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September 26th, 2013, 10:55 AM  #1 
Newbie Joined: Sep 2013 Posts: 2 Thanks: 0  HELP! probability that a 7 digit number is divisible by 7?
The seven digits {1; 2; 3; 4; 5; 6; 7} are written down in a random order. What is the chance that the resulting number is divisible by 7? For instance, 1234576 works, but 1234567 doesn’t. 
September 26th, 2013, 11:48 AM  #2 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  Re: HELP! probability that a 7 digit number is divisible by
The answer to the question is the number of numbers with digits 1,2,3,4,5,6,7 that are divisible by 7 divided by 7!. I don't know how you determine all such numbers divisible by 7 but here are a few: 1427356 6427351 1357426 6357421 
September 26th, 2013, 12:33 PM  #3  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: HELP! probability that a 7 digit number is divisible by Quote:
14 27356 1 42 7356 1427 35 6 14273 56 The numbers in bold are all divisible by 7 So must be 27356 and 14273 (For obvious reason) Same with your last example 63 57421 6 35 7421 6357 42 1 63574 21 The numbers in bold are all divisible by 7 So must be 57421 and 63574 (For obvious reason) I want to add 1234765 1276534  
September 26th, 2013, 07:53 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Re: HELP! probability that a 7 digit number is divisible by
There's 720 of those critters: 1: 1234576 2: 1234765 ... 719: 7653142 720: 7654213 720/5040 = 1/7 
September 26th, 2013, 08:41 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond  Re: HELP! probability that a 7 digit number is divisible by
Yes, and the first seven appear as the 2nd, 6th, 9th, 37th, 41st, 44th and 50th members of the sequence of ascending numbers that have the unrepeated digits 1, 2, 3, 4, 5, 6 and 7. Is it mere coincidence that 6! = 720?

September 27th, 2013, 10:37 AM  #6  
Newbie Joined: Sep 2013 Posts: 2 Thanks: 0  Re: HELP! probability that a 7 digit number is divisible by Quote:
Can you please tell me?  
September 27th, 2013, 07:22 PM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Re: HELP! probability that a 7 digit number is divisible by
In a tremendously shrewd, clever and highly advanced manipulation: loop a,b,c,d,e,f,g from 1 to 7 keeping 'em all different n = a*10^6 + b*10^5 + c*10^4 + d*10^3 + e*10^2 + f*10 + g if n@7 = 0 then count = count + 1 : print n Would you like my autograph? 
September 30th, 2013, 01:51 AM  #8 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: HELP! probability that a 7 digit number is divisible by
If you look at the value of 10^n (mod 7), you see that for any permutation of the digits 17, the remainder mod (7) is: a + b + 2c + 3d + 4e + 5f + 6g Where ag is a permutaion of the digits 17. Note that the digits here are not in the same order as they appear in the decimal expansion. It appears that if you keep the digits in the same order, but cycle them round, then you get each value 06 (mod 7) precisely once. This would explain why you get 6! numbers divisible by 7 and would also imply that you get 6! numbers that are n (mod 7) n = 06. I can't see how to prove this. 
October 4th, 2013, 02:10 PM  #10 
Newbie Joined: Oct 2013 Posts: 1 Thanks: 0  Re: HELP! probability that a 7 digit number is divisible by
I believe that everyone may have been "overthinking" this one. The probability that a random number is divisible by 7 is obviously 1 in 7. This is just due to the fact that every seventh number is divisible by 7 (starting at 7, 14, 21....). So given a set of random numbers, a sequential set of numbers, or in this case 7! (5,040) numbers... The randomness applies and does not skew the probability of 1/7. Just another thought on the topic for what it is worth. I also wonder how "lincoln40113" was able to determine 720 favorable outcomes. Was this brute force? and if so... cudos on that effort!!! 

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