IIT JEE , AIEEE Forum - Mathematics , Algebra

titun

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

nunoxic

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

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

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

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

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)

avb1989

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

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

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

I am thankful to Edison. The correct option is 3

supriyasingh

good question as well as good answer too.

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