Close Divisors

  1. Find four 3-digit numbers which have the same first digit, such that the sum of all four numbers is divisible by three of them.
  2. For each n (4<=n<=7), can you find a set of n+1 integers, all the same length and beginning with the same digit, such that their sum is divisible by n of them? Try to minimize the sum.
  3. For each n (2<=n<=7), find the smallest integer M, such that when you divide M evenly by k, k+1, ..., k+n-1, you get n integers, all beginning with the same digit. You can choose k in each case.
Source: Recent puzzle reading (still looking for the actual source.)
Solutions were received from Denis Borris, Sandy Thompson, and Philippe Fondanaiche. The answers are:
  1. 108 + 117 + 135 + 180 = 540. 540 is evenly divisible by 108 (5), 135 (4) and 180 (3).
  2. For n=4, (from Denis Borris): a = 1020 b = 1190 c = 1428 d = 1785 e = 1717 and S = 7140 where S=a+b+c+d+e, and S=7*a=6*b=5*c=4*d.
    I don't yet have solutions for n>4. Denis postulates there are no solutions.
  3. Answers (from Denis Borris): 
     N            M     K
 2           30     2
 3           60     4
 4          840     5
 5        10080     6
 6       110880     6
 7      1441440     8
Bonus!:
 8     15135120     8
 9     21162960    12 (low = 1007760, high = 1627920)
 9     21162960    13 (low = 1058148, high = 1763580) 
10     21162960    12 (low = 1007760, high = 1763580)
Note: the others (between high and low)in the 
      "9" and "10" cases are repetitions.
Update 4/3/00. Al Petterson provides this proof for no solutions for n>4: 
I think I can prove there are no solutions for n>4 (problem part 2).  The
proof lacks some rigor, but I don't think it's flawed.

For n=5, we have S = a+b+c+d+e+x, where x need not be a divisor but a..e
are.  
We also have some N, where at the least N<=a 2*e, S/a > 2S/e.
This implies S/e is at least 5.
S/a >= 9, thus S/9 >=a.

For the smallest possible x, we want to minimize divisors, so setting a=S/9
we have
S/5+S/6+S/7+S/8+S/9+x=S. 
We solve for x: x=S*643/2520. (x>S*643/2520 if we use divisors greater than
9.)
But  a <= S/9 = S*280/2520.

Therefore x > 2*a, and they cannot start with the same digit and be the same
length.

In general, for n>4, the minimum possible multiplier for the largest of the
divisors is n, and the minimum value for non-divisor x must therefore be S -
(1/n - 1/(n+1) - ... - 1/(2n-1)).
This will always imply x > 2S/(2n-1) if n>4 and x cannot start with the same
digit and be the same length as S/(2n-1).
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