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s(n) is a n-digit number formed by attaching the first n per [#permalink]

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31 Jul 2014, 22:32

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s(n) is a n-digit number formed by attaching the first n perfect squares, in order, into one integer. For example, s(1) = 1, s(2) = 14, s(3) = 149, s(4) = 14916, s(5) = 1491625, etc. How many digits are in s(99)?

A. 350 B. 353 C. 354 D. 356 E. 357 What is the best way to approach this problem?

s(n) is a n-digit number formed by attaching the first n perfect squares, in order, into one integer. For example, s(1) = 1, s(2) = 14, s(3) = 149, s(4) = 14916, s(5) = 1491625, etc. How many digits are in s(99)?

350

353

354

356

357

What is the best way to approach this problem?

Focus on the points where the number of digits in squares change:

1, 2, 3 - Single digit squares. First 2 digit number is 10.

4 , 5,...9 - Two digit squares. To get 9, the last number with two digit square, think that first 3 digit number is 100 which is 10^2. so 9^2 must be the last 2 digit square.

10, 11, 12, ... 31 - Three digit squares. To get 31, think of 1000 - the first 4 digit number. It is not a perfect square but 900 is 30^2. 32^2 = 2^10 = 1024, the first 4 digit square.

32 - 99 - Four digit squares. To get 99, think of 10,000 - the first 5 digit number which is 100^2.

So number of digits in s(99) = 3*1 + 6*2 + 22*3 + 68*4 = 3 + 12 + 66 + 272 = 353
_________________

Can you please explain how you drove to the final evaluation - So number of digits in s(99) = 3*1 + 6*2 + 22*3 + 68*4 = 3 + 12 + 66 + 272 = 353

I'm not sure of how you got 3,6,22,68, etc.

Thanks in advance.

V

First two digit number is 10 and first 2 digit square is 16. 16 is the square of 4. This means the first 3 numbers have a square which is a single digit. First 3 digit number is 100 which is the square of 10. So numbers from 4 till 9 must have 2 digit squares. This gives us a total of 9-4+1 = 6 numbers First 4 digit number is 1000 and first 4 digit square is 1024 (which is 2^10 or 32^2). So numbers from 10 to 31 must have 3 digit squares. This gives us a total of 31 - 10 + 1 = 22 numbers. First 5 digit number is 10000 which is the square of 100. So numbers from 32 to 99 must have 4 digit squares. This gives us 99 - 32+1 = 68 numbers
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s(n) is a n-digit number formed by attaching the first n per [#permalink]

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18 Mar 2015, 08:19

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We need to find the total number of digits in all of the square numbers from 1 - 99.

1, 4, 9 <-- all have one digit. 16, 25, 36, ...., 81 <-- all have two digits.

Each 'n' will acquire a new digit when n^2 exceeds a power of ten.

\(sqrt(10)\) is more than 3 and less than 4, so n =1,2,3 will have 1 digit when you square n (1,4,9). This makes 3 numbers. \(sqrt(100) = 10\), so n = 4,5,6,7,8, and 9 will have two digits when you square n. (16, 25, ...,81). This makes 9 - 3 = 6 numbers. \(sqrt(1000)\) is more than 30 (30^2 = 900) and a quick check will show that \(31^2 = 961\), while \(32^2 = 1024\), so n = 10,11,...,31 will have 3 digits. This makes 31 - 9 = 22 numbers. \(sqrt(10000) = 100\), so all of the rest of the numbers from 32-99 will each have 4 digits. This makes 99-31 = 68 numbers.

Therefore, the total number of digits is 3(1) + 6(2) + 22(3) + 68(4) = 3 + 12 + 66 + 272 = 353.

s(n) is a n-digit number formed by attaching the first n per [#permalink]

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12 Sep 2015, 10:47

know the numbers wherein the count increases i.e at 10...all 3 digit squares come....at 31 all 4 digit squares come...till 99 so it is 18 digits till 10, next 21 numbers to go till 32, & lastly 68 numbers till 99...so 18+(21*3)+(68*4)=353

Re: s(n) is a n-digit number formed by attaching the first n per [#permalink]

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20 Sep 2017, 18:57

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