Finding the Largest Six-Digit Square Number

To determine the largest square number that has six digits, we first need to understand the range of numbers with six digits.

• The smallest six-digit number is $100,000$.
• The largest six-digit number is $999,999$.

A square number (also known as a perfect square) is an integer that results from multiplying an integer by itself. For instance, $16$ is a square number because $16 = 4^2$.

Calculating the Largest Six-Digit Perfect Square

To find the largest square number within the six-digit range, we should look for the largest integer whose square does not exceed $999,999$. We can do this by finding the square root of the largest six-digit number and then taking the floor (the largest integer less than or equal to) of that result.

The largest six-digit number is $999,999$.

Let's calculate its square root:

$\sqrt{999,999} \approx 999.9995$

The largest whole number (integer) that is less than or equal to $999.9995$ is $999$. This means that $999$ is the largest integer whose square will be a six-digit number.

Now, we will square this integer to find the largest square number of six digits:

$999^2 = (1000 - 1)^2$

Using the algebraic identity for squaring a binomial, $(a - b)^2 = a^2 - 2ab + b^2$, where $a = 1000$ and $b = 1$:

$999^2 = 1000^2 - (2 \times 1000 \times 1) + 1^2$

$999^2 = 1,000,000 - 2,000 + 1$

$999^2 = 998,000 + 1$

$999^2 = 998,001$

Verifying the Six-Digit Square Number

The calculated value, $998,001$, is indeed a six-digit number, falling within the range of $100,000$ to $999,999$. To ensure it is the largest, let's consider the square of the next integer after $999$, which is $1000$:

$1000^2 = 1,000,000$

This number, $1,000,000$, has seven digits, which means it falls outside the range of six-digit numbers. Therefore, $999^2 = 998,001$ is confirmed as the largest six-digit square number.

Comparing this result with the given options, $998001$ matches the derived largest six-digit square number.