My Math Forum A=abc=a !+b !+c ! (A is a 3-digit number)

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 November 4th, 2012, 06:47 PM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 A=abc=a !+b !+c ! (A is a 3-digit number) A=abc=a !+b !+c ! (A is a 3-digit number) find A Ans :145
 November 4th, 2012, 09:08 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,981 Thanks: 994 Re: A=abc=a !+b !+c ! (A is a 3-digit 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 - (a-1)!] = b[(b-1)! - 10] +c[(c-1)! - 1] ...and that's enuff for me to quit! Early Xmas gift, Albert: A 5digiter example: 40585
 November 5th, 2012, 12:58 AM #3 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: A=abc=a !+b !+c ! (A is a 3-digit 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 four-digits 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, 04:42 AM #4 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: A=abc=a !+b !+c ! (A is a 3-digit number) Denis :according to my program 40585 is also the only solution with 5digit numbers thanks for your Early Xmas gift
November 5th, 2012, 08:02 AM   #5
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Re: A=abc=a !+b !+c ! (A is a 3-digit number)

Quote:
 Originally Posted by Albert.Teng Denis :according to my program 40585 is also the only solution with 5digit numbers
Agree.
Btw, since 9! * 8 = 2903040 (7 digits; 8 required)), then a 7 digit number is maximum : 9! * 7 = 2540160

 November 5th, 2012, 07:39 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: A=abc=a !+b !+c ! (A is a 3-digit 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, 02:31 AM #7 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: A=abc=a !+b !+c ! (A is a 3-digit number) Ah, thank you, [color=#00BF40]MarkFL[/color], that's very kind.
November 6th, 2012, 05:18 AM   #8
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Re: A=abc=a !+b !+c ! (A is a 3-digit number)

Quote:
 Originally Posted by MarkFL @[color=#0000FF]Hoempa[/color], good work! You too, [color=#0000FF]Denis[/color], but I wanted to also acknowledge the other contributor.
You trying to soften us up so we send you a Xmas gift?

November 8th, 2012, 06:25 AM   #9
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Re: A=abc=a !+b !+c ! (A is a 3-digit number)

Quote:
 Originally Posted by Albert.Teng A=abc=a !+b !+c ! (A is a 3-digit number)
Factorions. There are only four such numbers in base 10.(and 5 such numbers in base 17. The largest one is quite impressive!)

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 3-digit number N with digits a,b,c satisfying N = a# + b# + c# where "#" is denoting the primorial function.

 November 8th, 2012, 07:33 AM #10 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: A=abc=a !+b !+c ! (A is a 3-digit 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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# abc=a^3 b^3 c^3

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