Question

A jar contains some number of pennies. When pennies are removed 2, 3, 4, 5, 6 or 8 at a

time, one penny is left over. There are no pennies left over when they are removed 7 at a
time. What is the least number of pennies that could be in the jar?

Answer 1

721.

Explanation:

The total number must be a multiple of 7

One less than the number is the LCM of 2 3 4 5 6 and 8.

The LCM is worked out as follows

2

3

4 = 2*2

5

6 = 2*3

8 = 2*2*2

LCM = 2*2*2* 3*5 = 120

But 121 is not divisible by 7 so it must be a multiple of 120 + 1 which is divisible by 7.

240 + 1 = 241 - not divisible by 7

480 + 1 = 481 - not divisible by 7

720 + 1 = 721

721/7 = 103

Answer 2

Explanation:

Let say n the number of pennies in the jar.

n is a multiple of 7 : n=7*k

n-1 is a multiple of 2,3,4,5,6,8 so a multiple of the lcm= 120

n-1=120*t

n-1=120*t

n=7*k

120*t=7k-1

7k=120*t+1

`k=(120*t+1)/(7)\\k=17*t+(t+1)/(7)\\t=6+7*a\\k=17*(6+7a)+(6+7*a+1)/(7) =102+119*a+1+a=103+120*a\\n=7*k=7*(103+120*a)=721+840a\\\\`

The least number of pennies is 721 (when a=0)

Proof:

n-1=720

720/2=360

720/3=240

720/4=180

720/5=144

720/8=90

721/7=103