Question 252284
<pre><font size = 4 color = "indigo"><b>
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</pre>