Divisibility by 11

Use the divisibility by $11$ criterion: The decimal number $N$ is divisible by $11$ if and only if the sum of the 1st, 3rd, 5th, $\dots$ digits minus the sum of the 2nd, 4th, 6th, $\dots$ digits is divisible by $11$ .

The criterion shows that the given $N$ is divisible by $11$ . A simple division yields $122\dots 21 \div 11=111\cdots1$ , a number of $2002$ ones. Again by the criterion, this number is divisible by $11$ . Now $11\dots11\div11=1010\cdots101$ , a number with $1001$ ones and $1000$ zeros. $1001$ is divisible by $11$ , hence $N$ is divisible by $11^3$ . Now $1010\dots01=10^{2000}+10^{1998}+\cdots+10^2+1\equiv1+1+\cdots+1=1001\pmod {11}$ . Also $1001\div11=91$ , which is not divisible by $11$ . So $11^3$ is the largest power of $11$ that divides $N$ .

$11^3$ = $\boxed{1331}$

If you add $\frac{11}{9} = 1.22222222222...$ to the number, the result is $\frac{11}{9}$ shifted over by $2002$ places. That is, $x + \frac{11}{9} = 10^{2002} \frac{11}{9}$ So $x = \frac{11}{9} (10^{2002} - 1)$ It remains to find out how many times $11$ divides $10^{2002} - 1$ . By binomial theorem, $\begin{aligned} 10^{2002} - 1 &= \sum_{i=0}^\infty {2002 \choose i} 11^i (-1)^{2002-i} - 1\\ &= \sum_{i=0}^\infty (-1)^i {2002 \choose i} 11^i - 1\\ &= \sum_{i=1}^\infty (-1)^i {2002 \choose i} 11^i \\ &= - {2002 \choose 1} 11^1 + {2002 \choose 2} 11^2 - {2002 \choose 3} 11^3 + \cdots + {2002 \choose 2002} 11^{2002} \end{aligned}$ But we need only look at the first few terms. $2002$ contains exactly one factor of $11$ , which means $- {2002 \choose 1} 11^1$ contains two factors of $11$ and ${2002 \choose 2} 11^2$ contains three. All the remaining terms are clearly divisible by $11^3$ , so everything but the first term is divisible by $11^3$ . This means the sum is divisible by $11^2$ but not by $11^3$ .

Coming back to the original number, [ \frac{11}{9} $10^{2002} - 1$ ] is therefore divisible by $11$ exactly $3$ times; the answer is $11^3 = \boxed{1331}$ .

Let's use the notation $(abc)_{n}$ to denote a string of $n$ repetitions of $abc$ . So, $N=1(2)_{2001}1$ .

$N = 1(2)_{2001}1 = 11 \times (1)_{2002}$ .

Since $2002 = 2 \times 7 \times 11 \times 13$ , the string of 2002 1's can be split in several ways. To mention a few:

$(1)_{2002} = (11)_{1001} = 11 \times (01)_{1001}$ , formula (I)

$(1)_{2002} = (1111111)_{286} = 1111111 \times (0000001)_{286}$ , formula (II)

$(1)_{2002} = (1)_{1001} (1)_{1001} = (1)_{1001} \times 1(0)_{1000}1$ , formula (III).

The 11-test looks like the 9-test. Where the 9-test adds all digits repeatedly and shows divisibility by 9 if the result equals 0, the 11-test alternates adding and subtracting digits: e.g. $3723764$ gives $4-6+7-3+2-7+3$ , or $(4+7+2+3) - (6+3+7) = 0$ so $3723764$ is divisable by 11.

If a number exists of 1's (and 0's) only, in the 11-test, all 1's in odd positions will cancel out 1's in even positions.

I choose formula (I) to progress:

$(01)_{1001}$ is divisable by 11 because it consists of 1001 ones on the '+1 position' and the 11-test gives $1001 \equiv 0 \pmod {11}$ .

$(01)_{1001} = (01)_{11} \times (0000000000000000000001)_{91}$ . The number $(00 00 00 00 00 00 00 00 00 00 01)_{91}$ cannot be divisable by 11, using the 11-test.

So, possible extra divisors of 11 have to be found in $(01)_{11}$ .

$(01)_{11} = 11 \times (9 \: 18 \: 27 \: 36 \: 45 \: 54 \: 63 \: 72 \: 81 \: 91)$ , where I added blanks to enhance reading.

Now using the 11-test, $918273645546372819$ is symmetrical, cancelling out everyting, leaving the trailing 1 to give an 11-test resulting in 1.

Concusion: $N = 1(2)_{2001}1 =$

$= 11 \times (1)_{2002}$

$= 11 \times 11 \times (01)_{1001}$

$= 11 \times 11 \times (01)_{11} \times (0000000000000000000001)_{91}$ $= 11 \times 11 \times 11 \times 9182736455463728191 \times (0000000000000000000001)_{91}$ $= 11^3 \times 9182736455463728191 \times (0000000000000000000001)_{91}$ $= \boxed{N = 1331 \times 9182736455463728191 \times (0000000000000000000001)_{91}}$ .