Write each number as the product of prime faetors using LADDER method AN determine the LCM and HCF of each set of numbers $$ 54: 36: 60 $$

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1. **Prime Factorization of Each Number**

We start by writing the numbers $$54$$, $$36$$, and $$60$$ side by side and then “laddering” by dividing with primes.

**Step 1: Divide by $$2$$**  
   Since all three numbers are even, we can divide each by $$2$$:

$$
   \begin{array}{cccc}
   2 & | & 54 & 36 \quad 60 \
     &   & \downarrow & \downarrow \quad \downarrow \
     &   & 27 & 18 \quad 30 \
   \end{array}
   $$

**Step 2: Divide by $$3$$**  
   Now, look at the new row: $$27$$, $$18$$, and $$30$$. All are divisible by $$3$$:

$$
   \begin{array}{cccc}
   2 & 3 & | & 54 \quad 36 \quad 60 \
     &   &   & \downarrow \quad \downarrow \quad \downarrow \
     &   &   & 27 \quad 18 \quad 30 \
     &   & 3 & \downarrow \quad \downarrow \quad \downarrow \
     &   &   & 9 \quad 6 \quad 10 \
   \end{array}
   $$

**Step 3: Divide by $$2$$**  
   In the row $$9$$, $$6$$, $$10$$, both $$6$$ and $$10$$ are divisible by $$2$$ (while $$9$$ is not), so we divide those numbers:

$$
   \begin{array}{cccc}
   2 & 3 & 2 & | \; 54 \quad 36 \quad 60 \
     &   &   & \downarrow \quad \downarrow \quad \downarrow \
     &   &   & 27 \quad 18 \quad 30 \
     &   & 3 & \downarrow \quad \downarrow \quad \downarrow \
     &   &   & 9 \quad 6 \quad 10 \
     &   &   &  \quad (\text{divide }6\text{ and }10\text{ by }2)\
     &   &   & 9 \quad 3 \quad 5\
   \end{array}
   $$

**Step 4: Divide by $$3$$**  
   Now, in the row $$9$$, $$3$$, $$5$$, the first two numbers are divisible by $$3$$:

$$
   \begin{array}{cccc}
   2 & 3 & 2 & 3\, | \; 54 \quad 36 \quad 60 \
     &   &   &   & \downarrow \quad \downarrow \quad \downarrow \
     &   &   &   & 27 \quad 18 \quad 30 \
     &   & 3 &   & \downarrow \quad \downarrow \quad \downarrow \
     &   &   &   & 9 \quad 6 \quad 10 \
     &   &   &   &  \quad (\text{divide }6\text{ and }10\text{ by }2)\
     &   &   &   & 9 \quad 3 \quad 5\
     &   &   &   & (\text{divide }9\text{ and }3\text{ by }3)\
     &   &   &   & 3 \quad 1 \quad 5\
   \end{array}
   $$

**Step 5: Divide by $$3$$**  
   In the row $$3$$, $$1$$, $$5$$, only the first number is divisible by $$3$$:

$$
   \begin{array}{cccc}
   2 & 3 & 2 & 3\, 3\, | \; 54 \quad 36 \quad 60 \
     &   &   &      & \downarrow \quad \downarrow \quad \downarrow \
     &   &   &      & 27 \quad 18 \quad 30 \
     &   & 3 &      & \downarrow \quad \downarrow \quad \downarrow \
     &   &   &      & 9 \quad 6 \quad 10 \
     &   &   &      & (\text{after dividing by }2)\
     &   &   &      & 9 \quad 3 \quad 5\
     &   &   &      & (\text{after dividing by }3)\
     &   &   &      & 3 \quad 1 \quad 5\
     &   &   &      & (\text{divide }3\text{ by }3)\
     &   &   &      & 1 \quad 1 \quad 5\
   \end{array}
   $$

**Step 6: Divide by $$5$$**  
   Finally, in the row $$1$$, $$1$$, $$5$$, the third number is divisible by $$5$$:

$$
   \begin{array}{cccc}
   2 & 3 & 2 & 3\, 3\, 5\, | \; 54 \quad 36 \quad 60 \
     &   &   &          & \downarrow \quad \downarrow \quad \downarrow \
     &   &   &          & 27 \quad 18 \quad 30 \
     &   & 3 &          & \downarrow \quad \downarrow \quad \downarrow \
     &   &   &          & 9 \quad 6 \quad 10 \
     &   &   &          & (\text{after dividing by }2)\
     &   &   &          & 9 \quad 3 \quad 5\
     &   &   &          & (\text{after dividing by }3)\
     &   &   &          & 3 \quad 1 \quad 5\
     &   &   &          & (\text{after dividing }3\text{ by }3)\
     &   &   &          & 1 \quad 1 \quad 5\
     &   &   &          & (\text{divide }5\text{ by }5)\
     &   &   &          & 1 \quad 1 \quad 1\
   \end{array}
   $$

2. **Expressing the Prime Factorizations**

The primes on the left represent the factors extracted in order. For each original number, the prime factors are:

- For $$54$$:  
     $$
     54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3
     $$

- For $$36$$:  
     $$
     36 = 2 \times 3 \times 2 \times 3 = 2^2 \times 3^2
     $$

- For $$60$$:  
     $$
     60 = 2 \times 3 \times 2 \times 5 = 2^2 \times 3 \times 5
     $$

3. **Finding the HCF (Highest Common Factor)**

To find the HCF, we take the common prime factors with the smallest exponents in all three numbers.

- For prime $$2$$: Exponents are $$1$$ (in $$54$$), $$2$$ (in $$36$$), and $$2$$ (in $$60$$). The smallest exponent is $$1$$.
   - For prime $$3$$: Exponents are $$3$$ (in $$54$$), $$2$$ (in $$36$$), and $$1$$ (in $$60$$). The smallest exponent is $$1$$.
   - The prime $$5$$ does not appear in all three.

Hence, the HCF is:
   $$
   \text{HCF} = 2^1 \times 3^1 = 6
   $$

4. **Finding the LCM (Least Common Multiple)**

To find the LCM, we take each prime factor with the highest exponent present in any of the numbers.

- For prime $$2$$: The highest exponent among the numbers is $$2$$.
   - For prime $$3$$: The highest exponent is $$3$$ (from $$54$$).
   - For prime $$5$$: The highest exponent is $$1$$ (from $$60$$).

Hence, the LCM is:
   $$
   \text{LCM} = 2^2 \times 3^3 \times 5 = 4 \times 27 \times 5 = 540
   $$

5. **Final Answer**

$$
   \begin{aligned}
   54 &= 2 \times 3^3,\[1mm]
   36 &= 2^2 \times 3^2,\[1mm]
   60 &= 2^2 \times 3 \times 5,\[2mm]
   \text{HCF} &= 6,\[1mm]
   \text{LCM} &= 540.
   \end{aligned}
   $$
Prime factors: - 54: $$2 \times 3^3$$ - 36: $$2^2 \times 3^2$$ - 60: $$2^2 \times 3 \times 5$$ HCF: 6 LCM: 540

Write each number as the product of prime faetors using LADDER method AN determine the LCM and HCF of each set of numbers \( 54: 36: 60 \)

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