Question

When a number is a product of 6 and 225 and a multiplier is a perfect square number

Answer 1

To analyze the scenario where a number is the product of 6 and 225, we first calculate the product:

\[
6 \times 225 = 1350
\]

Next, we want to determine how this number relates to perfect square multipliers.

### Step 1: Factorization of 1350

We can factor 1350 to understand its prime components:

\[
1350 = 6 \times 225 = 2 \times 3 \times 15^2
\]

Further breaking it down:

\[
15 = 3 \times 5 \quad \Rightarrow \quad 15^2 = (3 \times 5)^2 = 3^2 \times 5^2
\]

Thus,

\[
225 = 15^2 = 3^2 \times 5^2
\]

Combining everything:

\[
1350 = 2 \times 3^3 \times 5^2
\]

### Step 2: Expressing 1350 in terms of primes

Now, we express 1350 as:

\[
1350 = 2^1 \times 3^3 \times 5^2
\]

### Step 3: Perfect Square Multiplier

A **perfect square** is a number that can be expressed as \(n^2\) for some integer \(n\). The prime factorization of a perfect square must have even powers for all prime factors.

To find a perfect square multiplier for 1350, we analyze the exponents in the factorization of 1350:

- \(2^1\) (needs to be raised to \(2^2\), so we need one more \(2\))
- \(3^3\) (needs to be raised to \(3^4\), so we need one more \(3\))
- \(5^2\) (already a perfect square)

### Step 4: Finding the multiplier

The necessary add-on to make the product a perfect square is:

\[
2^1 \times 3^1
\]

Thus, a multiplier of:

\[
2^1 \times 3^1 = 2 \times 3 = 6
\]

### Step 5: Conclusion

To find a perfect square number that can be multiplied with 1350 to result in a perfect square:

\[
\text{Perfect Square Multiplier} = 6
\]

Therefore, if we multiply 1350 by its multiplier of 6:

\[
1350 \times 6 = 8100 \quad \text{(which is a perfect square since } 90^2 = 8100)
\]

In summary, 1350 is the product of 6 and 225, and for it to become a perfect square through multiplication, we multiply it by the perfect square multiplier of \(6\).