In every numeration system (base b) there is a three-digit integer that is (b+1) times the sum of its digits. In base ten the number is 198. In base three the number is 121. Find the number in base seven that has this property
8 * (a + b + c) = 49a + 7b + c
8a + 8b + 8c = 49a + 7b + c
41a - b - 7c = 0
41a - b = 7c
a = 1
b = 6
c = 5
1657
Proof
8 ( 1 + 6 + 5 ) = 1(7)^2 + 6(7) + 5
8 * 12 = 49 + 42 + 5
96 = 49 + 47
96 = 96
This question is really intresting. I solved it myself recently.
