Squares make patterns

Algebra Level 2

Without using a calculator, find the value of

8100000000180000000001 . \large \sqrt{8100000000180000000001}.


The answer is 90000000001.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

7 solutions

Tanveer Alam
Aug 10, 2015

8100000000180000000001 \boldsymbol{\sqrt{8100000000180000000001}} = 8100000000000000000000 + 2 × 90000000000 × 1 + 1 \boldsymbol{=\sqrt{8100000000000000000000+2\times90000000000\times1+1}} = ( 90000000000 + 1 ) 2 \boldsymbol{=\sqrt{(90000000000+1)^2}} = 90000000001 =\boxed{90000000001}

Forgot that breaking into addition will make it that simple!

Tushar Pandeya - 5 years, 7 months ago
Joel Toms
Aug 10, 2015

Obviously, this would be easy with a calculator. But we shall never be ruled by machines.

To solve, we use the pattern

( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2ab+b^2 .

In this question, there are three numerical components to the square number (separated nicely by zeroes), corresponding to the three components of ( a + b ) 2 (a+b)^2 . W.l.o.g., take a a as the larger component of a + b a+b . So we reason as follows, and hope for the best:

  • a 2 = 81 × 1 0 20 a = ± 9 × 1 0 10 a^2=81\times10^{20}\Rightarrow a=\pm9\times10^{10} .

Since our answer is positive, this largest component must also be positive. So:

  • a = 9 × 1 0 10 a=9\times10^{10} .

Now:

  • b 2 = 1 b = ± 1 b^2=1\Rightarrow b=\pm1 .

Given these values:

  • 2 a b = ± 18 × 1 0 10 2ab=\pm18\times10^{10} .

Since we see + 18 × 1 0 10 +18\times10^{10} in our square number (and not 82 82 and loads of 9 9 s, as we would with a b ab negative), take

  • a = 9 × 1 0 10 a=9\times10^{10}
  • b = 1 b=1

and just make sure:

( 90000000001 ) 2 (90000000001)^2

= ( 9 × 1 0 10 + 1 ) 2 =(9\times10^{10}+1)^2

= 81 × 1 0 20 + 2 ( 9 × 1 0 10 ) + 1 =81\times10^{20}+2(9\times10^{10})+1

= 8100000000180000000001 =8100000000180000000001 ,

as it should be.

So 8100000000180000000001 = 90000000001 \sqrt{8100000000180000000001}=\boxed{90000000001\,} .

Chew-Seong Cheong
Aug 11, 2015

8100000000180000000001 = 81 × 1 0 20 + 18 × 1 0 10 + 1 = ( 9 × 1 0 10 ) 2 + 2 ( 9 × 1 0 10 ) ( 1 ) + 1 2 = ( 9 × 1 0 10 + 1 ) 2 = 9 × 1 0 10 + 1 = 90000000001 \begin{aligned} \sqrt{8100000000180000000001} & = \sqrt{81 \times 10^{20}+18\times 10^{10} + 1} \\ & = \sqrt{(9\times 10^{10})^2 + 2(9 \times 10^{10})(1) + 1^2} \\ & = \sqrt{(9\times 10^{10}+1)^2} \\ & = 9\times 10^{10}+1 \\ & = \boxed{90000000001} \end{aligned}

1
2
3
sqrt (9**2)(10**20) + 2(9)(10**10) + 1
sqrt ( 9(10**10) + 1 )**2
Answer: 9(10**10) + 1

Sir, idol ko po kau sir =) . ang lufet..

Keil Cerbito - 5 years, 10 months ago

Log in to reply

sisiw lang ito eh

Vincent Miller Moral - 5 years, 10 months ago
Sadasiva Panicker
Oct 29, 2015

(9^10 + 1)^2 = 90000000001^2 Therefore Root of 8100000000180000000001 = 90000000001

Hadia Qadir
Aug 30, 2015

99=81 9+9=18 so we will use " a^2 + 2ab + b^2 " , a ^2 =( 9 * 10^ 10 )^2 b^2 = 1 2ab = 2(910^101) so, by applying on the pattern : 8110^20 + 1810^10 + 1 = the number above by sqrt both the number and the pattern that gives us (9*10^10 + 1) 90000000001

Alaa Mohammed
Aug 11, 2015

9 9=81 9+9=18 so we will use " a^2 + 2ab + b^2 " , a ^2 =( 9 * 10^ 10 )^2 b^2 = 1 2ab = 2 (9 10^10 1) so, by applying on the pattern : 81 10^20 + 18 10^10 + 1 = the number above by sqrt both the number and the pattern that gives us

(9*10^10 + 1)

90000000001

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...