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Realize that the number \(\sqrt{64516}\) is slightly greater than \(62500\). hence it can be broken as \(= \sqrt{62500+2016} =\) \(\sqrt{(250^{2} + 2016)}\) Now this looks to me of the form \(\sqrt{(a+b)^{2}}\) so can we break \(2016\) as \(b^2 +2*a*b\). this can be done as \(2016 = 2*250*4 + (4)^2\) so our number becomes \(= \sqrt{(250 + 4)^2}\)\(= 254\)

Re: What is the value of (64516)^(1/2) [#permalink]

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12 Oct 2017, 07:36

Can´t you just simply recognize that 4x4 is 16 which rules out every answer choice that ends with something different than 6. Than apply the logic used above

Can´t you just simply recognize that 4x4 is 16 which rules out every answer choice that ends with something different than 6. Than apply the logic used above

JannikHeine , I can see how it would seem logical to think that way. The answer is:

No. (6 * 6) also ends in 6. This is multiplication. We don't know what those extra digits are going to do when multiplied by each other.

There are 2- 3- and x-digit numbers ending in 6 whose square ends in 16.

For example 246 * 246 = 60516 196 * 196 = 38416 46 * 46 = 2116

You can use shortcuts that aren't terribly difficult. I multiplied the first two digits of each answer and compared the 3-digit result with the first three digits of the prompt, i.e., 65416 --> 654xx

I knew that 25 * 25 = 625. (If not, multiply.) 625 is pretty close to 654. 25_ was still in.

Like fitzpratik : C and D, with first two digits of 23, were out.

If 25_\(^2\) = 625xx isn't quite 654xx, 23_\(^2\) would certainly not be closer.

(Proof: 23 * 23 = 529, so 23_\(^2\) would not even make it into the 60,000s.)

Then E: (26*26) = 676. Too large. Eliminate

Down to A and B. You only have to calculate one. I used B: 254 * 254 = 65416

But if I had chosen A: 256 * 256 = 65536. Too large. The right choice must be

Re: What is the value of (64516)^(1/2) [#permalink]

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12 Oct 2017, 13:08

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I think that it is also helpful to realize that it 64516 is divisible by 4 because the last 2 digits (16) are divisible by 4. To find out if it's divisible by 6, it has to be even and the sum of all the digits must be divisible by 3. It is even, but 6+4+5+1+6 = 22 which is not divisible by 3. This rules out all but B and D. Then using the 250^2 as mentioned already, you can rule out D, or you can just start multiplying the last two digits of B and D to use trial by elimination.

I think that it is also helpful to realize that it 64516 is divisible by 4 because the last 2 digits (16) are divisible by 4. To find out if it's divisible by 6, it has to be even and the sum of all the digits must be divisible by 3. It is even, but 6+4+5+1+6 = 22 which is not divisible by 3. This rules out all but B and D. Then using the 250^2 as mentioned already, you can rule out D, or you can just start multiplying the last two digits of B and D to use trial by elimination.

zflodeen -- Gah! I was so focused on unpredictable squares that I did not even think about divisibility rule for 6. That'll teach me.

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