Question

Find the greatest three-digit number which is a perfect square. Explain it...

Answer 1

We know, the largest three-digit number is 999.

Also, our knowledge of elementary mathematics reminds us that:

(a + b)^2 = a^2 + 2 * a * b + b^2.

We know, 30^2 = 900; which is not very far from 999.

Putting a = 30 in the above equation, we get:

(30 + b)^2 = 30^2 + 2 * 30 * b + b^2

Or, (30 + b)^2 = 900 + 60b + b^2

Now, we have to check out for what largest integer value of b, the value of (60b + b^2) remains less than or equal to 99.

Clearly b = 2 takes 60b to 120; so we can't take it.

For b = 1; (60b + b^2) = 61 and this meets our requirement.

So, the largest three-digit number which is a perfect square is (30 + 1)^2 = 31^2 = 961.

Answer 2

Answer:

961 is the correct answer

Step-by-step explanation:

because

$( {31})^{2} = 961 \\ {32}^{2} = 1024$