4!<100a+10b+c<7 The largest digit of abc must be less than 7. If one of them is 6, then the first digit is 7. So none of them is bigger than 6. Now the largest one can only be 5. 5!=120 If two of them are 5. We get 2*5!=240. The first digit is 2, and we assume that there are two 5 in it. 255 is not equal to 240+2=242. And we can easily prove that it can't be 555, either. So one of them must be 5, and another one is 1. 120+x!+1=100*1+10x+5 Obviously, x is 4. So abc is 145.
(It's much harde to write the solution down than to think of it...= = )
3 comments:
4!<100a+10b+c<7
The largest digit of abc must be less than 7.
If one of them is 6, then the first digit is 7.
So none of them is bigger than 6.
Now the largest one can only be 5.
5!=120
If two of them are 5.
We get 2*5!=240.
The first digit is 2, and we assume that there are two 5 in it.
255 is not equal to 240+2=242.
And we can easily prove that it can't be 555, either.
So one of them must be 5, and another one is 1.
120+x!+1=100*1+10x+5
Obviously, x is 4.
So abc is 145.
(It's much harde to write the solution down than to think of it...= = )
來ㄚ~~~
不論國文,英文,物理,化學,生物,音樂,幾何
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所以啦
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恆成立orz
吳～看英文是很吃力的...... XD
劉～這老套了～換一個吧～ XD
還有......
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這樣以後哪分得出來誰是誰啊...... XD
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