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Math Help - Finding 3 digit no. and factorial

  1. #1
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    Finding 3 digit no. and factorial

    Hi, i am having trouble with this problem. Even a bit confused about the validity of the question, though....

    A 3 digit no. ABC A>0 is such that A! + B! + C! becomes the no. itself. We have to find the value of B....
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  2. #2
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    Quote Originally Posted by math123x View Post
    Hi, i am having trouble with this problem. Even a bit confused about the validity of the question, though....

    A 3 digit no. ABC A>0 is such that A! + B! + C! becomes the no. itself. We have to find the value of B....
    It looks to me as though the only solution is 145 = 1! + 4! + 5!.
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  3. #3
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    Obviously, 7,8 and 9 can not be in the number since 9!>8!>7!>1000. So far it was easy. Next, we know that 6 can also not be in the number since 6!=720 which means that A\geq 7 but we already saw that that is impossible.

    So we already know that the number can only possibly have \{0,1,2,3,4,5\} as its digits. Next, we will observe that A \neq 5 since, if it would be, then B! + C! \geq 380 (= 500-120 = 500-5!), but since B,C \in \{0,...,5\} we conclude that that is impossible. Using similar arguments, we can easily show that A=4, A=3,A=2 are impossible as well. So we know for sure that A=1.

    Now, there are only two options for B,C such that 100 \leq A!+B!+C! <200. Can you see which ones they are, and why Opalg's answer is the only correct one?
    Last edited by Defunkt; November 10th 2009 at 10:24 AM.
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  4. #4
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    A great difficulty solved again..... Earlier i got till A that A =1 but not got the idea for B and C..... Thanks for the solution and great help....
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