How can the square root of a large number be found without a calculator?

[Quark Itself]

I stumbled upon a math question, at the glance of it, seemed easy.
One is supposed to find the exact value of this square root:
√30*31*32*33+1
They are all under the square root operator and Fundamental BEDMAS/BIDMAS applies of course.
The trick here is when the condition states that one cannot use a calculator.
Any suggestions, anyone?

 

[jgens]

I do not know any tricks to actually solve this, but you can get a fairly good estimate using basic calculus. In particular, note that 31² < (30*31*32*33 + 1)^1/2 < 32². We can model √ around 31⁴ linearly by the function defined by L(x) = 31² + (2*31²)^-1(x-31⁴). Take x = 31.5⁴ as an initial guess and we get something like L(x) ≈ 992. I think that actual answer is 991, so this gets you a pretty good estimate and all of the calculations are manageable.

I am sure someone else knows a nice method to tackle this problem though, so I would like to see that as well.

 

[D H]

You are looking for some number $n=\sqrt{30*31*32*33+1}$. Squaring both sides to get rid of the radical yields $n^2=30*31*32*33+1$. Writing 31, 32, and 33 as 30+1, 30+2, and 30+3 yields $n^2=30^4+6*30^3+11*30^2+6*30+1$. This suggests writing n in a similar form: $n=30^2+a*30+b$. Squaring yields $n^2 = 30^4+2a*30^3+(a^2+2b)*30^2+2ab*30+b^2$. Equating the expansion of 30*31*32*33+1 with this readily yields a=3, b=1, and thus n=991.

 

[D H]

We can model √ around 31⁴ linearly by the function defined by L(x) = 31² + (2*31²)^-1(x-31⁴). Take x = 31.5⁴ as an initial guess and we get something like L(x) ≈ 992. I think that actual answer is 991, so this gets you a pretty good estimate and all of the calculations are manageable.

This is another approach. The answer has to be fairly close to 992. The word "exact" in the question indicates the number has to be an integer. The last digit of this number must be either one or nine, and the answer certainly isn't 999 or 981. The only viable choices are 991 and 989. Just square each (not hard) and compare to 30*31*32*33+1 (which you will have to expand for this approach). Since the original guess was 992, starting with 991 will eliminate the need to try 989 as the second choice since the answer is indeed 991.

 

[jgens]

You are looking for some number $n=\sqrt{30*31*32*33+1}$. Squaring both sides to get rid of the radical yields $n^2=30*31*32*33+1$. Writing 31, 32, and 33 as 30+1, 30+2, and 30+3 yields $n^2=30^4+6*30^3+11*30^2+6*30+1$. This suggests writing n in a similar form: $n=30^2+a*30+b$. Squaring yields $n^2 = 30^4+2a*30^3+(a^2+2b)*30^2+2ab*30+b^2$. Equating the expansion of 30*31*32*33+1 with this readily yields a=3, b=1, and thus n=991.

That is a nice solution! Much better than my thoughts on the problem.

 

[Quark Itself]

Thanks for the quick replies ! Really appreciate it !
I was thinking along the lines of having a number n which would be that exact value but as I worked it through I got stuck, where you presented the n = 30^2 + 30a + b part was where I got stuck, but your explanation was really good so thanks again.

 

[lurflurf]

30*31*32*33+1=x^2
since
31*32-30*33=(x+1)-(x-1)
and
30*31*32*33=(x+1)*(x-1)
we can see that
30*33=x-1
31*32=x+1
and x is easy to find