Puzzle:

Replace each letter by a digit. Each letter must be represented by the same digit and no beginning letter of a word can be 0.

O N E

O N E

O N E

+ O N E

-------

T E N

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Solution:

Use trial and error. 0 =1, N = 8 ,E = 2, T = 7

1 8 2

1 8 2

1 8 2

+ 1 8 2

------

7 2 8

Use trial and error. 0 =1, N = 8 ,E = 2, T = 7

1 8 2

1 8 2

1 8 2

+ 1 8 2

------

7 2 8

## 5 comments:

4E=N

4N=E

4O=T

MEANS N=E=0, AND 4O=T MEANS YOU CAN CHOSE O AS 1 OR 2 THEN T WILL BE 4 OR

8

E=2

N=8

O=1

T=7

182

182

182

+ 182

-----

728

Cheers!!!

Given:

(1) 4E=N+10A

(2) 4N+A=E+10B

(3) 4O+B=T

(4) -1 < E < 10

(5) -1 < N < 10

(6) 0 < O < 10

(7) 0 < T < 10

(8) N,E,O,T,A,B are integers.

(9) N!=E!=O!=T

From (3) & (7) => 4O+B<10 =>

if O=1 then B < 6 & if O=2 then B < 2 =>

(10) if B>1 then O=1

From (2) => A=E+10B-4N

Combining this with (1) =>

4E=N+10(E+10B-4N) =>

4E=N+10E+100B-40N =>

39N=6E+100B

Using (5) we have 10 cases for N. In each case I don't consider large B's because E becomes negative:

1. N=0 => B=0, E=0 => conflict with (9)

2. N=1 => B=0, E=39/6 => conflict with (8)

3. N=2 => B=0, E=13 => conflict with (4)

4. N=3 =>two cases: B=0, B=1

4.1 B=0, E=117/6 => conflict with (4) for all N > 2, so we don't consider B=0 case anymore.

4.2 B=1, E=17/6 => coflict with (8)

5. N=4, B=1, E=56/6 => coflict with (8)

6. N=5 => B=1, E=95/6 => conflict with (4) for all N > 4, so we don't consider B=1 case anymore.

7. N=6, B=2, E=17/3 => conflict with (8)

8. N=7, B=2, E=73/6 => conflict with (8)

9. N=8 =>two cases: B=2, B=3

9.1 B=2, E=112/6 => conflict with (4) for all N>7, so we don't consider B=2 case anymore.

9.2 B=3, E=2 => NO CONFLICT!

Let's check the rest to make sure this is the only answer:

10. N=9, B=3, E=51/6 => conflict with (8)

We got: N=8, E=2, B=3

Because B=3 & (10) => O=1

Using the original equasion: T=7

The answer is:

N=8, E=2, O=1, T=7

and we proved that this is the only correct answer.

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