Answer: 192

Finding common factors and products

Factors of 15 and 21 will also be factors of the number

Products of the prime factors will also be factors of the number

Image

 

The list already has 8 numbers in it so the number is 105.

1 + 3 + 5 + 7 + 15 + 21 + 35 + 135 = 192

Considering numbers which have 15 and 21 as factors

15 and 21 are both factors of the number we are looking for.

The first few numbers in the 15 times table are 15, 30, 45, 60, 75, 90, 105, 120, ...

The first few numbers in the 21 times table are 21, 42, 63, 84, 105, 126, 147, 168, ...

So 105 is the smallest number that has both 15 and 21 as factors.

The factors of 105 are 1 and 105, 3 and 35, 5 and 21, and 7 and 15. That gives 8 factors, so 105 is the number we were looking for.

The sum of all the factors of 105 is 1 + 105 + 3 + 35 + 5 + 21 + 7 + 15 = 192.

Considering what the other factors of the number must be

15 = 3$\times$5, so 3 and 5 must also be factors of the number.

21 = 3$\times$7, so 7 must also be a factor of the number.

Since 5 and 7 are both factors of the number, 5$\times$7 = 35 must also be a factor of the number.

So 3, 5, 7, 15, 21, 35, 1 and the number itself must all be factors of the number - which makes 8 factors. 

So the number must be 3$\times$5$\times$7 = 105, and the sum of all of the factors is 1 + 3 + 5 + 7 + 15 + 21 + 35 + 105 = 192.

Considering the lowest common multiple of 15 and 21

15 and 21 are both factors of the number we are looking for, so the lowest common multiple of 15 and 21 must also be a factor of the number we are looking for.

15 = 3$\times$5 and 21 = 3$\times$7, so the lowest common muliple of 15 and 21 is 3$\times$5$\times$7 = 105.

The factors of 105 are 1 and 105, 3 and 35, 5 and 21, and 7 and 15.

That gives 8 factors, so 105 is the number we were looking for (all other numbers in the 105 times table are larger than 105, and have themself as a factor, so have more than 8 factors).

The sum of all the factors of 105 is 1 + 105 + 3 + 35 + 5 + 21 + 7 + 15 = 192.