數迷(2question)help!!!

Question 1:
Considering SEVEN, and SIX to be largest: i.e. 98786 + 98786 + 954= 198526 < 200000
TWENTY cannot exceed 200000
Hence T must be 1

Moreover, S >=4, if not the sum wont be a six-digit number

For S = 4, SEVEN and SIX becomes 4EVEN and 4IX respectively
Which yields E = 9, if E < 9, the sum of three number cannot be six-digit again.
Now consider if E =9
49V9N + 49V9N + 4IX = 1W9NTY
Which clearly gives W = 0
49V9N + 49V9N + 4IX = 109NTY
However, Even for the largest number of 49V9N + 49V9N + 4IX, i.e. 49897 + 49897 + 465, the sum <109000, therefore since E can only be 9 when S is 4, and now E cannot 9, hence S =/= 4, S >= 5

Now consider the addition of the last 4 digits,
i.e. the number EVEN + EVEN + SIX = ?ENTY
Consider the thousand-digit, unit-digit of E + E + carry (進位) from (VEN + VEN + SIX) = E

Moreover, VEN + VEN + SIX <= 3000, let the thousand-digit of VEN+VEN+SIX be A. Now consider the thousand-digit of the addition. EVEN + EVEN + SIX = ?ENTY. We can see that the thousand-digit are all E. That means E+E+A = ?E

Hence there are only 3 cases
1)E = 0, A = 0
2)E = 8, A = 2
3)E = 9 A = 1

For the case E = 0, A = 0
It simplies that VEN + VEN + SIX < 1000.
Hence the carry from the addition of last three digits cannot affect the thousand-digit
But, in this case. W is forced to be 0 (since VEN + VEN + SIX < 1000.) which contradicts with E = 0. So case 1 has no solution

For the case E = 8, A = 2:
E = 8, A = 2, which gives 2000< VEN + VEN + SIX <3000
That means for the addition of last 4 digits, it only gives 1 carry to the ten-thousand-digit. That implies S cannot be 5 in this case. Since if S = 5, W is forced to be 1, but T is also 1. Moreover, if S = 9, W will be forced to be 9 too. Hence S=/= 9

Therefore V can only be 5,6,7,9. If not, the addition cannot satisfy gives 2000< VEN + VEN + SIX <3000

Consider if V = 5, S can only be 9. However, as proved, S =/= 9. So V =/=5
Consider if V = 6, S can only be 7. Carry from addition of last 2 digit is 2, so N = 0
Which yields X= y in unit digit, so contradiction found, V =/=6
Consider if V = 7. S can be 6 only. And W is forced to be 3. Moreover, carry from last 2 digits is 2. So N =2. Which gives:
6 8 7 8 2
6 8 7 8 2
6 I X
-+-----------
1 3 8 2 1 Y
It only leaves 4,5,9,0. by trial and error: E=8 I=5 N=2 S=6 T=1 V=7 W=3 X=0 Y=4, is one of the solution

Consider If V = 9
2000< VEN + VEN + SIX <3000, S can only be 6 and 7
If S = 6
6898N
6898N
6IX
+---------
138N1Y
the carry from ten-digit can only be 1 or 2, hence N can only be 5 or 6. But S = 6, so N = 5. Now consider unit-digit, N+N+X = Y where N = 5. It will give X = yY contradiction found. Hence S =/=6 if V= 9
Now consider S = 7 when V = 9, W is forced to be 5.
Also, carry from last two digits must be 2 since 8N + 8N + IX = ?1Y
Therefore N is forced to be 7 too. Contradiction. Hence V =/= 9

For the case E =9, A =1
S cannot be 5 since W will be forced to be 1.
S cannot be 9 since w will be forced to be 9
Hence S can only be 6,7,8
Carry from last 2 digits must be 2 (as proved in above cases)
For S = 6, W is forced to be 3. V can only be 2,4,5
If V = 2, N = 2, contradiction.
If V = 4, N = 6, contradiction
If V = 5, N = 8. I can only be 2 (due to carry from last two digits = 2). That means carry from unit-digit must be 1. That means 20>N+N+X =>10. Which yields no solution for X. Hence S cannot be 6

For S = 7, W = 5. V can only be 2,3,4
If V = 4, N = 7, contradiction
If V = 3, N = 5, contradiction
If V = 2, N = 3, I is forced to be 2. Contradiction

For S = 8, V can only be 2,3,4.
When V = 2, N = 4. I can only be 3 due to carrying problem (mentioned in above cases). That gives N+N+X < 10. Contradiction
When V = 3, N = 6, I = 2. (X,Y) will have no solution.
When V = 4, N = 8. Contradiction.

Hence the only solution is E=8 I=5 N=2 S=6 T=1 V=7 W=3 X=0 Y=4,

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2007-04-05 13:01:44 補充：
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