Finding the Least Perfect Square Divisor

The problem asks for the smallest perfect square number that is exactly divisible by 2, 5, 7, 8, and 10.

Prime Factorization and LCM Calculation

First, find the prime factorization of each number:

• $2 = 2^1$
• $5 = 5^1$
• $7 = 7^1$
• $8 = 2^3$
• $10 = 2^1 \times 5^1$

To find the Least Common Multiple (LCM), take the highest power of each prime factor present in the numbers:

LCM = $2^3 \times 5^1 \times 7^1 = 8 \times 5 \times 7 = 280$.

The number we are looking for must be divisible by 280.

Determining the Perfect Square

We need the smallest perfect square that is a multiple of the LCM (280). Let the required number be $N$.

The prime factorization of the LCM is $280 = 2^3 \times 5^1 \times 7^1$.

For a number to be a perfect square, all the exponents in its prime factorization must be even.

To make the LCM a perfect square, we need to adjust the exponents to the next highest even number:

• The exponent of 2 is 3. The next even number is 4.
• The exponent of 5 is 1. The next even number is 2.
• The exponent of 7 is 1. The next even number is 2.

So, the smallest perfect square multiple is $N = 2^4 \times 5^2 \times 7^2$.

Calculating the Final Value

Calculate the value of N:

$N = 2^4 \times 5^2 \times 7^2 = 16 \times 25 \times 49$

$N = 400 \times 49 = 19600$.

The number 19600 is a perfect square ($\sqrt{19600} = 140$) and is divisible by 2, 5, 7, 8, and 10 because it's a multiple of their LCM.

Comparing with Options

Comparing 19600 with the given options, we find it matches Option 4.