The solution of Part 1 is left to the reader (hint: use long division).
We are looking
for a number that is divisible by 3333...333 (100 3's). Such a
number would have to be divisible by 3 and by 1111...111 (100 1's)
(it can be shown by using the divisibility rule for 3 that the
latter itself is not divisible by 3). By the result of Part 1, this
number will have to be in the form of 1111...111 (100k
1's). This number will also have to be divisible by 3, and the only
numbers of such form that are are those where k is a
multiple of 3 (again, this can be shown by using the divisibility
rule for 3). The smallest such k is 3, so our number
1111...111 (300 1's).
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