Math

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Help please on how to solve the following problem: If E,N,O,T, and W each represent a different integer, can you find more than one solution to this problem?

Thanks in advance!

  • One factor i forgot -

    I don't know if this is imperative but at the top of the problem it says one + one = two .... not sure if that is a hint or just a title so to speak for the problem...

  • Math -

    This is a cryptrythm. Each letter in the one is a numberal, as in 123.

    ONE
    ONE
    ____
    TWO

    Well, O+O is nine or less, when means O is four or less. If O is four, E is 2. O cant be three, because two E's add to it. IF O is 2, E is one.

    Lets try both
    4N2
    4N2
    TW4
    Now W has to be even, and it cant be 2 or 4, they are used. If W is six, N is three, and T is eight.
    W cannot be4 or 2 as they are used. This is solution one.

    Solution two. E=1
    2N1
    2N1
    TW2 Again, W has to be even, it cannot be 4 (why?). Try w=six, then n is three, and T is four. Try w=8, N=4, which cannot be, as T will be 4. Try W=(1)0, or N=5. That makes T=5 also.

    We have two solutions.

  • Math -

    Now here is another for you to work:

    SEND
    +MORE
    -----
    Money

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