\begin{aligned} ext{If } x &= 2^3 imes 3^2 imes 5^3 imes 7^3 \ y &= 2^2 imes 3^3 imes 5^4 imes 7^3 ext{, and } \ z &= 2^4 imes 3^4 imes 5^2 imes 7^5 ext{, } \end{aligned}
then H.C.F. of x, y and z is

Q: \begin{aligned} \text{If } x &= 2^3 \times 3^2 \times 5^3 \times 7^3 \\ y &= 2^2 \times 3^3 \times 5^4 \times 7^3 \text{, and } \\ z &= 2^4 \times 3^4 \times 5^2 \times 7^5 \text{, } \end{aligned} then H.C.F. of x, y and z is

\begin{aligned} \text{If } x &= 2^3 \times 3^2 \times 5^3 \times 7^3 \\ y &= 2^2 \times 3^3 \times 5^4 \times 7^3 \text{, and } \\ z &= 2^4 \times 3^4 \times 5^2 \times 7^5 \text{, } \end{aligned} then H.C.F. of x, y and z is

Question

Accepted Answer

To determine the H.C.F. (Highest Common Factor) of the numbers $x$, $y$, and $z$, we need to find the smallest power of each prime factor that appears in all three numbers.

Given:
$$ x = 2^3 \times 3^2 \times 5^3 \times 7^3 $$
$$ y = 2^2 \times 3^3 \times 5^4 \times 7^3 $$
$$ z = 2^4 \times 3^4 \times 5^2 \times 7^5 $$

**Step-by-step Explanation:**

1. **Prime Factor 2:**
   - In $x$, the power of 2 is 3.
   - In $y$, the power of 2 is 2.
   - In $z$, the power of 2 is 4.
   - The smallest power of 2 among these is 2.

2. **Prime Factor 3:**
   - In $x$, the power of 3 is 2.
   - In $y$, the power of 3 is 3.
   - In $z$, the power of 3 is 4.
   - The smallest power of 3 among these is 2.

3. **Prime Factor 5:**
   - In $x$, the power of 5 is 3.
   - In $y$, the power of 5 is 4.
   - In $z$, the power of 5 is 2.
   - The smallest power of 5 among these is 2.

4. **Prime Factor 7:**
   - In $x$, the power of 7 is 3.
   - In $y$, the power of 7 is 3.
   - In $z$, the power of 7 is 5.
   - The smallest power of 7 among these is 3.

Thus, the H.C.F. of $x$, $y$, and $z$ is:
$$ 2^2 \times 3^2 \times 5^2 \times 7^3 $$

Now, let's express this in terms of the options given:

- $2^2 \times 3^2 \times 5^2 = (2 \times 3 \times 5)^2 = 30^2$

Therefore, the H.C.F. is:
$$ 30^2 \times 7^3 $$

**Justification for Option 1:**
- Option 1 states $ (30)^2 \times 7^3 $, which matches exactly with our calculated H.C.F. Therefore, Option 1 is correct.

**Explanation for Incorrectness of Other Options:**

- **Option 2: $ (15)^3 \times 7^4 $**
  - This option suggests a different combination of powers that does not match the calculated H.C.F. The powers of 2, 3, and 5 do not align with the smallest powers found in $x$, $y$, and $z$.

- **Option 3: $ (30)^3 \times 7^3 $**
  - This option has the correct power of 7 but incorrectly suggests a higher power for the combination of 2, 3, and 5. The power should be 2, not 3.

- **Option 4: $ 30 \times 7^5 $**
  - This option has incorrect powers for both the combination of 2, 3, and 5, and for 7. The power of 7 should be 3, not 5.

In conclusion, Option 1 is the correct answer as it accurately represents the H.C.F. of the given numbers.