Euler problem 43

The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.

Let d₁ be the 1^st digit, d₂ be the 2^nd 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 0 to 9 pandigital numbers with this property.

analyse:


We can create a list with all 0-9 pan-digitals, see problem 41, then extract the 3 digit 
subgroups and check for divisibility by apropiate prime. However this will yield to a hell 
of a lot loops (about 10! fac). So we have to think of something smarter.
more analyse lead to the following axioma's:


axioma 1: a 10 digit pandigital can't start with a zero. d1 <> 0.
axioma 2: d4 must be even. d4 = {0,2,4,6,8}
axioma 3: d456 must be div by 5. d6 = {0,5}
axioma 4: d678 must be div by 11. With d6=0 we get 011,022,033. This is div by 11 but contains dual digits.
So d6 must be a 5. 


Solution:


Create a array with digit d1 to d10.
Increment each digit starting with digit one.
Check for unique digit and apply dividible rules recursive.

Problem43 = 16695334890 elapsed time:    8 ms. Test Passed.