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

165_{7}

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.