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 S.o.d. classification (Posted on 2013-07-15)
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 Sodd - Seven=0 mod 11, where Sodd represents the sum of the digits on odd numbered places in the number and Seven represents the sum of the digits on the even-numbered places.
Example:
for 1234563
Sodd=1+3+5+3=12=
Seven=2+4+6

Let's assign a number N to the set of all 7-digit numbers divisible by 11 in which
N=MAX(Sodd , Seven).
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 Sodd = Seven=1, namely:
1100000,1001000,1000010.

EX2. Set #36 includes all the numbers having a pattern 9a9b9c9 (so that Sodd 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.)

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 computer solution (spoiler) Comment 1 of 1

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             01             32             243             1004             3005             7356             15687             30248             54009             907510            1379711            1953012            2606513            3310514            4027515            4713216            5317517            5785518            6058519            6075020            6026721            5745622            5308823            4746024            4094525            3397226            2700627            2052828            1501529            1092030            756031            509632            318533            182034            91035            36436            91`

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

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