RPS #003 - Nested Radical

Algebra Level 5

Solve for $x$ $\displaystyle\sqrt{1089 + (2^{11} - 66)\sqrt{1089 + (2^{11} - 33)\sqrt{1089 + 2^{11}\sqrt{\dots}}}} = x$

The answer is 2015.

2 solutions

Aareyan Manzoor
Jan 17, 2015

by Ramanujan identity, $\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{\dotsm\dotsm}}}=a+x+n$ now insert $a=0,x=2^{11}-66,n=33$ $\displaystyle\sqrt{33^2 + (2^{11} - 66)\sqrt{1089 + (2^{11} - 33)\sqrt{33^2+ 2^{11}\sqrt{\dots}}}} = 33+2^{11}-66=\boxed{2015}$

Typo there in the identity on LHS! no $x + 2n$ , it should be $x + n$

Kartik Sharma - 6 years, 4 months ago

It is a very easy problem, if you know " Ramanujan's Identity". But, how many people in the world will know " Ramanujan 's Identity".

Panya Chunnanonda - 6 years, 4 months ago

Didn't know Ramanujan's identity! But, still managed to solve it. :)

Atomsky Jahid - 4 years, 7 months ago

Most overrated (400 point) problem that I have solved on brilliant. This should be level-2 problem Anyways +1.

Shreyansh Mukhopadhyay - 2 years, 10 months ago

Abdullah Ahmed
Feb 24, 2017

x=√{1089+(2^11-66)√{1089+(2^11-33)....

or, x^2-33^2 = (2^11-66)√{1089+(2^11-33)....

now it is clear that

(2^11-66)<√{1089+(2^11-33)....

so, x-33=2^11-66

x=2015

RPS #003 - Nested Radical

The answer is 2015.

2 solutions

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