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
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