 1

61/6 = 10 remainder 1