Tokorev S.i.

Can we select from the sequence $1, 1/2, 1/3, \ldots$ (preserving the order)

a) a hundred numbers,

b) an infinite subsequence of numbers,

such that each, starting from the third, is equal to the difference of the two preceding ones $\left(a_{k}=a_{k-2}-a_{k-1}\right)$?
a) Such a subsequence can be constructed, for example, as follows. We write down the first hundred different Fibonacci numbers $1,2,3,5,8,13, \ldots$. We divide all the numbers by their least common multiple and write down the results in reverse order. All fractions will simplify, and the resulting numbers will be from our sequence, written in descending order, with each number, starting from the third, being the difference of the two preceding ones.

b) The denominator of each fraction divides the LCM of the two preceding ones. Therefore, the denominators of all fractions divide the LCM of the first two. Consequently, there can only be a finite number of different denominators, which means that our subsequence is finite.

## Answer

a) Yes; b) No.