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One way to attack this problem is to consider squaring the answer choices. When squared, the right answer will be 157609. We want to minimize calculation, so we want to look for shortcuts.

First, if the square of a number ends in 9 then the number itself will end in 3 or 7. So B and C are gone. We could have also noticed that both B and C are even, so their squares will also be even.

E, 403, is pretty close to a number that's easy to work with, 400, so let's consider E next. 400^2 = 160000, so 403^2 will be bigger still. Our number is smaller, so E is gone.

If we start multiplying A, 323, out the long way we find that the tens digit is 2, not 0. Eliminate.

So, The number must be Greater than 350 and the units digit of the number must end in 3 and 7... Among the given options only (D) fits in perfectly , hence, correct answer must be (D)

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Abhishek....

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157609 is odd so its sq rt cannot be even. Eliminate B,C

Option E is 403, 400 is sq rt of 160000 which is greater than 157609. We are left with 323 and 397. Now 350 square is 122500 so 323 can be eliminated. Hence answer is d 397

Since the units digit has to be 9 so only 323 and 397 will remain, then we can take a square of 320 and determine that 323 will not be our answer coz 320^2= 102400 so therefore D is our answer

\(400^2 = 160000\) Also, for the last digit 9, the square root could be 3 or 7 because \(\sqrt{49} = 7\) and \(\sqrt{9} = 3\) Since the number in question is slightly below 160000, the square root cannot be 403.

Therefore, the number must be 397(Option D)

Another method to tests our answer is \((397)^2 = (400 - 3)^2 = 160000 + 9 - 2(400)(3) = 160009 - 2400 = 157609\)
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