### Nine Digit Number - Math Puzzle

**Source:**Sent to me by Sudeep Kamath via http://puzzletweeter.com/2013/06/21/884/

**Problem:**

Find a nine digit number abcdefghi that uses all the digits 1,2,3,...9 exactly once and satisfies:

a is divisible by 1

ab (2 digit number with digits a and b) is divisible by 2

abc (3 digit number with digits a, b and c) is divisible by 3

.......

.......

abcdefghi is divisible by 9.

Update (29/06/2013):

**Solution:**Posted by JDGM, Anti, Meghana Kolan, Shwetabh Sameer and Unknown in comments! A more detailed solution provided by me in comments!

ReplyDelete381654729I noticed e must be 5 and b, d, f, h each even. There are 24 permutations of those evens and 24 permutations of the remaining odds. That already narrows us down to 24² possible abcdefghi and instead of pushing on by hand I wrote a quick and dirty program to check which of those it was.

381654729

ReplyDeleteThe divisibility for 7 was a mess though. Does anybody have an elegant method for that?

381654729

ReplyDeleteLet Dn be the digit n

D5 = 5

D2, D4, D6, D8 - even

D1, D9 , D3, D7 - odd

D1 can be any odd number so is D9 since the sum of the digits is always a multiple of 9

D3-D4 should be a multiple of 4 and owing to the fact that D3 should be odd we have D4 = 2 or 6

Assuming D1-D2-D3 is a multiple of 3 (for it to be divisible by 3), we need D4-D5-D6 to be an even nultiple of 3 leaving us with D4-D5-D6 = 258 or 654

D6-D7-D8 has to be a multiple of 8 which gives us D6-D7-D8 = 432 or 472 or 816

Using ---65432 , ---65472, ---25816 and accounting for its divisibility by 3 and 7 we get the above result.

This comment has been removed by the author.

ReplyDelete381654729 is the answer. Write a simple code and you'll get the same.

ReplyDelete381654729

ReplyDeleteYeah. 7 was a pain. :-)

Thanks for the solution. Just providing a more detailed solution.

ReplyDeletee is 5.

a, c, g, i are odd - 1, 3, 7, 9

b, d, f, h are even - 2, 4, 6, 8

cd is divisible by 4 and c is odd, implies d is either 2 or 6

abc, abcdef, abcdefghi is divisible by 3

So, abc, def, ghi are divisible by 3

def is divisible by 3 would mean it is one out of

258

654

abcdefgh is divisible by 8, means

fgh is divisible by 8

Since f is either 8 or 4, gh is divisible by 8

gh is one of 16, 32, 72, 96

ghi is divisible by 3. Hence possible options for ghi: 321, 327, 723, 729, 963

In summary:

def is 258 or 654

ghi is 321, 327, 723, 729 or 963

a, c are odd - 1, 3, 7, 9

b is even - 2, 4, 6, 8

abc is divisible by 3

abcdefg is divisible by 7

Assuming def is 258:

def is 258

ghi is 963

a, c are odd - 1, 7

b is even - 4

abcdefg is divisible by 7

1472589 mod 7

7412589 mod 7 are both not zero

No solution

Assuming def is 654:

def is 654

ghi is 321, 327, 723, 729

a, c are odd - 1, 3, 7, 9

b is even - 8

abc is divisible by 3

abcdefg is divisible by 7

Possible options:

a8c654321 - a/c are 7/9

a8c654327 - a/c are 1/9

a8c654723 - a/c are 1/9

a8c654729 - a/c are 1/3

8 options to be checked for divisibility by 7

a8c6543 mod 7 = a + 5 + 4*c + 5 = a+4*c+3

a8c6547 mod 7 = a + 5 + 4*c + 2 = a+4*c

Possible options:

a8c654321 - a/c are 7/9 ( 0 mod 7 for abcdefg gives no solution )

a8c654327 - a/c are 1/9 ( 0 mod 7 for abcdefg gives no solution )

a8c654723 - a/c are 1/9 ( 0 mod 7 for abcdefg gives no solution )

a8c654729 - a/c are 1/3 ( 0 mod 7 for abcdefg gives a=3, c=1 )

So, the number is

381654729

Brilliant Explanantion ...

ReplyDelete