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

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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?
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 Aug 2014, 00:21, edited 1 time in total.
Renamed the topic and edited the question.

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

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31 Jul 2014, 22:46
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cs2209 wrote:
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
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17321 [4], given: 232 Intern Joined: 04 Sep 2014 Posts: 1 Kudos [?]: [0], given: 0 Re: s(n) is a n-digit number formed by attaching the first n per [#permalink] ### Show Tags 04 Sep 2014, 19:27 Hi Karishma, 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 Kudos [?]: [0], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7668 Kudos [?]: 17321 [0], given: 232 Location: Pune, India Re: s(n) is a n-digit number formed by attaching the first n per [#permalink] ### Show Tags 04 Sep 2014, 20:26 fishfulthinking wrote: Hi Karishma, 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 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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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.

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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

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

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Re: s(n) is a n-digit number formed by attaching the first n per   [#permalink] 20 Sep 2017, 18:57
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