Question

The greatest 6 digit number, which is a perfect square is…..
    (A) 998001
    (B) 995001
    (C) 997001
    (D) 996001

Answer 1

$Hint $: In this Question we are asked to find among 4 six-digit numbers which is the largest six-digit number with a perfect square. So we can check each of these numbers by using a long division method to find out the perfect Square. Since we have to find the largest number so we’ll start with the largest among these four.

Complete step-by-step solution -
Starting with 998001 which is the largest among these 4
Start grouping them in two starting from the unit place and these groups are called pairs
So, we got 4 pairs, 99, 80 and 01
Now make these pairs be dividend and find the divisor whose square is just less than or equal to the first pair, make that number the quotient and divisor as well, so we get 9 as divisor and quotient as \[{9^2}\]=81

	9
9	\[\begin{gathered} \overline {99} \overline {80} \overline {01} \\ 81 \\ \end{gathered} \]

Now this 81 is subtracted from the first pair if there is any and then bring down the next pair on the right side of the remainder obtained before, together they make the new remainder just like we do in normal division

9	9
	99800181
	1880

Now , take the sum of previous divisor with itself i.e. 9+9 = 18 and write at place of divisor with some space on its right side to add another digit , say x and so 1880 becomes the new dividend

9	9
	99800181
18x	1880

Now , find this x such that when 18x when multiplied with x gives a number which is just less than 1880 or equal to it , and write this x by the side of the quotient as well.

	9x
9	99800181
18x	1880

This x=9 which makes 18x = 189,
Such that 189 × 9 = 1701 which is to be subtracted from 1880

	99
9	99800181
189	18801701

Like we did earlier if find the remainder bring down the other pair, find new divisor just like we found 189 and repeat till we get reminder zero
So, our new remainder is 17901, new divisor is found by adding the previous divisor i.e. 189 to the very first divisor i.e. 9
So, the new divisor is 198x, again x is to be a number which when multiplied by 198x gives us a number less than or equal to 17901 and write this number by the side of quotient too.

	999
9	99800181
189	18801701
1989	1790117901
	0

So, the quotient is 999 which is the perfect Square of 998001
So, the correct option is ‘A’

$Note:$ This question can also be solved by using a trick. Let \[{x^2}\]be the largest six-digit perfect square. \[{x^2}\]≤999,999. You know that \[{1000^2} = 1000000\] which is only 1 more than 999,999, so, \[{x^2}\] must be next smaller square, which is ${(1000-1)}^2 \times {(1000-1)}^2$.