Each letter in this addition problem stands for a different digit, but, for example, all of the
H’s stand for the same digit. Reconstruct the problem to find the value of THERE.


  1. Answer:
      9641 + 9641 = 19282
    Step-by-step explanation:
    You want to assign a different digit to each of the letters H, A, L, T, E, R such that …


    The T in the most significant digit of the sum must be T = 1, because a carry from the sum of two digits cannot exceed 1. That means E = 2. In order for H+H = TH, we must have …
      H + H + 1 = 10 + H
      H = 9
    In order for there to be a carry into the 4th column (from the right), the value of A must be 5 more than the value of T:
      A = 6
    Since E in the third sum digit is even, there can be no carry from the sum L +L = R. This leaves R an even digit not 0, 2, or 6. It cannot be 4, because that would require L=2, but we already have E=2. This leaves R = 8 and L = 4.
    The sum is …
      HALT +HALT = THERE   ⇒   9641 +9641 = 19282
    The value of THERE is 19282.


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