Replace each letter by a digit. The same digit must represent each letter, and no beginning letter of a word can be zero. No two letters can be the same number. Find the digits represented by the letters 'O', 'N', 'E', 'T'.
O N E
O N E
O N E
O N E
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T E N
I have tried this, but I can't seem to crack it. It doesn't make sense to me. Can anyone help with this?
On
If we assume $E$ and $N$ represent different digits, we need $E\ne0$.
$$400O+40N+4E=100T+10E+N$$ Therefore $$ 400O+39N=100T+6E $$ If we reduce modulo $100$, we see that $39N\equiv 6E\pmod{100}$, so $13N\equiv 2E\pmod{100}$. Since the inverse of $13$ modulo $100$ is $77$, we have $N\equiv77\cdot2E\equiv54E\pmod{100}$.
Compute $54x$ modulo $100$ for $1\le x\le9$ to see when the remainder is between $0$ and $9$.
It's only possible for $x=2$. Thus $E=2$ and $N=8$. Then $400O+312=100T+12$, $O=1$ and $T=7$.
On
A slightly different method: EN has to be a multiple of 4 and EN+NE is a multiple of 5 (last two digits match 5×NE), therefore E+N is a multiple of 5. For any choice of E we can pick N forcing the sum to be a multiple of 5 and N to be even, leaving only one possible N for each E. The only such candidates for EN that are also multiples of 4: 00, 28, 32, 64, 96. Of these only EN=00 and EN=28 are consistent with NE×4 ending with EN. Then 00 violates the usual rule that different letters represent different digits and is just plain inelegant (100+...+100=400? Come on!). So E=2 and N=8. We then can't begin ONE with 0 (convention) or greater than 1 (ONE<1000/4=250), so 182+...+182=728 is all there is.
'ONE' represents the number $100 \times O + 10 \times N + E$, just as $781$ represents $7 \times 100 + 8 \times 10 + 1$. This is what the decimal position system is.
So 'ONE' added 4 times to itself is just $400 \times O + 40 \times N + 4E$ and this should represent the same number as 'TEN' = $100 \times T + 10\times E + N$.
This means 'N' is a the final digit of a multiple of $4$, so $2,4,6,8,0$ are the options for 'N'. For more hints, see the comments.