There are 818,181 7-digit numbers divisible by 11, the smallest being 1,000,010 and the largest 9,999,990.
In each of them S_odd - S_even=0 mod 11, where S_odd represents the sum of the digits on odd numbered places in the number and S_even represents the sum of the digits on the even-numbered places.
Example:
for 1234563
S_odd=1+3+5+3=12=
S_even=2+4+6


Let's assign a number N to the set of all 7-digit numbers divisible by 11 in which
N=MAX(S_odd , S_even).
We shall call this set set #N , and denote by q(N) the number of its members.

Examples:

EX1. q(1)=3, since there are 3 numbers possesing S_odd = S_even=1, namely:
1100000,1001000,1000010.

EX2. Set #36 includes all the numbers having a pattern 9a9b9c9 (so that S_odd is 36 and a+b+c=3 mod 11); so q(36)=number of compositions of number 3 into 3 non-negative integers, triplets like 111,102....003,030,300 +number of compositions of number 14 into 3 non-negative integers, triplets like 059,068,,,167,158,149,...770,...950 +number of compositions of number 25 into 3 non-negative integers i.e. 799,889,898,979,988,997.


a. Which is the most numerous set?
b. List q(i) for 1=1 to 36.

(Inspired by KS's "Divisibility and Digit Sum",
revised and vetted by Charlie.)

DEFDBL A-Z
DIM ct(36)

FOR n = 1000010 TO 9999999 STEP 11
   ns$ = LTRIM$(STR$(n))
   todd = 0: teven = 0
   FOR i = 1 TO 7 STEP 2
     todd = todd + VAL(MID$(ns$, i, 1))
   NEXT
   FOR i = 2 TO 6 STEP 2
     teven = teven + VAL(MID$(ns$, i, 1))
   NEXT
   IF teven > todd THEN SWAP todd, teven
   ct(todd) = ct(todd) + 1
NEXT n
FOR i = 0 TO 36: PRINT i, ct(i): NEXT

finds

i             q(i)
0             0
1             3
2             24
3             100
4             300
5             735
6             1568
7             3024
8             5400
9             9075
10            13797
11            19530
12            26065
13            33105
14            40275
15            47132
16            53175
17            57855
18            60585
19            60750
20            60267
21            57456
22            53088
23            47460
24            40945
25            33972
26            27006
27            20528
28            15015
29            10920
30            7560
31            5096
32            3185
33            1820
34            910
35            364
36            91

making set#19, with q(19) = 60750, the answer to part a.

Posted by Charlie on 2013-07-15 18:40:55