4. Photographs are archived in the order of their numbering in identical albums, with exactly 4 photographs per page. In this case, the 81st photograph in sequence landed on the 5th page of one of the albums, and the 171st photograph landed on the 3rd page of another. How many photographs can each album hold?

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# Answer: 32.

Solution. Let $x, y$ be the album numbers in which the 81st and 171st photos are placed, respectively, and $n>4$ be the number of pages in the album. Then $4 n(x-1)+16<81 \leqslant 4 n(x-1)+20, 4 n(y-1)+8<171 \leqslant 4 n(y-1)+12$, i.e., $61 \leqslant 4 n(x-1)<65, 159 \leqslant 4 n(y-1)<163$. Therefore, $n(x-1)=16, n(y-1)=40$. From the first inequality, it follows that $n$ can be $1,2,4,8,16$, from the second $1,2,4,5,8,10,20$, 40. Thus, $n=8, 4 n=32$.

Answer to variant 2: 54.

Answer to variant 3: 40.

Answer to variant $4: 42$.