Aa) We need to find the maximum number of pages Ai Lin can have for the scrapbook such that every page contains the same number of photographs and newspaper cuttings.
Let the number of photographs and newspaper cuttings per page be x.
Total number of photographs = 24
Total number of newspaper cuttings = 42
To find the maximum number of pages, we need to divide the total number of photographs and newspaper cuttings by the number of photographs and newspaper cuttings per page respectively and take the ceiling function to round up to the nearest integer, since we need an integer number of pages.
Number of pages = ceil(24/x) + ceil(42/x)
We want to maximize the number of pages, so we need to minimize x.
For x = 6, we get:
Number of pages = ceil(24/6) + ceil(42/6) = 4 + 7 = 11
For x = 5, we get:
Number of pages = ceil(24/5) + ceil(42/5) = 5 + 9 = 14
For x = 4, we get:
Number of pages = ceil(24/4) + ceil(42/4) = 6 + 11 = 17
For x = 3, we get:
Number of pages = ceil(24/3) + ceil(42/3) = 8 + 14 = 22
For x = 2, we get:
Number of pages = ceil(24/2) + ceil(42/2) = 12 + 21 = 33
For x = 1, we get:
Number of pages = ceil(24/1) + ceil(42/1) = 24 + 42 = 66
Therefore, the maximum number of pages Ai Lin can have for the scrapbook is 17.
b) For each page of the scrapbook, there will be 3 photographs and 6 newspaper cuttings.
We found this by dividing the total number of photographs and newspaper cuttings by the maximum number of pages, which is 17:
Number of photographs per page = 24/17 ≈ 1.41 ≈ 3 (rounded up)
Number of newspaper cuttings per page = 42/17 ≈ 2.47 ≈ 6 (rounded up)
Therefore, for each page of the scrapbook, there will be 3 photographs and 6 newspaper cuttings.