Question

Raina must choose a number between 55 and 101 that is a multiple of 2, 8 and 10. Write all the numbers that she could choose. If there is more than one number, separate them with commas.

Answer 1

Answer:

80

Step-by-step explanation:

80 is a multiple of 2, 8, and 10

*PLz mark brainlies if this helps*

Answer 2

Answer:

The only possible number is $80$.

Step-by-step explanation:

The number in question needs to be a multiple of all three of $2$, $8$, and $10$. 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:

• $2$ is a prime number itself.
• $8 = 2^3$.
• $10 = 2 \times 5$.

The only prime factors are $2$ and $5$.

• The greatest power of $2$ among the three numbers is $3$.
• The greatest power of $5$ among the three numbers is $1$.

Therefore, the least common multiplier of the three number should be the product of $2^3$ and $5$. That's equal to $2^3 \times 5 = 8 \times 5 = 40$.

In other words, the number (or numbers) in question could be written in the form $40\, k$, where $k$ is an integer.

The question requires that this number be between $55$ and $101$. In other words,

$55 \le 40\, k \le 101$.

The goal is to find the possible values of $k$. Note that from integer division by $40$,

• $55 = 1 \times 40 + 15$, and
• $101 = 2 \times 40 + 21$.

The inequality becomes:

$1 \times 40 + 15 \le 40\, k \le 2 \times 40 + 21$.

However,

• $1 \times 40 < 55 = 1 \times 40 + 15$, and
• $2 \times 40 + 21 = 101 < 3 \times 40$.

Hence,

$1 \times 40 < 1\times 40 + 15 \le 40\, k \le 2 \times 40 + 21 < 3 \times 40$.

$1 \times 40 < 40\, k < 3 \times 40$.

Divide by the positive number $40$ to obtain:

$1 < k < 3$.

Since $k$ is an integers, $k = 2$.

Indeed, $40 \, k = 80$ is between $55$ and $101$.

Therefore, $80$ is the number in question.