What is the greatest six digit number which is a perfect square
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The greatest 6-digit number is 999 999. Since it isn't a perfect square, it follows that the greatest 6-digit number which is one (let's call it G), can't be other than the square of the integer part of the square root of 999 999:
G = { integer [sqrt (999 999) ] }² . We have:
999 999 = 10⁶ - 1 = (10³ +1)(10³ - 1) = 1001 * 999
The next integer to 999 is 1000, whose square is 1 000 000 > 999 999, so 999 is the greatest 3-digit number with a 6-digit square. It follows that:
G = 999² = 998 001
I just don't get what you needed Long Division for!
G = { integer [sqrt (999 999) ] }² . We have:
999 999 = 10⁶ - 1 = (10³ +1)(10³ - 1) = 1001 * 999
The next integer to 999 is 1000, whose square is 1 000 000 > 999 999, so 999 is the greatest 3-digit number with a 6-digit square. It follows that:
G = 999² = 998 001
I just don't get what you needed Long Division for!
Answered by
1
Answer:
998001 = (999)²
Step-by-step explanation:
1. Do the long division method of square root for highest six digit number
i.e 999999
2. Subtract the remainder from the number
999999 - 1998
=998001
This is our answer
(You can apply this for same kind of questions for finding greatest number only)
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