The answer is 101010101.

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Denote this number $N$ . Note that since $N$ contains only 1's and 0's, it can be written as the sum of various powers of 10, or, more formally,

$N=\sum _{ i=0 }^{ n }{ { a }_{ i }\cdot { 10 }^{ i } }$

where $a_i$ is equal to either 1 or 0. Now, consider the first few values of ${ 10 }^{ i }\pmod {22}$ .

${ 10 }^{ 0 }\pmod {22}=1\\ { 10 }^{ 1 }\pmod {22}=10\\ { 10 }^{ 2 }\pmod {22}=12\\ { 10 }^{ 3 }\pmod {22}=10\\ { 10 }^{ 4 }\pmod {22}=12$

After a few trials a pattern quickly emerges; when $i$ is even, it leaves a remainder of 12, and when $i$ is odd, it leaves a remainder of 10. Therefore, $N$ is some combination of powers of 10 leaving remainders of 1, 10 and 12. Since $N\equiv 5\pmod {22}$ , the sum of the remainders of this combination must also be. Let $a$ represent the number of ${10}^{1}$ 's in $N$ , $b$ represent the number of ${10}^{2k+1}$ 's and $c$ represent the number of ${10}^{2k}$ 's. We have that:

$a+10b+12c\equiv 5\pmod {22}$ .

After trialing a few values for $a$ , $b$ and $c$ , we see that $a=1$ , $b=0$ , $c=4$ satisfies the conditions and gives us the least value of $N$ .

Since $a=1$ , we have one ${10}^{0}$ .

Since $b=0$ , we have no ${10}^{2k+1}$ 's.

Since $c=4$ , we have four ${10}^{2k}$ 's, and we choose $k=1,2,3,4$ to give us the least number of digits of $N$ .

Hence, $N={ 10 }^{ 0 }+{ 10 }^{ 2 }+{ 10 }^{ 4 }+{ 10 }^{ 6 }+{ 10 }^{ 8 }$

$N=\boxed { 101010101 }$ .