+ P S U L A R A T A N U N E R U T N S S
In the above cryptarithm, different letters stand for different non-negative integers. What is the value of the sum
S A T U R N + U R A N U S + P L A N E T S ?
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I have done exactly like that but it required 4 trials. @Sharky Kesa: Do you have a leaner solution?
f r o m i t e r t o o l s i m p o r t p e r m u t a t i o n s f o r i i n p e r m u t a t i o n s ( ′ 0 1 2 3 4 5 6 7 8 9 ′ ) : \(str = '' . j o i n ( i ) ) [S,A,T,U,R,N,P,L,E,ANYNUMBER]\(= l i s t ( s t r ) ) i f i n t ( S + A + T + U + R + N ) + i n t ( U + R + A + N + U + S ) = = i n t ( P + L + A + N + E + T + S ) : p r i n t ( i n t ( S + A + T + U + R + N ) + i n t ( U + R + A + N + U + S ) + i n t ( P + L + A + N + E + T + S ) )
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if n + s = s then n = 0 ; r + u = t + ? with ? = 0 or ? = 10 ; if u + n = e then ? = 10 because n = 0 ; r + u = 10 + t ; u + 1 = e ; t + a = 10 ; if a + r + 1 = 10 + a then r = 9 ; s + u + 1 = l + 10 ; s + u = l + 9 ; p = 1.;
r + u = 10 + t and r = 9 then u = 1 + t ;
t + 1 = u ; u + 1 = e ; t + a = 10 ; s + u = l + 9 ; n = 0 ; p = 1 ; r = 9 ;
t, u and e are 3 consecutive numbers, t =/= 5 otherwise t = a ; t can be 2, 3, 4 or 6 ;
Trial and error:
if t = 2, u = 3, e = 4, a = 8 ; s = l + 6, possible values for l are 5, 6 or 7 but then s would be greater than 10 so t =/= 2 ;
if t = 3, u = 4, e = 5, a =7 ; s = l + 5, values for l are 2, 6, 8, for 6 and 8 s would be greater than 10, for l = 2 s = a then t =/= 3 ;
if t = 4, u = 5, e = 6, a = 6, e = a then t =/= 4 ;
last trial is t = 6: t = 6, u = 7, e = 8, a = 4 s = l + 2, values for l are 2, 3 and 5, l =/= 2 otherwise s = a, l =/= 5 otherwise s = u then l = 3 and s = 5.
Answer: n = 0 p = 1 r = 9 t = 6 u = 7 e = 8 a = 4 l = 3 s = 5
The sum is trivial.