Answer:
[tex]33[/tex].
Step-by-step explanation:
There are a total of [tex](100 - 1) + 1 = 100[/tex] integers between [tex]1[/tex] and [tex]100[/tex]. The goal is to find the number of integers in that range that are divisible by neither [tex]2[/tex] nor [tex]3[/tex].
The least common multiple of [tex]2[/tex] and [tex]3[/tex] is [tex]6[/tex]. When dividing that integer by [tex]6[/tex], the remainder would be an integer between [tex]0[/tex] and [tex]5[/tex], inclusive. However, if this integer is divisible by neither [tex]2[/tex] nor [tex]3[/tex]. the remainder must also be divisible by neither [tex]2[/tex] nor [tex]3[/tex], meaning that [tex]1[/tex] and [tex]5[/tex] are the only possible remainders.
Hence, finding the number of integers between [tex]1[/tex] and [tex]100[/tex] (inclusive) divisible by neither [tex]2[/tex] nor [tex]3[/tex] is equivalent to finding the number of integers in that range for which the remainder is [tex]1[/tex] or [tex]5[/tex] when divided with [tex]6[/tex].
The list of integers between between [tex]1[/tex] and [tex]100[/tex] (inclusive) where the remainder is [tex]1[/tex] (mod [tex]6[/tex]) forms an arithmetic procession with first term [tex]1[/tex], last term [tex]96 + 1 = 97[/tex], and common difference [tex]6[/tex]. The number of items in this sequence would be:
[tex]\displaystyle \frac{97 - 1}{6} + 1 = 16 + 1 = 17[/tex].
Similarly, the integers in this range where remainder is [tex]5[/tex] (mod [tex]6[/tex]) would forms an arithmetic procession with first term [tex]5[/tex], last term [tex]90 + 5 = 95[/tex], and common difference [tex]6[/tex]. Number of items in this sequence:
[tex]\displaystyle \frac{95- 5}{6} + 1 = 15 + 1 = 16[/tex].
Note that these two sets of numbers are disjoint from one another. Hence, the number of elements in their union would be equal to the sum of elements in each set:
[tex]17 + 16 = 33[/tex].
Hence, there would be a total of [tex]33[/tex] numbers between [tex]1[/tex] and [tex]100[/tex] (inclusive) that are divisible by neither [tex]2[/tex] nor [tex]3[/tex].
