getting started continued
3 consider the expression (2 cdot 2 cdot 2 cdot 3)
(a) how does (2 cdot 2 cdot 2 cdot 3) relate to the factors determined by cherise and jemma?
(b) how does it relate to 24?
4 when you know a factor of a given whole number, how can you determine another factor?
5 what is the remainder when you divide a whole number by any of its factors? explain your reasoning.
6 create a factor tree to show the prime factorization of each number.
(a) 66
(b) 210

Question

Accepted Answer

### Question 3 (a)
# Explanation:
## Step1: Analyze the expression
The expression is $2 \cdot 2 \cdot 2 \cdot 3$. We assume Charise and Jemma were finding factors (probably prime factors) of a number (likely 24, as in part (b)). This expression is a product of prime factors (2 is prime, 3 is prime). So it should be the prime factorization of the number they were factoring (24).
## Step2: Relate to factors
If Charise and Jemma were finding factors of 24, this expression is the prime factorization of 24, breaking it down into its prime components.
# Answer:
The expression $2 \cdot 2 \cdot 2 \cdot 3$ is the prime factorization of 24, so it represents the product of the prime factors determined by Charise and Jemma (if they were factoring 24).

### Question 3 (b)
# Explanation:
## Step1: Calculate the expression
Calculate $2 \cdot 2 \cdot 2 \cdot 3$. First, $2 \cdot 2 = 4$, then $4 \cdot 2 = 8$, then $8 \cdot 3 = 24$.
## Step2: Relate to 24
So the expression $2 \cdot 2 \cdot 2 \cdot 3$ equals 24, meaning it is the prime factorization of 24 (expressing 24 as a product of its prime factors).
# Answer:
The expression $2 \cdot 2 \cdot 2 \cdot 3$ equals 24 (since $2\times2\times2\times3 = 24$), and it is the prime factorization of 24.

### Question 4
# Explanation:
## Step1: Recall factor definition
A factor $f$ of a whole number $n$ means $n = f \times k$ for some whole number $k$.
## Step2: Solve for the other factor
To find the other factor $k$, we can divide the whole number $n$ by the known factor $f$. So $k=\frac{n}{f}$, as long as $f
eq0$ (and since $f$ is a factor of $n$, $f$ is a positive whole number, so division is valid here).
# Answer:
If $f$ is a factor of whole number $n$, then the other factor is $\frac{n}{f}$ (found by dividing $n$ by $f$).

### Question 5
# Explanation:
## Step1: Recall factor and division
By definition, a factor $f$ of a whole number $n$ means $n = f \times q$ for some whole number $q$ (the quotient).
## Step2: Analyze the remainder
When we divide $n$ by $f$, we have $n\div f = q$ with a remainder $r$. But from $n = f \times q + r$, and since $n = f \times q$ (because $f$ is a factor), substituting gives $f \times q = f \times q + r$, so $r = 0$.
# Answer:
The remainder is 0. Reason: If $f$ is a factor of $n$, then $n = f \times q$ for some whole number $q$. When dividing $n$ by $f$, $n\div f = q$ with remainder $r$, so $n = f\times q + r$. Substituting $n = f\times q$ gives $f\times q = f\times q + r$, so $r = 0$.

### Question 6 (a) - Prime Factorization of 66
# Explanation:
## Step1: Start with 66
We start with 66. We can divide 66 by 2 (the smallest prime factor of 66, since 66 is even). $66\div2 = 33$.
## Step2: Factor 33
Now factor 33. 33 can be divided by 3 (since $3\times11 = 33$). $33\div3 = 11$.
## Step3: Check 11
11 is a prime number (it has no factors other than 1 and itself).
## Step4: Write the prime factorization
So the prime factorization of 66 is $2\times3\times11$. The factor tree would be:
- 66
  - 2 (prime)
  - 33
    - 3 (prime)
    - 11 (prime)
# Answer:
The prime factorization of 66 is $2\times3\times11$. The factor tree: 66 splits into 2 and 33; 33 splits into 3 and 11 (both 3 and 11 are prime).

### Question 6 (b) - Prime Factorization of 210
# Explanation:
## Step1: Start with 210
Start with 210. It's even, so divide by 2: $210\div2 = 105$.
## Step2: Factor 105
105 ends with 5, so divide by 5: $105\div5 = 21$.
## Step3: Factor 21
21 can be divided by 3: $21\div3 = 7$.
## Step4: Check 7
7 is a prime number.
## Step5: Write the prime factorization
So the prime factorization of 210 is $2\times3\times5\times7$. The factor tree would be:
- 210
  - 2 (prime)
  - 105
    - 5 (prime)
    - 21
      - 3 (prime)
      - 7 (prime)
# Answer:
The prime factorization of 210 is $2\times3\times5\times7$. The factor tree: 210 splits into 2 and 105; 105 splits into 5 and 21; 21 splits into 3 and 7 (all 3, 5, 7 are prime).

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QUESTION IMAGE

Question

Explanation:

Question 3 (a)

Step1: Analyze the expression

Step2: Relate to factors

Step1: Calculate the expression

Step2: Relate to 24

Step1: Recall factor definition

Step2: Solve for the other factor

Answer:

Question 3 (b)