112. Factorization. Into what factors does the number 1000000000001 decompose? This question is easy to answer if you know something about numbers of this particular form. It is no less easy to specify two factors if, instead of 11 zeros, you insert, for example, 101 zeros between the two ones.

There is one curious, simple, and beautiful rule for all such cases. Can you find it?
112. If the number of zeros enclosed between two ones is equal to any number that is a multiple of 3 plus 2, then the two factors can always be written down immediately using the following curious rule: \(1001=11 \times 91 ; 1000001=101 \times 9901 ; 1000000001=\) \(1001 \times 999001 ; 1000000000001=10001 \times 99990001\). In the last case, we get the required answer, and \(10001=73 \times 137\). The multiplicity of 3 in 11 is \(3(11=3 \times 3+2)\). Therefore, in each factor, we insert three zeros and add an extra nine.

If our number, as I assumed, contained 101 zeros, then the largest number by which 3 can be multiplied without exceeding 101 would be 33, and the factors would contain 33 zeros and 34 nines and would have the form indicated above. If the number of zeros in our number were even, then you could find the two factors as follows: \(1001=11 \times 91\); \(100001=11 \times 9091 ; 10000001=11 \times 909091\) and so on.