In this alphametic equation, each of the small letters in bold denotes a different base b digit from 0 to b-1. Neither t nor n can be zero.
(ten)*(ten) - ten = ninety
Determine the minimum value of b, for which there exists at least one solution to the above equation.
The next base with a solution is base 66. T=60, E=51, N=55, I=64 and Y=0. In base 10, TEN=264781 and NINETY=70108713180.
I used the following in UBASIC, and weeded out a few bad solutions as I did not put in vigorous checking.
10 B_min=5
11 B_max=11
12 for B=B_min to B_max
20 for T=ceil(sqrt(B)) to B-1
30 N_min=floor(T*T/B)
31 for N=N_min to N_min+2
32 if N=B then 97
33 if N=T then 97
39 ' print B,T,N
40 for E=0 to B-1
41 if E=N then 96
42 if E=T then 96
50 Ten=B*(B*T+E)+N
51 Noneto=B*(B*(B*(B*B+1)*N+E)+T)
52 Ioooy=Ten*(Ten-1)-Noneto
55 Y=Ioooy@B
56 I=floor(Ioooy/(B*B*B*B))
60 Zero=Ioooy-Y-I*B*B*B*B
70 if Zero=0 then print B,N;E;T;I;Y,Ten;Noneto+Ioooy
96 next E
97 next N
98 next T
99 next B
Next larger solution
