Problems & Puzzles: Puzzles

Puzzle 946. The largest prime that divides...

Paul Cleary sent the following nice puzzle:

Show that the sum of all the permutations of any 39 digit number will always be divisible by the Prime 900900900900990990990991.

or

Show that the sum of all the permutations of any 47 digit number will always be divisible by the Prime 316362908763458525001406154038726382279.

BTW, these primes are the largest prime divisors in each case.

Q. Send your proof for each prime.

 

Contributions came from Giovanni Resta, Emmanuel Vantieghem, John Arioni and Jan van Delden..

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Giovanni wrote on March 16, 2019:

If we consider a number of n digits (d1,d2,...,dn) there are n! permutations (if the digits are repeated some permutations will be the same).
 

If we focus on a single digit dk we see that it will appear on every position (1st, 2nd, .. n-th) from the end exactly  n!/n = (n-1)! times for symmetry reasons.  The digit dk when in position i from the end has weight 10^(i-1). So the contribution of digit dk to the overall sum is 
 

dk * (n-1)! * (1+10+100+1000+...+10^(n-1))  = dk * (n-1)! * (10^n-1)/9.

Hence, adding the contributions of all the digits we have that the sum is

(d1+d2+...+dn) * (n-1)! * (10^n-1)/9.

If n=39 the sum is a multiple of 900900900900990990990991 because that is a divisor of 10^39-1. The same for the other prime which is a divisor of 10^47-1.

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Emmanuel wrote on March 19, 2019:

It is very easy to show that the sum of all the permutations of an  n-digit number  m  is divisible by  R(n), the repunit with  n  digits.
Simply : write down all permutations under each other.
It is then easy to see that every 'column' is a permutation of the first one.
Thus, if  s  is the sum of the digits in the first column, then  s  is also the sum of all the digits in any other column.
So, when you make the sum of all permutations of  m  by 'posponing carry', you obtain the number
   s*10^(n-1) + s*10^(n-2) + ... + s*10+s = s*R(n).

 
I give an example when  m = 1223.
Write all the permutations of  m  as follows :
     1*1000 + 2*100  + 2*10 + 3
     1*1000 + 2*100  + 3*10 + 2
     1*1000 + 3*100  + 2*10 + 2
     2*1000 + 1*100  + 2*10 + 3
     2*1000 + 1*100  + 3*10 + 2
     2*1000 + 2*100  + 1*10 + 3
     2*1000 + 2*100  + 3*10 + 1
     2*1000 + 3*100  + 1*10 + 2
     2*1000 + 3*100  + 2*10 + 1
     3*1000 + 1*100  + 2*10 + 2
     3*1000 + 2*100  + 1*10 + 2
     3*1000 + 2*100  + 2*10 + 1
  -------------------------------------- + (summing with postponed carry)
  24*1000 + 24*100 + 24*10 + 24 = 24*1111 (= 26664  after carry).
 
If  m  is a 39-digit number then the sum of all permutations will be of the form  s*R(39).
Factorization of  R(39)  is :
3*37*53*79*265371653*900900900900990990990991
This proves the first assertion.

 
If  m  has 47 digits we find :
R(47) = 35121409*316362908763458525001406154038726382279
Which proves the second assertion.

 
Now we shall prove that  s  cannot have a prime divisor > 900900900900990990990991  in case 1  and
                                 that  s  cannot have a prime divisor > 316362908763458525001406154038726382279  in case 2

 
When  m  is composed of  n1  times the digit  d1
                                          n2  times the digit  d2
                                           ............................
                                           nk  times the digit  dk
then it is an easy combinatorial exercise to compute that :
     s = A*(n1*d1 + ... + nk*dk)  with  A = ((n1 + ... +nk - 1)!) / ((n1!)*...*(nk!)).

 
Now, to prove the assertion about the greatest prime divisor of  s*R(n), it suffices to look at the primefactors of  s.
Well, the factor  A  (which can be a fraction instead of an integer) cannot have a prime divisor > 38.
The value of  n1*d1 + ... + nk*dk  is at most  9*n.  Therefore, the greatest prime factor in the cases  n = 39  or  47  cannot be else than those that are given.

 
Thus, all assertions of the puzzle are correct.

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John wrote on March 18, 2019:

900900900900990990990991

is the largest prime divisor of A=111... 111 (39 digits), while

316362908763458525001406154038726382279

is the largest prime divisor of B=1111... 1111 (47 digits).

Now, let's consider any 39-digits number: such a number will be

made up of some quantities of digits 0, digits 1, digits 2

and digits 9.

If we sum all the permutations of our 39-digits number just the

way we learned in primary school, on each of the 39 columns,

each of the digits 0, 1, 2, ..9 will appear with the same

own frequency. This means that the sum of the values in any

of the 39 columns is regurarly the same (say C). Then the sum of

all the permutations is C+10*C+ .. + 10^38*C=A*C which is

obviously divided by 900900900900990990990991 as well as A.

How large is C:

let's define the following:

- qn= quantity of the digit n in the 39-digits number

- vn = value of the digit n

- P = number of permutations

then:

C= sum (qn*vn*P/39) n= 0,1,2,3,.....9, that is:

C= q0*0*P/39 + q1*1*P/39 + q2*2*P/39+ .....+ q9*9*P/39=

= P/39*(q1+2*q2+3*q3+......+9*q9); q0+q1+q2+ ....+q9=39

It's not a problem to check that the largest possible prime factor

of C is less than or equal to 347, so we can state that the largest

prime factor of A (900900900900990990990991) is also the

largest factor of A*C

Mutatis mutandi, the same arguments hold for 47-digits numbers.

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Jan wrote on March 23, 2019:

The question implies that digit 0 is not allowed (or we allow a number starting with extra 0s).

 

Given a number N with n digits, a[0]..a[n-1], with all digits a[j]>0, consisting of k<=n distinct digits.

Suppose there are M permutations. We could add the sum of all these permutations in two ways:

  • By permutation
  • By location j (and multiplying with 10^j)

These sums should be equal.

 

Furthermore the sum of all digits over all permutations can be split in two ways:

  • By permutation (total sum of digits =M*sum of digits of N
  • By location (sum of digits by location), equal for every location.

 

The sum of digits by location is therefore equal to M/n*sum of digits of N.
To find the sum of all M permutations we should multiply with sum(10^j,j=0..n-1).
The sum of all M permutations is therefore (M/n*sum of digits of N)*(10^n-1).

 

We have n|M by construction and find 10^n-1| sum of all permutations of N.

 

And the given primes are the largest prime factors of 10^n-1, with n=39 respectively n=47.

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