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!
3 answers
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...
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.
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.
Now here is another for you to work:
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