從老頭的網誌上看來的數學謎題。
我試解了一下,覺得很有趣,貼在這邊與大家分享。
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A very interesting rillde told by a friend.

forty + ten + ten = sixty

Each of the character in the above equation represents a digit from0 to 9. Find out which character is which numerical digit. Notethat there is only one solution to this puzzle.

Rewrite the equation in the following way so that you can see itclearer.

 forty
   ten
+  ten
------
 sixty

You can start solving the puzzle now. If you still don't understandthe rule of the riddle, keep on reading, and I 'll show you anexample.

For example, from the right most side of the equation, we know thaty+n+n=y, or maybe y+n+n=y+10. Hence, we can say thatn is either 0 or 5.

Alright, that's all. Go on finish the rest. If you know thesolution, just response and let me know that you have the answer.Please don't give us your answer in your response, let others havesome fun.
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可不要被上面的小提示騙到了,以為這個謎題很簡單
其實還挺花腦筋的喲!

Consider the carry-out and rewrite theequation:

 C4 C3 C2 C10
 f o r t y
     t e n
+    t e n
--------
 s i x t y
  1. (y + n + n = y + c1*10)
    → (n = 0, c1 = 0), or (n = 5, c1 = 1)

  2. if (n = 5, c1 = 1), then
    (1 + t + e + e = t + c2*10)
    → No possible solution for "e"!
    if (n = 0, c1 = 0), then

    (t + e + e = t + c2*10)
    → since "0" has been used fot "n", so we have (e = 5, c2 = 1)
    (n = 0, e = 5)

  3. (o + c3 = i + c4*10), and
    (f + c4 = s)
    "c3" might be 1 or 2 since summation of three digits cannot exceed 27.
    "c4" must be 1 since summation of two digits cannot exceed 18.
    so we have
    (f + 1 = s).........................................(1)

  4. if (c3 = 1), then "o" must be "9" to make (c4 = 1).
    → but that will make "i" equal to "0", which is occupied by "n"!!
    → so we have
    (c3 = 2) and (o = 9, i = 1)

  5. Now there are 6 number remained: (2, 3, 4, 6, 7, 8).
    since (c2 + r + t + t = x + c3*10) and (c2 = 1, c3 = 2)
    (1 + r + t + t = x + 20)..........................(2)
    → both "t" and "r" must be greater than 6 to make (c3 = 2).

    if (t = 7), then equation(2) becomes (r + 15 = x + 20)
    → "r" must be 8 to make (c3 = 2), but
      in that case, eqation(1) cannot be satidied.
    → Wrong answer!

    if (t = 8), then equation(2) becomes (r + 17 = x + 20)
    → "r" can be 6 or 7

     → if (r = 6, x = 3), eqation(1) cannot be satisfied.
     → Wrong answer!

     → the last possible solution is (r = 7, x = 4)
     → from equation(1), (f = 2, s = 3)
     → and the last symbol is (y = 6)
     → Correct answer!

    → The original equation is: 29786 + 850 + 850 = 31486

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