An alphametic puzzle about Jules Verne

Question

Jules Verne was born on February 8,1828.

https://en.wikipedia.org/wiki/Jules_Verne

Answer 1

We see that V/8 = J, with a remainder. There are no zeroes shown after the second subtraction, so I assume that the remainder must be non-zero. This means V = 9 and J = 1.

Next, 1E/8 = U and leaves no remainder. Hence 1E = 16 (so E = 6) and U = 2.

It seems we cannot subtract a multiple of 8 from R but we can from RN. Hence R must be less than 8 and L is 0, so R = 3, 4, 5, or 7. Also, RN/8 in particular is 6 with a remainder, so ?? = 6 x 8 = 48 and 48 < RN < 56. Hence R = 4 or 5, but RN =/= 49 since that would imply 9 = N = V. So R = 5. Moreover, N = 3, 4, 5 (not 7 because RN cannot be 47 or 57 based on our bounds), and since R = 5, N = 3 or 4.

Finally we look at the bottom subtraction. ?4 must be a multiple of 8, meaning ?4 = 64. But that implies the minuend is ?6 = 66 and S = 8. The former implication makes 5N - 48 = 6, so 5N = 54 (so N = 4).



To double-check, 12068 x 8 = 96544, so 96546/8 will indeed give a remainder of 2.

In conclusion, L = 0, J = 1, U = 2, N = 4, R = 5, E = 6, S = 8, and V = 9.