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Algebra

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11 Jan 2007 13:48:57 IST

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Let a be the 81-didit number all digits of which are equal to 1. Then the number a is

1) divisible by 9 but not divisible by 27

2) divisible by 27 but not divisible by 81

3) divisible by 81 but not divisible by 243

4) divisible by 243

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Marshall Eriksen

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11 Jan 2007 14:11:00 IST

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If There Will Be No Repurcussions :
My Answer Will Be (C)
I Doubt It Is Correct....But I Applied All The Concepts I Knew...

Asmita Bhattacharya

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11 Jan 2007 16:10:35 IST

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The correct answer is the option(3).

As 9,27,81,243 are some powers of 3 Therefore for a no. to be divisible by them the sum of the digits should be divisible by them.

Titun

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11 Jan 2007 16:28:47 IST

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Asmita,

I don't agree with your solution.

Because, 54, 108 are divisible by 27 but the sum of the digits in either case is not divisible by 27.

So, can u provide another solution?

rahul_c

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11 Jan 2007 16:40:30 IST

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a = 111111...............81 times

a = (1/9)9999999.............81 tmes

a = (10^81 - 1)/ 9

a = (10^81 -1)/ (10 -1)

a is the sum of G.P whose common ratio is 10 consisting of 81 terms

hence a is divisible by 9

the correct option is a)

i may be wrong

amey

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11 Jan 2007 17:01:01 IST

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I completely agre with Rahul and would like to give the same solution as given by him.
i.e. a is divisible by 9 and not by 27
the CORRECT option hence is (a)

Ajinkya Bapat

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11 Jan 2007 17:09:13 IST

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it is obviously divisible by 81 as the no. consists of 9 sets of nine 1s which is divisible by nine.it yields(division)123456790 this repeated 9 times sum is also divisible by 9.

edison

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11 Jan 2007 20:22:34 IST

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Here we can write the no. 'a' as follows

a = 111111...............81 times

a = (1/9)9999999.............81 times

a = (10^81 - 1)/ 9

a = (10^81 -1)/ 9

a = (10^27x3- 1³)/9

a = (10²⁷ - 1)(10⁵⁴ + 10²⁷ + 1)/9

a = (10⁹ - 1)(10¹⁸ + 10⁹ + 1)(10⁵⁴ + 10²⁷ + 1) / 9

a = 9 * 111111111*(10¹⁸ + 10⁹ + 1)(10⁵⁴ + 10²⁷ + 1) / 9

a = 111111111*(10¹⁸ + 10⁹ + 1)(10⁵⁴ + 10²⁷ + 1) .......(1)

now any no. is divisible by 3 if sum of the digits is 3 or integral multiple of 3.

Further for divisibility by 9 the sum of digits should be 9.

Thus, 111111111 = 9 * m = 3²(m) ........(2)

10¹⁸ + 10⁹ + 1 = 3*n ( However, this no.is not divisible by 9) .....(3)

10⁵⁴ + 10²⁷ + 1 = 3*p ( However, this no.is not divisible by 9) .....(4)

Where m,n and p are +ve integers not divisible by 3

Combining (1), (2), (3) and (4) we obtain

a = 3⁴ (mnp) = 81mnp

so a is divisible by 81 but not by 243 = 3⁵

so option (3) is correct

Asmita Bhattacharya

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11 Jan 2007 20:38:09 IST

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I just mean to say that just calculate the sum of the digits upto a 2 digit no. if the no. by which u r dividing is a 2 digit no.This can be generalised for 1 digit, 3 digits & ... no.

Titun

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12 Jan 2007 20:20:42 IST

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I am thankful to Edison. The correct option is 3

supriyasingh

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12 Jan 2007 20:56:56 IST

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good question as well as good answer too.

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