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Staggered divisibility (Posted on 2013-09-16) Difficulty: 3 of 5
The number, 1406357289, is a pandigital number (i.e. containing all of the digits - 0 to 9 - once, in some order), but it also has a rather interesting sub-string divisibility property.
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, 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

Find the sum of all pandigital numbers with this property.

Source: Euler Project.

No Solution Yet Submitted by Ady TZIDON    
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spoiler | Comment 1 of 3
d4d5d6 is divisible by 5 so d6=0 or 5 

d6d7d8 is divisible by 11 so d6-d7+d8 = 0 mod11.  d6=0 requires d7=d8 which is not allowed so d6=5.
Possibles are:
506
517
528
539
550  ** dupe digit
561
572
583
594

d7d8d9 is divisible by 13

Possibles are:
065  ** 5 already assigned
17_  ** none possible
286
390
611  ** dupe digit
728
832
949  ** dupe digit

d8d9d10 is divisible by 17
Possibles are:
86_  ** none possible
90_  ** none possible
289
323  ** dupe digit

d6 to d10 are therefore 5,7,2,8,9 leaving 0,1,3,4,6 to be assigned.

d3d4d5 divisible by 3 requires some combination of 0,3,6 but only 357 is divisible by 7 so d5=3 giving possibles 063 and 603 leaving 1,4 unassigned.

Then 4106357289, 4160357289, 1406357289, 1460357289 are solutions and their sum is 11133429156.

Did I miss any?


  Posted by xdog on 2013-09-16 11:35:56
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