What is the largest five-digit which is a perfect square ?

Question

(A) 97344 
(B) 98596
(C) 99856 
(D) 99956 
(E) 99996

aggarwal1994 · Answer

The largest 5 digit number is 99,999.
Finding the square root of this number gives a number greater than 316 but less than 317

So, the largest 5 digit number which is a perfect square is 316*316 = 99,856

Ans: C

nbrn226 · Answer

Is there any easier way than to actually find out square roots?

Posted from my mobile device

JerryAtDreamScore · Answer

Breaking Down the Info:

The largest 5-digit square must be below 100,000 which is  10^5 .

Let us set  x^2  so that  x^2 < 100,000 . Square rooting gives us  x < 100\sqrt{10} .

\sqrt{10}  is bit bigger than 3, so we can start with  x = 100*3  to get  x^2 = 90000 .

Next try x^2 = 310^2 = (300 + 10)^2 = 90000 + 6000 + 100 = 96100 . We can go higher.

x^2 = 320^2 = (300 + 20)^2 = 90000 + 12000 + 400 = 102400  is too big, so x must be between 310 and 320.

Now try 315 since it is the midpoint of 310 and 320.  x^2 = (315)^2 = (300 + 15)^2 = 90000 + 9000 + 225 = 99225 . We are quite close!

An increment of 1 for x would increase 99225 by 315 + 316, which gives us  99225 + 631 = 99856 .

Answer: C

IanStewart · Answer

Yes, there are a lot of alternatives here. It can occasionally be helpful to know that 2^10 = 1024, because that helps us estimate large powers of 2 (using 2^10 ~ 10^3= 1000). But 2^10 = (2^5)^2 = 32^2, so for this question, 320^2 = (32^2)(10^2) = 102,400, and we can see that we're squaring something close to, but less than, 320.

You can then think about how far apart perfect squares this large will be. If you look at 320^2 - 319^2 = (320 + 319)(320 - 319) = 639, you can see that squares this big are roughly 640 apart. So 319^2 is roughly 640 less than 320^2 = 102,400, and 318^2 is roughly 640 less than 319^2, and we can see we'll want to subtract ~640 exactly four times to get below 100,000, from which we can tell the answer is 316^2. And now you can use the difference of squares again to avoid computing the square by long multiplication: 320^2 - 316^2 = (320 + 316)(320 - 316) = 636*4 = 2544, so 316^2 = 320^2 - 2544 = 102,400 - 2,544 = 99,856.

Or once you know the answer is 316^2, you can pick the right answer in a different way, if you know remainder theory. When we divide 316 by 9, the remainder is 1 (which you can see by summing the digits of 316 and taking the remainder). So when we divide 316^2 by 9, the remainder must be equivalent to 1^2 = 1. Now by summing the digits of the only plausible answer choices (C, D and E are the only ones close enough to 100,000 to be possible) we can see only answer C gives us a remainder of 1.

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What is the largest five-digit which is a perfect square ?

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