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Note that the fraction IN.VERVEINVERVEIN...has a period of 7, so its multiplicative order is of the form kx+1=10^7, 9999 less than k less than 100000, giving {k = 239, x = 41841}, {k = 717, x = 13947}, {k = 2151, x = 4649}.
13947 has only the multiple 41841 under 100,000 with the correct sequence of letters, and 2151*5 is already over 10,000, too big for LENS. Hence x=41841, so S=4,A=1,L=8.
Now for LENS we have 8EN4, giving a repeating decimal aN.bEcbE*. Since the counterpart k of 41841 was 239, we know that (8000+100E+10N+4)=239y, -1 less than E less than 10, -1 less than N less than 10, with E=6, N=0, giving the decimal 20.5635620 recurring (8604/41841=956/4649) again with the correct sequence of letters, so we need not look further afield for a solution.
Hence 0=N, 1=A, 2=I, 3=R, 4=S, 5=V, 6=E, 7=?, 8=L, 9=?, with 2 candidates for C. INEZ breaks the deadlock since 2067 is clearly divisible by 3, meaning that 2069 is prime. So 9=Z, meaning that 7=C.
Hence CARNAVAL= 71301518.
Notes: Adapted from a similar New Scientist puzzle.
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