Divisibility of 10 Digit Numbers: 1234567890

[Ian Rumsey]

1234567890
Reading from left to right, if you take the first digit of the above number it may be divided by one.
If you take the first two digits, '12' this number is divisible by two.
If you take the first three digits '123' this number is divisible by three.
However if you take the first four digits '1234' this number is not divisible by four.
If we transpose the 4 and 6 and make the number 1236547890 we may proceed.
We can now take the first four digits '1236' and this number may be divided by 4 and the first five digits 12365 may be divided by 5 and the first six digits 123654 divided by 6.
Unfortunately 1236547 cannot be divided by 7 nor can we change the 7 for a 3 as it has already been used and the digits 8, 9, 0, will not fit.
This particular series of numbers does not satisfy the divisional requirement.
The question is-
What would be the order of the digits which would satisfy the divisional requirement already shown and also allow the divisional progression to proceed for the remaining part of the 10 digit number.
Pocket calculators and slide-rules may be used.

 

[keffi]

9876543210

 

[Rogerio]

9876543210

Sorry, but 9876543 is not divisible by 7.

What about 3816547290 and 9872541630 ?

 

[TenaliRaman]

3816547290

There is simple analytical (and quite a longish method) for this one ... luckily i had posted a solution to this one in another forum so was able to do the details pretty quickly.

-- AI

 

[Rogerio]

... 3816547290 is the right ( and unique ) answer!

 

[BobG]

1234567890
Pocket calculators and slide-rules may be used.

Darn, now I really wish I would have checked this thread sooner.

 

[recon]

Rogerio, could you please share your reasoning with us? Thank you.

 

[Rogerio]

Rogerio, could you please share your reasoning with us? Thank you.

OK, let's go!
(I'm going to use the symbol '=' with the extra meaning 'belongs to')

The 10 digits number is divisible by 10 -> '0' is the last digit.

The 9 digits number is divisible by 9 -> as the first nine digits sum 45, don't care about the last digit.

The 5 digits number is divisible by 5 -> '0' is not available, so '5' is the last digit.

Our whole number is 'A B C D 5 E F G X 0' .
AB is divisible by 2 -> B is even
ABCD is divisible by 4 -> D is even , and so on.

B,D,E,G belong to [2,4,6,8]
A,C,F belong to [1,3,7,9]

C is odd and ABCD is div by 4 -> D = [2,6]
F is odd and ABCD5EFG is div by 8 -> G = [2,6]
So, D,G=[2,6] and B,E=[4,8]

ABC is div by 3, so A+B+C is div by 3
But ABCD5E is div by 6 and 3, too. So D+5+E is div by 3

Since ABC is div by 3, and B=[4,8]
if B=4 -> A,C=[1,7] So, E=8 and F=[3,9]. But D5E is div by 3 , and D=[2,6] , so D=2 . So G=6. But FG should be div by 8, so F=9 .
So, ABCD5EF = 1472589 or 7412589 which are not div by 7.
So B can't be '4' .
Then B=8 (and E=4).

E=4 and D5E is div by 3 -> D=6 . So G=2.


B=8 and ABC is div by 3 -> A,C=[1,3] or A,C=[7,9]

if A,C=[7,9] -> F=[1,3] but FG is div by 8 so F=3.
So ABCD5EF = 7896543 or 9876543 , which are not div by 7.

if A,C=[1,3] -> F=[7,9] but FG is div by 8 so F=7.
So ABCD5EF = 1836547 or 3816547 . And only 3816547 is div by 7.

So, the number is 3816547290 :-)