## October 22, 2015

### Peculiar Number Puzzle

I just found a number with an interesting property:
When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.

It's not a small number, but it's not really big, either.
Find the smallest number with such property.

Movin said...

Remainder on dividing by any K (K from 2 to 10) is K -1. So the number can be lcm(1,2,3,...10)-1 = 2519

Unknown said...

119

Josh D. said...

I believe the solution is 2,519.

From the first statement, having a remainder of 1 means it is one less than an number divisible by 2.
From the second statement, having a remainder of 2 means it is one less than a number divisible by 3.
From the third statement, having a remainder of 3 means it is one less than an number divisible by 4.
..etc... for the rest of the statements.

This means that it is one less than a number that is divisible by 2,3,4,5,6,7,8,9,10. That means we need the Least-Common-Multiple (LCM) of those nine numbers.

Getting the prime factorization of each of those numbers.
2 = 2^1
3 = 3^1
4 = 2^2
5 = 5^1
6 = 2^1 * 3^1
7 = 7^1
8 = 2^3
9 = 3^2
10 = 2^1 * 5^1

The LCM is the product of the highest exponents for each of those bases.
So the LCM in this case = 2^3 * 3^2 * 5^1 * 7^1 = 8*9*5*7 = 2,520.

Our answer is one less than that number, so our answer is 2,519.

You can do the division with each of those to check that the remainders are correct.

Suvinthra M said...

I go with Josh D's answer and explanation : - 2519.

Jerry Critter said...

Yes, 2519