I hope I counted correctly!

Algebra Level 5

Given a number x = 1234567891011121314...20142015 x= 1234567891011121314...20142015 , how many digits will there be in x 2 x^{2} ?


The answer is 13905.

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Hobart Pao by Hobart Pao , 23, USA

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1 solution

Henny Lim
Oct 27, 2015

First we should count how many digits are in x.

9 times 1 digit, then 90 times 2 digits, then 900 times 3 digits, and (2015-1000+1) times 4 digits (makes it 1016 times). We got 6953 digits.

Any square of n-digit number which begins with 1 should contains of 2n-1 digits, then the square of x should consists of (2x6953 - 1) digits = 13905 digits.

7 Helpful 0 Interesting 0 Brilliant 1 Confused

What's that rule you used?

Mr Yovan - 5 years, 7 months ago

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Even me, same question.. can you please elaborate your answer? In fact, I don't even get how you counted the number of digits "9 times 1 digit, then 90 times 2 digits, then 900 times 3 digits, and (2015-1000+1) times 4 digits".... please explain?

Jake Tricole - 1 year, 9 months ago

That's exactly what I did, well done!

Hobart Pao - 5 years, 7 months ago

is there any theorem like that ??

Ganesh Ayyappan - 5 years, 6 months ago

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I read it in Arthur Benjamin's "The Magic of Math: Solving for x and Figuring out wh y "

Hobart Pao - 5 years, 6 months ago

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can v use this in talent exams by jus stating this property as a fact?? ... or is there any proof??

Ganesh Ayyappan - 5 years, 6 months ago

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