Do you know my Phone number?

If you want the easy way, here's a simple Python code:

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from itertools import permutations

for permutation in permutations('0123456789'):
  if all(int(''.join(permutation[:i])) % i == 0 for i in range(1, 11)):
    print ''.join(permutation)

The code above is inefficient, because it goes through all the $10! = 3,628,800$ permutations. Note that you have to have even digits in all even places, because an odd number isn't divisible by an even digit, so you could use this information to optimize the code and get:

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from itertools import permutations

for odds in permutations('13579'):
  for evens in permutations('02468'):
    permutation = [x for pair in zip(odds, evens) for x in pair]
    if all(int(''.join(permutation[:i])) % i == 0 for i in range(1, 11)):
      print ''.join(permutation)

That goes through $5! \cdot 5! = 14,400$ permutations (252 times faster).

But it's still cheating, so let's do it manually.

Let [i] denote the i'th digit, and [ijk] denote the i'th and j'th and k'th digits.

(1) Note that the last digit [10] has to be 0. Otherwise the number won't be divisible by 10.

(2) [5] has to be 0 or 5, in order for [12345] to be divisble by 5. Since [10]=0, we get [5]=5.

(3) [123] has to be divisible by 3, and [123456] has to be divisible by 6. Note that this implies that [456] has to be divisible by 3. We know that [5]=5, and we also know that [4] and [6] have to be even.

So, [4]+5+[6] is divisible by 3, and [4] + [6] are even $\Rightarrow$ ([4] + [6]) % 3 = 1 and ([4] + [6]) = even $\Rightarrow$ [4] + [6] = 4 or 10 or 16.

4 is impossible because it can only be 2+2 (duplicate), and 16 is impossible because it can only be 8+8 (duplicate), so [4]+[6] = 10. So they can be either 2 and 8, or 4 and 6.

(4) [34] has to be divisible by 4. Also, 3 is odd. So [4] has to be 2 or 6 (try it: 12, 16, 32, 36, ...) So [456] have to be 258 or 654.

(5) [678] has to be divisible by 8. [7] is odd, and [6] is either 4 (and then [8] can only be 2 or 8) or 8 (and then [8] can only be 4 or 6).

So there's only a few options for [678]: 432, 472, 816, 896... (I made a mistake earlier, so ) I will consider only 432 and 472 here. The other options 816 and 896 are addressed in my comment below.

So [8]=2. Therefore, [456] cannot be 258, and can only be 654. Therefore, [678] is either 432 or 472. Also, there's only one option left for [2], and that is 8.

That's a lot, so let's recap what we know.

• [1]=1 or 3 or 7 or 9
• [2]=8
• [3]=1 or 3 or 7 or 9
• [4]=6
• [5]=5
• [6]=4
• [7]=3 or 7
• [8]=2
• [9]=1 or 3 or 7 or 9
• [10]=0

(6) Let's now take a look at [123]. It's divisible by 3, so [1]+8+[3] is divisible by 3, so [1]+[3] % 3 = 1. We also know that [1]+[3] is even because they are both odd. So there's only a few options for [1]+[3]: 4, 10, 16. So [13] has 6 options: 13, 31, 37, 73, 79, 97.

(7) Let's consider [1234567]. It has to be divisible by 7. There are 6 options for [13] and 2 options for [7]. So, 12 total - we can check these manually.

• 1836543 % 7 = 2
• 1836547 % 7 = 6
• 3816543 % 7 = 3
• 3816547 % 7 = 0
• 3876543 % 7 = 6
• 3876547 % 7 = 3
• 7836543 % 7 = 3
• 7836547 % 7 = 5
• 7896543 % 7 = 4
• 7896547 % 7 = 1
• 9876543 % 7 = 5
• 9876547 % 7 = 2

So [1234567] = 3816547. We also know that [8]=2 and [10]=0, so [9]=9 (only option left).

Therefore, the number is $\boxed{3816547290}$ .

10th digit must be 0, so the fifth digit can't be other than 5.

Even place digit mut be even. As a consequence odd place digit have to be odd.

It is a fact that a multiple of 4 (and hence a multiple of 8) must have as its last digit an even number, and if the digit just before the last is odd then the last digit must be 2 or 6.

We have one of the two following configurations

_ _ _ 2 5 _ _ 6 _ 0

_ _ _ 6 5 _ _ 2 _ 0

For the divisibility-by-3 criterion the groups of the first three and second three digit (and hence also the third group of 3 digit) must sum up to a multiple of 3.

We can add a digit on sixth place of each possibility, and the last even digit to the second to get:

_ 4 _ 2 5 8 _ 6 _ 0

_ 8 _ 6 5 4 _ 2 _ 0

Let's spend the divisibility by 8 criterion. Since

_ 4 _ 2 5 8 _ 6 = _ 4 _ 2 5 0 0 0 + 8 _ 6

_ 8 _ 6 5 4 _ 2 = _ 8 _ 6 5 0 0 0 + 4 _ 2

and any multiple of 1000 is divisible be 8, it must be for the first case

8 _ 6 divisible by 8

and for the second case

4 _ 2 divisible by 8

This leaves only the following possibilities

_ 4 _ 2 5 8 1 6 _ 0

_ 4 _ 2 5 8 9 6 _ 0

for the first combination and

_ 8 _ 6 5 4 3 2 _ 0

_ 8 _ 6 5 4 7 2 _ 0

for the second combination.

Since the first group of three digit sums up to a multiple of 3

we only have these chances

1 4 7 2 5 8 9 6 3 0 (fails 7 test) and 7 4 1 2 5 8 9 6 3 0 (fails 7 test)

for the first combination. while, for the second,

1 8 9 6 5 4 3 2 7 0 (fails 7 test) 9 8 1 6 5 4 3 2 7 0 (fails 7 test)

1 8 9 6 5 4 7 2 3 0 (fails 7 test) 9 8 1 6 5 4 7 2 3 0 (fails 7 test)

1 8 3 6 5 4 7 2 9 0 (fails 7 test) 3 8 1 6 5 4 7 2 9 0

7 8 9 6 5 4 3 2 1 0 (fails 7 test) 9 8 7 6 5 4 3 2 1 0 (fails 7 test)

So the only possibility left is 3816547290.

Why 381654729 is awesome

Here's my programming solution, but I used bit maths to make it better, by using the divisibility tests of $2,4,5,6,8$

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from itertools import permutations
for i in permutations('0123456789'):
    k=''.join(i)
    if k[9]=='0':
        a=k[0]
        b=k[1]
        c=k[2]
        d=k[3]
        e=k[4]
        f=k[5]
        g=k[6]
        h=k[7]
        if int(b)%2==0 and int(a+b+c)%3==0 and int(c+d)%4==0 and int(e)%5==0 and int(d+e+f)%6==0 and int(a+b+c+d+e+f+g)%7==0 and int(f+g+h)%8==0:
            print k

This printed for me, the answer as $\boxed{3816547290}$ , not too long time...