Answer:
The only possible number is [tex]80[/tex].
Step-by-step explanation:
The number in question needs to be a multiple of all three of [tex]2[/tex], [tex]8[/tex], and [tex]10[/tex]. As a result, it must also be a multiple of the least common multiplier (lcm) of the three number.
Start by finding the least common multiplier of the three numbers.
Factor each number into its prime components:
- [tex]2[/tex] is a prime number itself.
- [tex]8 = 2^3[/tex].
- [tex]10 = 2 \times 5[/tex].
The only prime factors are [tex]2[/tex] and [tex]5[/tex].
- The greatest power of [tex]2[/tex] among the three numbers is [tex]3[/tex].
- The greatest power of [tex]5[/tex] among the three numbers is [tex]1[/tex].
Therefore, the least common multiplier of the three number should be the product of [tex]2^3[/tex] and [tex]5[/tex]. That's equal to [tex]2^3 \times 5 = 8 \times 5 = 40[/tex].
In other words, the number (or numbers) in question could be written in the form [tex]40\, k[/tex], where [tex]k[/tex] is an integer.
The question requires that this number be between [tex]55[/tex] and [tex]101[/tex]. In other words,
[tex]55 \le 40\, k \le 101[/tex].
The goal is to find the possible values of [tex]k[/tex]. Note that from integer division by [tex]40[/tex],
- [tex]55 = 1 \times 40 + 15[/tex], and
- [tex]101 = 2 \times 40 + 21[/tex].
The inequality becomes:
[tex]1 \times 40 + 15 \le 40\, k \le 2 \times 40 + 21[/tex].
However,
- [tex]1 \times 40 < 55 = 1 \times 40 + 15[/tex], and
- [tex]2 \times 40 + 21 = 101 < 3 \times 40[/tex].
Hence,
[tex]1 \times 40 < 1\times 40 + 15 \le 40\, k \le 2 \times 40 + 21 < 3 \times 40[/tex].
[tex]1 \times 40 < 40\, k < 3 \times 40[/tex].
Divide by the positive number [tex]40[/tex] to obtain:
[tex]1 < k < 3[/tex].
Since [tex]k[/tex] is an integers, [tex]k = 2[/tex].
Indeed, [tex]40 \, k = 80[/tex] is between [tex]55[/tex] and [tex]101[/tex].
Therefore, [tex]80[/tex] is the number in question.
