The point of giving large numbers, like 100 7s, is to force analytical solution, rather than heuristic approach with actual multiplication or division. Let's construct a number consisting of just 5s and 7s, which is divisible by 100 7s.

First, we note that all 7s divided by 7 yields all 1s, which may be easier to work with. Although, we add a new restriction: multiplier must be divisible by 7.

When multiplying 111...111 by a number, the digits of the product are going to be sums of the digits of the multiplier plus carry, if any. For example, take 525 as a multiplier. The rightmost digit of the product is 5, next one is 5+2=7, after that 5+2+5 resulting in 2, then digit 3 will be repeated: 5+2+5+carry(1). At the left edge, the first 5 will drop out resulting in 5+2+carry(1) = 8, then 2 goes away, leaving 5 as the leftmost digit of the product.

However, we want only 5 and 7 as our product digits. If we carry the leftmost 5 outside 100th digit perimeter putting zeros in the middle, the rightmost 5 shall drop out when the leftmost 5 kicks in. That is, 100 1s multiplied by 101-digit multiplier 5000....00025 will result in 5 on the left and right with a whole bunch of 7s in the middle. Unfortunately, 101-digit number 5000...00025 is not exactly divisible by 7 (yields the remainder of 1.)

Building a number divisible by 7 is an interesting task in itself.

Note the sequence of remainders, the powers of 10 yield when divided by 7:

10^{0} ≡ **1** (mod 7)

10^{1} ≡ **3**** **(mod 7)

10^{2 }^{ }≡ **2**** **(mod 7)

10^{3} ≡ **6**** **(mod 7)

10^{4} ≡ **4**** **(mod 7)

10^{5} ≡ **5**** **(mod 7)

That sequence of six remainders (1,3,2,6,4,5) is periodic. Curiously, pairs of remainders at the intervals of 3 add up to 7: 1+6=3+4=2+5=7. That leads to many marvelous divisibility rules. E.g., any number with same digits at intervals of 3 is divisible by 7, like 532,532.

So let's build a 101-digit number of the form 500...00200...005, which is divisible by 7. We designate the positions in the large number as 012345,012345,... from right to left. Thus the rightmost 5 is in the position 0 and the leftmost 5 is in the position 5, since 101 ≡ 5 (mod 6). The two 5s add up to the remainder 5*1+5*5=30, or remainder of 2 from division by 7. Now, if place digit 2 in the position 3, we add 2*6=12 to the remainder of 2 for a total of 14, which is divisible by 7.

Thus, **101-digit number 500...002005 **is divisible by 7 and when multiplied by 100 1s will result in 3 5s on the right, followed by a whole bunch of 7s, and a bunch of 5s on the left.

Now I want to go back to the sequences of remainders from division of powers of 10 by some prime number greater than 5. Those sequences seem to be a kin to the Little Fermat's Theorem.

Number 13 gives a similar sequence, also of a period 6: (1,10,9,12,3,4). Again, pairs of remainders at interval of 3 add up to 13. And so we know, that a number like 532,532 is divisible by 13 in addition to being divisible by 7. For similar reasons that number is divisible by 11, which has its periodic sequence of length 2: (1,10). However, the fact that 532,532 is also divisible by 19 is purely coincidental.

After a quick study of some other prime number periodic sequences, I can come up with some examples, like 227,772 is divisible by 37. Periodic sequence for 37 being: (1,10,26).

Using those sequences one could come up with a multitude of number tricks.

Since we have been running out of the unproved theorems in Number Theory of late, let me introduce this conjecture:

**Prove (disprove) that for any prime number P > 5 as described above, its periodic sequence of remainders is a multiple of P.**