149162536496481100121144...........

Reading left to right, determine the 2010th digit in the abovementioned string.

__Source__: Modified and amended version of an Olympiad problem which I saw in a math periodical in 2004.

- Thread starter K Sengupta
- Start date

149162536496481100121144...........

Reading left to right, determine the 2010th digit in the abovementioned string.

1² to 3² have 1 digit.

4² to 9² have 2 digits.

10² to 31² have 3 digits.

32² to 99² have 4 digits.

100² to 316² have 5 digits.

317² to 999² have 6 digits.

1000² to 3162² have 7 digits.

Ty CRG.CRGreathouse said:

1² to 3² have 1 digit.

4² to 9² have 2 digits.

10² to 31² have 3 digits.

32² to 99² have 4 digits.

100² to 316² have 5 digits.

317² to 999² have 6 digits.

1000² to 3162² have 7 digits.

# 1-digit perfect squares = 3 --> 3*1 =3

# 2-digit perfect squares = 6 --> 6*2 =12

# 3-digit perfect squares = 22 --> 22*3 =66

# 4-digit perfect squares = 68 --> 68*4 =272

# 5-digit perfect squares = 217 --> 217*5 =1085

# 6-digit perfect squares = 683 --> 683*6 = 4098

It is now obvious that, Sum (i=1 to 6) (# i-digit perfect squares)*i > 2010

Now, Sum (i=1 to 5) (# i-digit perfect squares)*i = 3+12+66+272+1085= 1438

Also, (2010-1438)/6 = 95 + 2/6

Since 316+95 = 411, it follows that the required digit must be the second digit reading from the left in the base ten expansion of 412^2 = 169744.

In other words, the required 2010th digit is 6.

Brute force checks out:K Sengupta said:Is this result accurate?

Code:

`find(k)=my(d);for(n=1,k,d=digits(n^2);if(d<k,k-=d,return(Vec(Str(n^2))[k])))`

but then how do you know the number of perfect squares in each set? i get confused after the part where you add up the numbers to reach 2010 digits and find that it's somewhere in the vicinity of 6-digit perfect squares X.X What is this step (2010-1438)/6 = 95 + 2/6? and the one after it...

I had never done that before, but it seemed to fit.Solarmew said:to begin with, how did you come up with the step sqrt(10) = 3.16227766 ...? I take it it's some predefined rule/law that i dunno about >.>

It's easy to see that 10^n is the first number that, when squared, has 2n+1 digits. To find the first number with 2n digits, find sqrt(10^(2n-1)) which is 10^(n-1) * sqrt(10). The integer below this is the last with 2n-1 digits; the integer above this is the first with 2n digits.

It's sqrt(10) because we're looking at the number of base-10 digits. If you wanted to do this in base 8, you'd use sqrt(8).Solarmew said:And is it always a sqrt(10) or is there more to it?

Well, considering that I've never read about it before, I'd say this thread is the only place I can suggest.Solarmew said:Could you point me in the direction where could read up on this technique pls?

I wrote thatSolarmew said:but then how do you know the number of perfect squares in each set?

100² to 316² have 5 digits.

since 100 = sqrt(10000) and floor(sqrt(100000)). It's easy to count the number of members: 316 - 100 + 1 = 217 (the +1 is because it includes its endpoints).

K Sengupta might be able to answer this better than I.Solarmew said:i get confused after the part where you add up the numbers to reach 2010 digits and find that it's somewhere in the vicinity of 6-digit perfect squares X.X What is this step (2010-1438)/6 = 95 + 2/6? and the one after it...

The squares from 1^2 to 316^2 have 1438 digits, so you have 2010-1438 = 572 digits left. 95 * 6 = 570, so the first 95 6-digit squares use up 570 of the 572 digits. You then need the 2nd digit of the 96th 6-digit square, (316+96)^2.

Can you figure out what the smallest and largest natural numbers are that, when squared, have exactly 10 decimal digits?Solarmew said:omg this is way beyond me X.x ... but thank you anyways, this is very informative, i just gotta try to figure it out a lil bit more >.<