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Divisibility Property (Posted on 2016-08-31) Difficulty: 3 of 5
1406357289, a 0 to 9 pandigital number has a rather interesting sub-string divisibility property.
Let d1 be the 1st digit, d2 be the 2nd digit, and so on.
In the above number we note the following:

• d2d3d4=406 is divisible by 2
• d3d4d5=063 is divisible by 3
• d4d5d6=635 is divisible by 5
• d5d6d7=357 is divisible by 7
• d6d7d8=572 is divisible by 11
• d7d8d9=728 is divisible by 13
• d8d9d10=289 is divisible by 17

List all 0 to 9 pandigital numbers with all those properties.

D4, if solved without software.

No Solution Yet Submitted by Ady TZIDON    
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Solution Without Software Comment 2 of 2 |
d6 must be either 0 or 5 so d4d5d6 is divisible by 5
but if d6 is 0 then d7 must equal d8 for d6d7d8 to be divisible by 11, which is forbidden. so d6 = 5

There are then only these combinations of d6d7d8 that are divisible by 11 and have distinct digits:

506
517
528
539
561
572
583
594

In each case, there are one or two possible d5s such that d5d6d7 is divisibly by 7, so these are the possible d5-d8s:

3506
6517
2528 X duplicates
9528
5539 X duplicates
7561 
3572
6583
2594
9594 X duplicates


We can add valid d9s in the same way, since for a given d7d8 there's at most one d9 such that d7d8d9 is divisible by 13

35065 X duplicates
6517* X no solution  
95286 
75611 X duplicates 
35728
65832
25949 X duplicates

So at this point there are only three possibilities for d5-d9:

35728 (like the one given in the problem)
65832
95286 

Well, we know there can be at most one possible d10 to make d8d9d10 divisible by 17, so these possibilities are:

357289
658323 X duplicates
952867 

Now look at d3d4 together, which must be divisible by 3 (since d5 is already divisible by 3 in all cases)

357289 has (0,1,4,6) left for selection, and only 60 and 06 work
952867 has (0,1,3,4) left for selection, and only 03 and 30 work, BUT we must exclude 03 because d4 must be even for d2d3d4 to be even. 

That leaves three choices:

06357289
60357289
30952867

In all cases the choice of d1d2 doesn't matter because d2d3d4 will always be even, so there are six total solutions:

1406357289
1460357289
1430952867
4106357289
4160357289
4130952867



  Posted by Paul on 2016-08-31 13:59:45
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