integer part of S

Albert.Teng · September 22nd, 2012, 06:51 PM

[attachment=0:1henah5g]integeral part of S.jpg[/attachment:1henah5g]

MarkFL · September 22nd, 2012, 08:22 PM

Squaring gives:

$S^2=1991+S$

$S^2-S-1991=0$

Hence, taking the positive root, and using the floor function to return the integral portion, we find:

$\left\lfloor \frac{1+3\sqrt{885}}{2} \right\rfloor=45$

mathbalarka · September 23rd, 2012, 03:46 AM

If the question is to find the integral part of $\sqrt{1991+\sqrt{1991+\sqrt{1991 . . . + \sqrt{1991}}}}$ , then it is impossible to find it.

Since we do not know how many times 1991 is nested under the square root sign.

The question should be asking to find the integral part of $\underbrace{\sqrt{1991+\sqrt{1991+\sqrt{1991 + . . . }}}}_{\text{infinite times}}$

September 22nd, 2012, 08:22 PM	#2
MarkFL Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs	Re: integer part of S Squaring gives: $S^2=1991+S$ $S^2-S-1991=0$ Hence, taking the positive root, and using the floor function to return the integral portion, we find: $\left\lfloor \frac{1+3\sqrt{885}}{2} \right\rfloor=45$
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September 23rd, 2012, 03:46 AM	#3
mathbalarka Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory	Re: integer part of S If the question is to find the integral part of $\sqrt{1991+\sqrt{1991+\sqrt{1991 . . . + \sqrt{1991}}}}$ , then it is impossible to find it. Since we do not know how many times 1991 is nested under the square root sign. The question should be asking to find the integral part of $\underbrace{\sqrt{1991+\sqrt{1991+\sqrt{1991 + . . . }}}}_{\text{infinite times}}$
$mathbalarka is offline$

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