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Question 252284: Each letter in the problem to the right represents a different digit
All digits in NINE are odd. What is the smallest possible value of NINE?
FOUR
+FIVE
_________
NINE
a. 3135 b. 3537 c. 5153 d. 5351 e. 7173
Found 2 solutions by drk, Edwin McCravy:
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! It appears that 5153 will be the smallest.
Step 1: start with F + F. This creates and even number, so we add 1 to get an odd. That means that O + I is 10 + I.
step 2: In order to get O + I = 10 + 1 there needed to be a carry over from u + v.
step 3: R + E = E, this means that r = 0.
we have
(i) R + E = E - -> 0 + E = E
(ii) U + V = 10 + N
(iii)O + I + 1 = 10 + I - - > 9 + I + 1 = 10 + I
(iv) F + F = N - - > 1 + F + F = N
Let F = 2. N = 5.
U + V = 15; 8 + 7 = 15. I want to use numbers that don't over lap.
1 + 9 + I = 10 + 1 - -> I = 1.
N=5, I = 1, N = 5, E must be 3 to answer our question.
So, we get 5153.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
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
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