the sum of the integers from 0 to b-1 is b(b-1)/2
so, b(b-1)/2 = 3b+4
b^2-b = 6b+8
b^2-7b-8 = 0
(b-8)(b+1) = 0
b=8 since b>1
check:
sum(0..7) = 28 = 3*8+3
Vinny wrote down all the single-digit base-b numbers and added them in base b, getting 34_b.
What is b?
1 answer