问题1781: 完美平方数

Let's analyze the problem step by step:

Given:
The product of 8 and \( p \) is a perfect square. That is, \( 8p \) is a perfect square.

We need to find which of the given options for \( p \) makes \( 8p \) a perfect square.

First, factorize 8:
\( 8 = 2^3 \)

So, \( 8p = 2^3 \times p \)

For \( 8p \) to be a perfect square, all exponents in its prime factorization must be even.

Therefore, \( 2^3 \times p \) must have even exponents for all primes.

The exponent of 2 is currently 3 (odd). To make it even, we need to multiply by at least one more factor of 2. So, \( p \) must contain at least one factor of 2 to make the exponent of 2 even: \( 2^3 \times 2 = 2^4 \) (which is even).

Additionally, \( p \) might have other primes, but their exponents must be even (if they are already present) or if not present, it doesn't matter. However, since we are only multiplying by \( p \), we need to ensure that the entire product has even exponents.

So, let \( p = 2 \times k \), where \( k \) is a perfect square (so that its prime exponents are even). Then:
\( 8p = 2^3 \times 2 \times k = 2^4 \times k \), which is a perfect square since \( 2^4 \) is a perfect square and \( k \) is a perfect square.

But more generally, \( p \) must be of the form \( 2 \times \) (a perfect square). Because:

• The factor 2 in \( p \) will make the exponent of 2 in \( 8p \) become \( 3+1=4 \) (even).
• The rest of \( p \) (which is a perfect square) ensures that all other primes have even exponents.

Now, let's check the options:

A. \( p = 3 \)
\( 8p = 8 \times 3 = 24 = 2^3 \times 3^1 \). Exponents are 3 and 1 (both odd). Not a perfect square.

B. \( p = 4 \)
\( 8p = 8 \times 4 = 32 = 2^5 \). Exponent 5 is odd. Not a perfect square.

C. \( p = 12 \)
\( 12 = 2^2 \times 3^1 \)
So, \( 8p = 2^3 \times (2^2 \times 3^1) = 2^{5} \times 3^1 \). Exponents 5 and 1 are odd. Not a perfect square.

D. \( p = 18 \)
\( 18 = 2 \times 3^2 \)
So, \( 8p = 2^3 \times (2 \times 3^2) = 2^{4} \times 3^2 \). All exponents are even (4 and 2). Therefore, it is a perfect square.

Indeed, \( 8 \times 18 = 144 = 12^2 \), which is a perfect square.

Thus, the correct option is D. 18.

Answer:
\[
\boxed{18}
\]

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