consider $x\in[0,3)$ we find 22,$x\in[0,1)$,we find 8. And $x\in[k,l)$ is the same number with $x\in[k+3,l+3)$ So $22\times33+8=734$

Excuse me, but what does a [x] mean here? Does it mean the nearest integer to x?

wangsacl wrote: Excuse me, but what does a [x] mean here? Does it mean the nearest integer to x? $ [x]$ is meant as the biggest integer smaller than $ x$. So $ [x]\leq x<[x]+1$

jin wrote: And $x\in[k,l)$ is the same number with $x\in[k+3,l+3)$ How did you get this?

Bumping this thread to see if anybody knows?

Jin's solution is correct, but let me formalize it. Let $ g(a, b) $ be the number of distinct integer values $ f $ takes on the interval $ [a, b) $. Lemma 1: $ g(a, b) = g(a + 3, b + 3) $ for all $ a, b \in \mathbb{R} $. Proof: Let $ S = \{z \in \mathbb{Z} : \exists r \in [a, b)\:s.t.\:f(r) = z\} $ and let $ T = \{z \in \mathbb{Z} : \exists r \in [a + 3, b + 3) \:s.t. \:f(r) = z\} $. Letting $ p(x) = x + 35 $ I shall show that $ p: S \longrightarrow T $ is a bijection which will imply the desired result. Clearly if $ p(j) = p(k) $ then $ j = k $ so we have an injection. Now to show a surjection it suffices to show that for all $ t \in T $ we have that $ t - 35 \in S $ which is clear since if $ f(r) = t $ then $ f(r - 3) = t - 35 $. Therefore $ |S| = |T| $ which implies the lemma. Lemma 2: $ g(0, 3) = 22 $. Proof: By counting we get that letting $ x \in [0, 3) $ the integers that are reachable (from $ f $) are $ 0, 1, 2, 4, 5, 6, 7, 11, 12, 13, 14, 16, 17, 18, 19, 23, 24, 25, 26, 28, 29, 30 $ Lemma 3: $ g(0, 1) = 7 $ Proof: By counting we get that letting $ x \in [0, 1) $ the integers that are reachable (from $ f $) are $ 0, 1, 2, 4, 5, 6, 7 $ Now the answer to the APMO question itself is given by $ \sum_{i = 0}^{32}g(3i, 3i + 3) + g(99, 100) + 1 = 33g(0, 3) + g(0, 1) + 1 = 734 $ as desired.