## $2520$ is the smallest number that can be divided by each of the numbers from $1$ to $10$ without any remainder.

What is the smallest positive number that is evenly divisibledivisible with no remainder by all of the numbers from $1$ to $20$?

### The smallest number that is divisible by all numbers from 1 to N is the least common multiple (LCM) of this range of numbers. So, to solvethis problem, we need to find the LCM for numbers from 1 to 20.

The prime factors of the numbers from 1 to 20 are:

1: 1

2: 2

3: 3

4: 2^2

5: 5

6: 2*3

7: 7

8: 2^3

9: 3^2

10: 2*5

11: 11

12: 2^2*3

13: 13

14: 2*7

15: 3*5

16: 2^4

17: 17

18: 2*3^2

19: 19

20: 2^2*5

To find the least common multiple, you take each prime number and raise it to the highest power that it’s raised to in the list. Then, you multiply all these together.

So you’d get:

2^4 * 3^2 * 5 * 7 * 11 * 13 * 17 * 19

= 16 * 9 * 5 * 7 * 11 * 13 * 17 * 19

= 232,792,560

So, the smallest positive number that is evenly divisible (with no remainder) by all of the numbers from 1 to 20 is 232,792,560.

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