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 Add'em all ! (Posted on 2017-07-27)
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:

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.

 See The Solution Submitted by Ady TZIDON No Rating

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 work by hand | Comment 2 of 3 |
Since d6=0 or 5, I started there.

d6=0 is impossible since then 0d7d8 would require d7=d8 to be div by 11.

So d6=5.  I checked next div by 7 with middle digit 5 and followed with div by 11, 13, 17 in turn, discarding failed candidates.

e.g., the only 3-digit number beginning with 4 and having 5 in the middle and divisible by 7 is 455 which is not permissible.

e.g., 854 is divisible by 7 but there is no number 54x divisible by 11.

832 is the only number other than the 728 from the solution posted to make it to the final check, where it fails.  832 is divisible by 13 but no number 32x is divisible by 17.

So the last 7 digits are the same as the posted solution.  d3 must be 0 so that 063 is divisible by 3 and the only freedom is flipping the first two digits.

Thus the only other solution is 4106357289 and their sum is 5512714578.

Of course, all this presupposes there's no error in the above.

**edit**
Immediately after posting I saw Charlie's work which answers the question of errors I've made.

Edited on July 27, 2017, 11:01 am
 Posted by xdog on 2017-07-27 10:59:15

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