The following puzzle has just one solution.

  TOUGH
+ TOUGH 
 ------
 PUZZLE

Can you find it?

... and this one has two solutions.

  TOUGH
  TOUGH
+ TOUGH 
 ------
 PUZZLE

Can you find them?

We see our clues are at "U" and "Z" since both digits are present in the sum and the… uh… the other side of the equation.

T cannot be greater than 9 => P must be 1 (unless it is 0, which would just be rude)

if U < 5 then  2u + 10 = 2o
  therefore if U < 5 then O >= 5 and vice versa

if U >= 5 then  2u = 2o + 10 + 1
  therefore if U >= 5 then G >=5 and vice versa

if u = 0 then z = 0 => u <> 0

u <> 1  (P = 1)

if u = 2, then o = 7 and  t + t + 1 = 12.  t cannot be an integer in this case, thefore, u <> 2

if u = 3, then o = 6 and  t + t + 1 = 13 => t = 6, contradicting o =6.  u <> 3

if u = 4, then o = 8 and  t + t + 1 = 14.  t cannot be an integer in this case, thefore, u <> 4

if u = 5, then g >= 5 and z = 1, contradicting p = 1.  u <> 5

if u = 6, then o = 1, contradicting p = 1.  u<> 6

if u = 7, then o = 2 and  t + t = 17.  t cannot be an integer in this case, u <> 7

if u = 9, then o = 5 and t + t = 19, t cannot be an integer in this case, u <> 9

if u = 8, then o = 3, z = 7, t = 9.  g may be 5 or 6 but trial and error gives us 5

93852 + 93852 = 1877014

Posted by Ryan on 2007-08-09 19:10:35