Problem 2

Find the largest positive integer

N

such that the number of integers in

\{1,2,\dots,N\}

divisible by

3

equals the number of integers divisible by

5

7

(or both).

Step 4 of 6: Rule out all N at least 69

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D(N+105)=D(N)+2,\qquad \min_{69\le N\le173}D(N)=1\implies D(N)>0\text{ for all }N\ge69.

Detailed analysis

The floor expression has period 105 up to a constant: $D(N+105)=D(N)+2$ . A direct finite check of $69\le N\le173$ gives $\min D(N)=1$ . Adding 105 increases D by 2, so $D(N)>0$ for every $N\ge69$ ; hence every solution satisfies $N<69$ .