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November 4th, 2012, 05:47 PM  #1 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  A=abc=a !+b !+c ! (A is a 3digit number)
A=abc=a !+b !+c ! (A is a 3digit number) find A Ans :145 
November 4th, 2012, 08:08 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,308 Thanks: 935  Re: A=abc=a !+b !+c ! (A is a 3digit number)
Hmmm...well, I can confirm 145 is the only solution with 3digit numbers... wrote program... a = 1 to 5, b = 0 to 6, c = 0 to 6; only one can = 6...so max number = 565 100a + 10b + c = a! + b! + c! 100a  a! = b!  10b + c!  c a[100  (a1)!] = b[(b1)!  10] +c[(c1)!  1] ...and that's enuff for me to quit! Early Xmas gift, Albert: A 5digiter example: 40585 
November 4th, 2012, 11:58 PM  #3 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Re: A=abc=a !+b !+c ! (A is a 3digit number)
Here is another approach: max digit is 6 since 7! > 1000 would give at least four digits. If there is a digit 6, then the hundreds digit is at least 7 (or we have a fourdigits number...) which isn't possible. Now, the hundreds digit is equal to the amount of digits 5, since 5! = 120 and there is a 3 digit number. So it is 1 or 2. If it is two, then the number is 255 which doesn't work. So it must be 1, if such a number exists. 1! + 5! = 121. Leaves one more digit 1 to 4. Inspection (last digit check for a start) gives 4. Hence the only such number is 145. 
November 5th, 2012, 03:42 AM  #4 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: A=abc=a !+b !+c ! (A is a 3digit number)
Denis :according to my program 40585 is also the only solution with 5digit numbers thanks for your Early Xmas gift 
November 5th, 2012, 07:02 AM  #5  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,308 Thanks: 935  Re: A=abc=a !+b !+c ! (A is a 3digit number) Quote:
Btw, since 9! * 8 = 2903040 (7 digits; 8 required)), then a 7 digit number is maximum : 9! * 7 = 2540160  
November 5th, 2012, 06:39 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs  Re: A=abc=a !+b !+c ! (A is a 3digit number)
@[color=#0000FF]Hoempa[/color], good work! You too, [color=#0000FF]Denis[/color], but I wanted to also acknowledge the other contributor. 
November 6th, 2012, 01:31 AM  #7 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Re: A=abc=a !+b !+c ! (A is a 3digit number)
Ah, thank you, [color=#00BF40]MarkFL[/color], that's very kind. 
November 6th, 2012, 04:18 AM  #8  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,308 Thanks: 935  Re: A=abc=a !+b !+c ! (A is a 3digit number) Quote:
 
November 8th, 2012, 05:25 AM  #9  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: A=abc=a !+b !+c ! (A is a 3digit number) Quote:
An equivalent question would be to find all such numbers which are the sum of the primorials of there digits. Since if n is such a number then we can find the bound of 630. Though this is weak, we are now able to prove that these integers are finite. The sequence is (which I found by iterating) : 1,2,36. Prove that there are no such 3digit number N with digits a,b,c satisfying N = a# + b# + c# where "#" is denoting the primorial function.  
November 8th, 2012, 06:33 AM  #10 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Re: A=abc=a !+b !+c ! (A is a 3digit number)
possible digit's primorials 1# = 1 2# = 2 3# = 4# = 6 5# = 6# = 30 7# = 8# = 9# = 210 The [color=#0040FF]blue[/color] digits after a line are candidates for the hundreds digit so far. You have already that there is a max of 3 digits. And a max of 3 * 210 = 630. [color=#0040FF]1, 2, 3, 4, 5, 6, 7, 8, 9[/color] If it's a 3 digit number, 7, 8 or 9 is included and can't be the hundreds digit. [color=#0040FF]1, 2, 3, 4, 5, 6[/color] The hundreds digit can't be 5 or 6, else the sum of the primorials can't exceed 500. [color=#0040FF]1, 2, 3, 4[/color] The hundreds digit is even. (just noticed here, could use it earlier) [color=#0040FF]2, 4[/color] If the hundreds digit is 4 then the number ends in 6 which isn't possible since the other two must be larger than 6. [color=#0040FF]2[/color] If the hundreds digit is two then the tens digit is eighter 1 or 4. Then the units digit would be 2# + 1# = 3 or 2# + 4# = 8. Both fail so no candidates for hundreds digit. 

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