S denotes the sum of the digits and P denotes the product of the nonzero digits.

It is observed that for the year 2016, S = 9 and P = 12, so that:

S:P = 3:4

Determine all years from 2000 to 3000 such that:

S:P = 3:4

We are looking for a triplet ABC in number 2ABC, denoting the year in question.

After finding the triplet we will list all possible permutations (6,3, or 1).

Wlog we assume A being the lowest digit and A<3, 2*3*3*3=54 & (3/4*54- 2)> 9+9+9.

The candidate years begin: **20**,21**,22****

(+ permutations)

Trying A=0, 1 , 2 in:

(2+A+B+C)/2ABC=3/4 OR

2(2+A+B+C)=3*A*B*C

We get:

**20: 2004, 2022, 2040, 2106, 2160, 2202, 2220, 2400, 2601, 2610**

** ****21: 2118, 2181, 2811.**

**22: no ans.**

*Edited on ***August 23, 2016, 10:45 am**