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Consider a large number N = 1234567891011121314………979899100.

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Consider a large number N = 1234567891011121314………979899100. [#permalink] New post   18 Jan 2020, 22:07
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Consider a large number N = 1234567891011121314………979899100. What is the remainder when the first 100 digits of N is divided by 9?

A. 0
B. 1
C. 5
D. 8
E. 10
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Re: Consider a large number N = 1234567891011121314………979899100. [#permalink] New post   18 Jan 2020, 22:29
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DisciplinedPrep wrote:
Consider a large number N = 1234567891011121314………979899100. What is the remainder when the first 100 digits of N is divided by 9?

A. 0
B. 1
C. 5
D. 8
E. 10


THEORY
1. Divisibility by 9 depends on the sum of digits. The remainder too will be same as the sum of digits divided by 9
2. first 100 digits will depend on the single digits and 2-digit numbers.

First 100 digits
1-9..9 digits
Now 10 onwards we have 2 digits and we are looking for another 100-9=91 digits...91 means 91/2=45.5, that is 40 first 2-digit numbers ( 10 onwards --- 10+45-1=54) and first/tens digit of 41st number(55).......1 2 3 4 5 .....48 49 50 ... 53 54 5

SUM OF DIGITS
1+2+3+...4+8+4+9+5+0+..5+3+5+4+5
So units digit 0 to 4 are repeated 6 times(1-4, 11-14, 21-24, 31-34, 41-44, 51-54) and 5 to 9 are repeated 5 times = \(5*(\frac{9*10}{2})+(1+2+3+4)=235\)
tens digit \((0+1+2+3+4)*10+(5)*5=130\)
Sum = \(235+130=365=360+5\)

Divisibility by 9
360 is divisible by 9 and we have remainder as 5

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Re: Consider a large number N = 1234567891011121314………979899100. [#permalink] New post   18 Jan 2020, 23:38
chetan2u wrote:
DisciplinedPrep wrote:
Consider a large number N = 1234567891011121314………979899100. What is the remainder when the first 100 digits of N is divided by 9?

A. 0
B. 1
C. 5
D. 8
E. 10


THEORY
1. Divisibility by 9 depends on the sum of digits. The remainder too will be same as the sum of digits divided by 9
2. first 100 digits will depend on the single digits and 2-digit numbers.

First 100 digits
1-9..9 digits
Now 10 onwards we have 2 digits and we are looking for another 100-9=91 digits...91 means 91/2=45.5, that is 40 first 2-digit numbers ( 10 onwards --- 10+45-1=54) and first/tens digit of 41st number(55).......1 2 3 4 5 .....48 49 50 ... 53 54 5

SUM OF DIGITS
1+2+3+...4+8+4+9+5+0+..5+3+5+4+5
So units digit 0 to 4 are repeated 6 times(1-4, 11-14, 21-24, 31-34, 41-44, 51-54) and 5 to 9 are repeated 5 times = \(5*(\frac{9*10}{2})+(1+2+3+4)=235\)
tens digit \((0+1+2+3+4)*10+(5)*5=130\)
Sum = \(235+130=365=360+5\)

Divisibility by 9
360 is divisible by 9 and we have remainder as 5

C


Hi Chetan,

I fund this question to be very tough and time consuming.
First part is clear. Is there any other way of finding sum of the digits in a much faster manner than in your post.

Regards,
Arvind
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Re: Consider a large number N = 1234567891011121314………979899100. [#permalink] New post   19 Jan 2020, 01:07
Expert Reply
arvind910619 wrote:
chetan2u wrote:
DisciplinedPrep wrote:
Consider a large number N = 1234567891011121314………979899100. What is the remainder when the first 100 digits of N is divided by 9?

A. 0
B. 1
C. 5
D. 8
E. 10


THEORY
1. Divisibility by 9 depends on the sum of digits. The remainder too will be same as the sum of digits divided by 9
2. first 100 digits will depend on the single digits and 2-digit numbers.

First 100 digits
1-9..9 digits
Now 10 onwards we have 2 digits and we are looking for another 100-9=91 digits...91 means 91/2=45.5, that is 40 first 2-digit numbers ( 10 onwards --- 10+45-1=54) and first/tens digit of 41st number(55).......1 2 3 4 5 .....48 49 50 ... 53 54 5

SUM OF DIGITS
1+2+3+...4+8+4+9+5+0+..5+3+5+4+5
So units digit 0 to 4 are repeated 6 times(1-4, 11-14, 21-24, 31-34, 41-44, 51-54) and 5 to 9 are repeated 5 times = \(5*(\frac{9*10}{2})+(1+2+3+4)=235\)
tens digit \((0+1+2+3+4)*10+(5)*5=130\)
Sum = \(235+130=365=360+5\)

Divisibility by 9
360 is divisible by 9 and we have remainder as 5

C


Hi Chetan,

I fund this question to be very tough and time consuming.
First part is clear. Is there any other way of finding sum of the digits in a much faster manner than in your post.

Regards,
Arvind





Now when you know we are looking for 12345...515253545, and we are aware that the sum of digits can also tell us about divisibility by 9, we can just sum up all 54 numbers = 1=2+3+..53+54=54*55/2=27*11=9*3*11, which is divisible by 9.
So 12345...515253540 will be divisible by 9 or 1234..515253540=9x, but we are looking for 1234...515253545 = 1234..515253540+5=9x+5
Thus remainder is 5
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Re: Consider a large number N = 1234567891011121314979899100. [#permalink] New post   21 Jun 2023, 04:41
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Re: Consider a large number N = 1234567891011121314979899100. [#permalink]
21 Jun 2023, 04:41
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