Let a, b, c be the numbers "to carry", which can only be 0 or 1. a b c F O U R + F I V E ------------ N I N E in the leftmost column a=1 because if a=0, then N=2F, making N even, but N is odd. in the 2nd column from the left, b + O + I = 10 + I b + O = 10 b can only be 1, for if b were 0, O would be 10, not a digit. Thus O = 10-1 = 9 Now we have: 1 1 c F 9 U R + F I V E ------------ N I N E From the rightmost column, R + E = E + 10c Subtracting E from both sides, R = 10c c can't be 1, for R would be 10, not a digit, so c=0. Therefore R=0, and we have: 1 1 0 F 9 U 0 + F I V E ------------ N I N E E must be odd and less than 5, so as to carry only 0. So we have all the "carry"'s, and E is either 1 or 3, so we have 1 1 0 1 1 0 F 9 U 0 F 9 U 0 + F I V 1 or + F I V 3 ------------ ------------ N I N 1 N I N 3 The largest odd value N could be is 7, since 9 has been used. But if N were 7, then U+V would have to be 17, requiring U and V to be 9 and 8, but 9 has been used. Therefore N is either 5 or 3. But N can't be 3 because in either case above, that would make F be 1, and 1 has already been used. So N can only be 5. 1 1 0 1 1 0 F 9 U 0 F 9 U 0 + F I V 1 or + F I V 3 ------------ ------------ 5 I 5 1 5 I 5 3 and in both cases, F can only be 2, so now we have 1 1 0 1 1 0 2 9 U 0 2 9 U 0 + 2 I V 1 or + 2 I V 3 ------------ ------------ 5 I 5 1 5 I 5 3 Now the only possibilities for U and V are 8 and 7, but we can't tell which is which. They could go either way, so we have four possibilities: 1 1 0 1 1 0 1 1 0 1 1 0 2 9 7 0 2 9 8 0 2 9 7 0 2 9 8 0 + 2 I 8 1 or + 2 I 7 1 or + 2 I 8 3 or + 2 I 7 3 ------------ ------------ ------------ ------------ 5 I 5 1 5 I 5 1 5 I 5 3 5 I 5 3 All that's left is I, so we can easily fill that in in each case to make the addition correct, since all that's left for I in the the first two cases is 3 and 1 for the last two: 1 1 0 1 1 0 1 1 0 1 1 0 2 9 7 0 2 9 8 0 2 9 7 0 2 9 8 0 + 2 3 8 1 or + 2 3 7 1 or + 2 1 8 3 or + 2 1 7 3 ------------ ------------ ------------ ------------ 5 3 5 1 5 3 5 1 5 1 5 3 5 1 5 3 So those are the only four possibilities, and the smallest value for NINE is 5153 in the last two cases. So the correct choice is c. 5153. Edwin