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# Number Game: The Word Addition

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 Consider the summation of letters : $T \ A \ M \ I \ N \ G \\ + \\ H \ E \ L \ M \ E \ T \\ = \\ 9\ 2\ 5\ 7\ 6 \ 4$ In the addition shown above, the sum $925764$ represents a word. We have to find this word — on the conditions given below: Each letter represents a different digit. No letter represents zero.

So, let we go to find the word represented by 925764.

## Solution:

T+H cannot be 9. Otherwise, A & E both are 1 [which contradicts (1)] or one of A & E is zero [which contradicts (2)]. So, T+H is 8 and 1 is carried from A+E.

Then T cannot be:

0 –[[2]]

2 or 7 – then G & T would be the same digit, contradicting [1]

4 — since $T+H=8$, H would then be 4, contradicting [1].

So the possible digits for T & H are as follows:

${T : \ 1 \ 3 \ 5 \ 6}$

${H : \ 7 \ 5 \ 3 \ 2}$

Then, continuing the table with G, so that G+T is 4 or 14:

${G : \ 3 \ 1 \ 9 \ 8}$

${T : \ 1\ 3 \ 5 \ 6}$

${H : \ 7 \ 5 \ 3 \ 2}$

Because 1 is carried from A+E, A+E is either 11 or 12.

Suppose, A+E is 11. Then using [[1]] to continue the table with A and E: [${a, b, c, d, e, . . . }$ represent cases.]

$\rightarrow \ \, a \ b \ c \ d \ e \ f \ g \ h \ i \ j \ k \ l \\ {G : \ 3\ 3\ 3\ 3\ 1\ 1\ 1\ 1\ 9\ 9\ 8\ 8} \\ {T : \ 1 \ 1 \ 1\ 1\ 3\ 3\ 3\ 3\ 5\ 5 \ 6\ 6} \\ {H : \ 7\ 7\ 7\ 7\ 5\ 5\ 5\ 5\ 3\ 3\ 2\ 2} \\ {E : \ 2\ 9\ 5\ 6\ 2\ 9\ 4\ 7 \ 4 \ 7\ 4\ 7} \\ {A : \ 9\ 2\ 6\ 5\ 9\ 2 \ 7 \ 4 \ 7 \ 4 \ 7 \ 4}$

Suppose A+E is 12. Then by using [1] to continue the table with A and E

$\rightarrow m \ n \ o \ p \ q \ r \ s \ t \ u \ v \\ {G : \ 3\ 3\ 1\ 1\ 9\ 9\ 8\ 8\ 8\ 8} \\ {T : \ 1\ 1\ 3\ 3\ 5\ 5\ 6\ 6\ 6 \ 6} \\ {H : \ 7 \ 7\ 5\ 5\ 3\ 3\ 2\ 2\ 2\ 2} \\ {E : \ 4\ 8\ 4\ 8\ 4\ 8\ 3\ 9\ 5\ 7} \\ {A : \ 8\ 4\ 8\ 4\ 8\ 4\ 9\ 3\ 7 \ 5}$

Then, keeping [1] & [2] in mind, to continue the table with N, so that N+E is 6 or 16 ( if G+T 10; those case that are not listed are eliminated because no digit is possible for N):

$\rightarrow \ a \ e \ f \ g \ h \ i \ j \ k \ m \ o \ q \ r \\ {G : \ 3\ 1\ 1\ 1\ 1\ 9\ 9 \ 8 \ 3\ 1\ 9\ 9} \\ {T : \ 1\ 3\ 3\ 3\ 3\ 5\ 5\ 6\ 1\ 3\ 5\ 5} \\ {H : \ 7\ 5\ 5\ 5\ 5\ 3\ 3\ 2\ 7\ 5\ 3\ 3} \\ {N : \ 4\ 4\ 7\ 2\ 9\ 1\ 8\ 1\ 2\ 2\ 1\ 7} \\ {E : \ 2\ 2\ 9\ 4\ 7\ 4\ 7\ 4\ 4\ 4\ 4\ 8} \\ {A : \ 9\ 9\ 2\ 7\ 4\ 7\ 4\ 7\ 8\ 8\ 8\ 4}$

Again, using (1) and (2) to continue the table with M and L so that M+L is 4 or 5 (if A+E=12) , or 14 or 15 (if A+E =11; those cases are not listed are eliminated because no digits are possible for M and L):

$\rightarrow G \ T \ H \ N \ E \ A \ M \ L$

$a_1 \ \, 3\ 1\ 7\ 4\ 2\ 9\ 6\ 8 \\ a_2 \ \, 3\ 1\ 7\ 4\ 2\ 9\ 8\ 6 \\ e_1 \ \, 1\ 3\ 5\ 4\ 2\ 9\ 6\ 8 \\ e_2 \ 1\ 3\ 5\ 4\ 2\ 9\ 8\ 6 \\ e_3 \ \, 1\ 3\ 5\ 4\ 2\ 9\ 7\ 8 \\ e_4 \ \, 1\ 3\ 5\ 4\ 2\ 9\ 8\ 7 \\ f_1 \ \, 1\ 3\ 5\ 7\ 9\ 2\ 6\ 8 \\ f_2 \ \, 1\ 3\ 5\ 7\ 9\ 2\ 8\ 6 \\ g_1\ \, 1\ 3\ 5\ 2\ 4\ 7\ 6\ 9 \\ g_2 \ \, 1\ 3\ 5\ 2\ 4\ 7\ 9 \ 6 \\ g_3 \ \, 1\ 3\ 5\ 2\ 4\ 7\ 6\ 8 \\ g_4 \ \, 1\ 3\ 5\ 2\ 4\ 7\ 8\ 6 \\ h_1 \ \, 1\ 3\ 5\ 9\ 7\ 4\ 6\ 8 \\ h_2 \ \, 1\ 3\ 5\ 9\ 7\ 4\ 8\ 6 \\ i_1 \ \, 9\ 5\ 3\ 1\ 4\ 7\ 6\ 8 \\ i_2 \ \, 9\ 5\ 3\ 1\ 4\ 7\ 8\ 6 \\ k_1 \ \, 8\ 6\ 2\ 1\ 4\ 7\ 5\ 9 \\ k_2 \ \, 8\ 6\ 2\ 1\ 4\ 7\ 9\ 5$

Again continuing the table with I , so that I+M is 7 or 17 (if N+E 10; those cases not listed are eliminated because no digits are possible for I), there remains only one case $g$ with two sub-cases ${(g_2, g_4)}$:

$- G \ T \ H \ N \ E \ A \ I \ M \ L$

$g_2 \ 1 \ \, 3 \ \, 5 \ \, 2 \ \, 4 \ \, 7 \ \, 8 \ \, 9 \ \, 6 \\ g_4 \ 1 \ \, 3 \ \, 5 \ \, 2 \ \, 4 \ \, 7 \ \, 9 \ \, 8 \ \ 6$

Out of which only $g_4$ is correct since M+L = 14 .

 Substituting the letters for the digits $9 \, 2 \, 5 \, 7 \, 6 \, 4$ is ${I N H A L E}$.

## Note

• This problem requires handy exercise, and this should be done manually with great concentration.

• I am not good in latex, hence was unable to make perfect matrices. Sorry! Hope you’ll compromise with it.