EIGHT is not 8 (ii)

(1.) $E > I > G > H > 0$ and $T > 5$ , because a remainder is less than the divisor.

(2.) $T$ is odd. If $T$ were even, the number would be even, and dividing by an even number would leave an even remainder instead of 5.

(3.) $T = 7$ or $T = 9$ . (From 1. and 2.)

(4.) $H = 1$ or $H = 3$ . From 1. we know that $H \leq 6$ . We also rule out 5 and even values because the number does not end in 0, 5, or an even digit.

(5.) $G \not= 5$ . If it were, then the remainder in the third equation should be 2 or 4 (from 3.); this contradicts our conclusion in 4.

(6.) If $E$ is even, $I$ must be odd. If $I$ is even, $G$ must be odd. (Dividing an odd number by an even divisor results in an odd remainder.)

Call $m = \overline{EIG}$ .

Case 1. $HT = 39$

According to the fourth equation, the number is a multiple of 3. But then the remainder after division by 9 should be 0, 3, or 6; not 5. Contradiction.

Case 2. $HT = 37$

In this case $100m + 37 \equiv 0$ mod 3 and $\equiv 5$ mod 7. This implies $m \equiv 5$ mod 21. This may be reduced to $-5E + 10I + G \equiv 5$ mod 21.

Write $G = x + 4, I = x + y + 5, E = x + y + z + 6$ . We need $0 \leq x,y,z$ and $x + y + z \leq 3$ ; and $6x + 5y - 5z \equiv 2$ mod 21. There is no solution.

Case 3. $HT = 19$

Now $100m + 19 \equiv 5$ mod 9, which translates to $E + I + G \equiv 4$ mod 9.

Write $G = x + 2, I = x + y + 3, E = x + y + z + 4$ . We need $0 \leq x,y,z$ and $x + y + z \leq 4$ ; and $3x + 2y + z \equiv 4$ mod 9.

The solutions are:

• $x = y = 0, z = 4$ . This means $G = 2, I = 3, E = 8$ . This is the solution.

• $x = z = 0, y = 2$ . Then $G = 2, I = 5, E = 6$ . However, $65219\div 5$ has remainder 4, not 2.

• $x = 0, y = 1, z = 2$ . Then $G = 2, I = 4, E = 7$ . However, since $I$ is even, $G$ must be odd. (See 6. above.)

• $x = z = 1, y = 0$ . Then $G = 3, I = 4, E = 6$ . However, since $E$ is even, $I$ must be odd.

Case 4. $HT = 17$

Now $100m + 17 \equiv 5$ mod 7, which translates to $-3E - I + 2G \equiv 2$ mod 7.

Write $G = x + 2, I = x + y + 3, E = x + y + z + 4$ . We need $0 \leq x,y,z$ and $x + y + z \leq 5$ ; and $-2x + y + 2z \equiv 5\ (\text{or}\ -4)$ mod 9.

The solutions are:

• $x = 0, y = 1, z = 2$ . This means $G = 2, I = 4, E = 7$ . However, we already have $T = 7$ .

• $x = 0, y = 3, z = 1$ . Then $G = 2, I = 6, E = 8$ . But if $E$ is even, $I$ must be odd.

• $x = 0, y = 5, z = 0$ . Then $G = 2, I = 8, E = 9$ . But if $I$ is even, $G$ must be odd.

• $x = 1, y = 1, z = 3$ . Then $G = 3, I = 5, E = 9$ . But $95317\div 5$ has remainder 2, not 3.

• $x = 2, y = 0, z = 0$ . Then $G = 3, I = 4, E = 5$ . But $54317\div 5$ has remainder 2, not 4.

• $x = 3, y = 0, z = 1$ . Then $G = 4, I = 5, E = 7$ . However, we already have $T = 7$ .

• $x = 3, y = 2, z = 0$ . Then $G = 4, I = 7, E = 8$ . However, we already have $T = 7$ .

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Therefore we conclude that the only solution is $\boxed{EIGHT = 83219}$ .