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Counting Sextuplets (Posted on 2011-05-04) Difficulty: 3 of 5
Each of A, B, C, D, E and F is a positive integer with A ≤ B ≤ C ≤ D ≤ E ≤ F ≤ 25.

Determine the total number of sextuplets (A ,B, C, D, E, F) such that (A+B+C)*(D+E+F) is divisible by 75.

No Solution Yet Submitted by K Sengupta    
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Some Thoughts Beginning of Analytical Solution | Comment 2 of 7 |
For a number to be divisible by 75, it must be divisible by the prime factors of 75, which are 3,5,5.

However many of these factors is contained in the first triplet sum (a+b+c), the others must be factors of the other sum (d+e+f). By definition, a+b+c ≤ d+e+f.

The possibilities for the sums of either side are numerous, but not too numerous to list by hand:

3 ~ 5,5
3,25
6,25
9,25
12,25
15,25
18,25
21,25
24,25
3,50
6,50
9,50
12,50
15,50
18,50
21,50
24,50
27,50
30,50
33,50
36,50
39,50
42,50
45,50
48,50

5 ~ 3,5
5,15
10,15
15,15
5,30
10,30
15,30
20,30
25,30
30,30
5,45
10,45
15,45
20,45
25,45
30,45
35,45
40,45
45,45
5,60
10,60
15,60
20,60
25,60
30,60
35,60
40,60
45,60
50,60
55,60
60,60

3,5 ~ 5
15,15*
15,20
15,25
15,30*
15,35
15,40
15,45*
15,50
15,55
15,60*
15,65
15,70
30,30*
30,35
30,40
30,45*
30,50
30,55
30,60*
30,65
30,70
45,45*
45,50
45,55
45,60
45,65
45,70
60,60*
60,65
60,70

5,5 ~ 3
25,27
25,30
25,33
25,36
25,39
25,42
25,45*
25,48
25,51
25,54
25,57
25,60*
25,63
25,66
25,69
25,72
50,51
50,54
54,57
50,60*
50,63
50,66
50,69
50,72
* item is repeated

For reasons I won't go into here, if p[n] is the number of possible values of a≤b≤c which add up to n, and s n/3, then:
p[n] = p[n-1] + n/6 + ((n/3)%2 && n%2)

For example:
n    p[n]
0    0
1    0
2    0
3    1
4    1
5    2
6    3
7    4
8    5
9    7
10    8
11    10
12    12
13    14
14    16
15    19
16    21
17    24
18    27
19    30
20    33
21    37
22    40
23    44
24    48
25    52
...

The same number of possibilities apply for the second set, d≤e≤f, but the values are limited based on the values a,b,c.
Each possibility of the d,e,f triplet corresponds to a possibility of a,b,c, where a=d and b=e; naturally these possibilities must be eliminated from the total.
The exception would be the case where a=b=c=d=e=f, which can be easily accounted for because whenever a+b+c = d+e+f, there is only this one possibility anyway.

Specifically, if x = a+b+c and y = d+e+f, using the above definition, of p[], the number of *valid* possibilities for d+e+f should be at most p[y]-p[x].
However, this doesn't eliminate all invalid possibilities. A deeper analysis will be necessary to determine how many valid possibilities for all six variables there are in this problem -- this conjecture may not even be useful at all.
  Posted by DJ on 2011-05-04 21:55:22
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