come(20點)

F=2
O=9
R=7
T=8
Y=6
E=5
N=0
S=3
I=1
X=4

-------------------------------------------
列式:
29786
850
+ 850
----------
31486

2007-03-12 17:22:00 補充：
先看個位,Y N N=Y,得知N必是0或5.再看十位,T E E=T,得知T必是0或5,但因首T跟答案中的十位相同,T只可是5,N只可是0.再看萬位,F ? ?=S,得知千位須進一.再看千位,O ? ?=I,得知百位須進一.再看百位,R T T=X,設X要進二,那T須要6以上.再看回萬位,F ? ?=S,因為設X要進二,F必是2以上,S是F的加1.

2007-03-12 17:22:14 補充：
再看回千位,O ? ?=I,得知X要進二,O如是8,那I是0,不能成立,所以O是9,那I是1,再看萬位,F ? ?=S,得知0,1,5,9已用,最大可能F是2,S是3.剩下4,6,7,8,即R,T,X,Y.看看百位,R T T=X,3個均是Unknown,但我們得知十位進一給百位,T必是雙數,R是單數那X一定是雙數.R=7,7 T T=X,試7 8 8 1(進位)=4成功!最後Y=6,完成!

2007-03-12 17:24:20 補充：
空格是加號!

F=2
O=9
R=7
T=8
Y=6
E=5
N=0
S=3
I=1
X=4

-------------------------------------------
列式:
29786
850
+ 850
----------
31486

先看個位,Y+N+N=Y,得知N必是0或5.
再看十位,T+E+E=T,得知T必是0或5,但因首T跟答案中的十位相同,T只可是5,N只可是0.
再看萬位,F+?+?=S,得知千位須進一.
再看千位,O+?+?=I,得知百位須進一.
再看百位,R+T+T=X,設X要進二,那T須要6以上.
再看回萬位,F+?+?=S,因為設X要進二,F必是2以上,S是F的加1.
再看回千位,O+?+?=I,得知X要進二,O如是8,那I是0,不能成立,所以O是9,那I是1,
再看萬位,F+?+?=S,得知0,1,5,9已用,最大可能F是2,S是3.
剩下4,6,7,8,即R,T,X,Y.看看百位,R+T+T=X,3個均是Unknown,但我們得知十位進一給百位,
T必是雙數,R是單數那X一定是雙數.
R=7,
7+T+T=X,
試7+8+8+1(進位)=4
成功!最後Y=6,完成!

29786 + 850 + 850 = 31486

F=2, O=9, R=7, T=8, Y=6, E=5, N=0, S=3, I =1, X=4

First, we have Y + N + N = Y and T + E + E = T. That means N and E must be 0 and 5. If N was 5, there would be a 1 carried over, and there is no solution for E then. So, N=0 and E=5.

Second, we have FOR + T + T + carry = SIX. Since both of the top digits changed, that means there is a carry to the column with the O and the F. Since the 0 is already used, and the maximum carry is 2 (9+9+9 = 27 for the first column, 9+9+9+2=29 for any other column), that mean we must be adding enough to roll over from 9 to 1 in the 1000's column, a carry of 2 from the 100's column. So O=9, I = 1, and S = F+1. This also forces S to be 3,4,7, or 8, since it can't be either a used number or one more than a used number. Likewise, F must be 2,3,6, or 7.

Third, to have a carry of 2, that means R+T+T must be greater than 20. Since 0 and 1 are both used, that means x is at least 2, so R+T+T is between 22 and 28. If T was the highest unused value (8), and since we have a carry of 1 (T + 5 + 5 = T, carry 1), R would have to be 6 or 7 (5 is already used). Likewise, if R was 8, T would have to be 7. So R is 6, 7, or 8; and T is 7 or 8.

Now, two out of 6, 7, and 8 are used for R and T, so since S and F must be consecutive, they now must be 2, 3, or 4. Since either S or F will have be be 3, that raises the new minimum for R+T+T to be 24, forcing R to be 7, T to be 8, and X to be 4. This then forces F to be 2, and S to be 3.

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