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
4 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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