[amar_kashyap]

Originally Posted by khanna_sumit

1111111 is one of the factors, so is 1111111111111

An extension of the same problem:
1111.......n times is a composite number for sure only when n is composite number.

[khanna_sumit]

Originally Posted by amar_kashyap

An extension of the same problem:
1111.......n times is a composite number for sure only when n is composite number.

we know divisibility rule for 9. sum of digits mut be divisible by 9.
does that work for 81?

is 11111......81 times divisible by 81?

[vineet.nitd]

Originally Posted by khanna_sumit

we know divisibility rule for 9. sum of digits mut be divisible by 9.
does that work for 81?

is 11111......81 times divisible by 81?

yes that is also divisible by 81 ...

any number whose disgit sum will be 81 or the multiple of same comes out to be divisible by 81 ...

[khanna_sumit]

Originally Posted by vineet.nitd

yes that is also divisible by 81 ...

any number whose disgit sum will be 81 or the multiple of same comes out to be divisible by 81 ...

How come it is divisible by 81?

and how can u say that rule is true?

[amar_kashyap]

Originally Posted by khanna_sumit

How come it is divisible by 81?

and how can u say that rule is true?

1111...81 times= 10^80+10^79+10^78+............1
= (10^80-1)+(10^79-1)+(10^78-1)+......(10^1-1)+ 81
=9(11..80times+11..79 times+11..78 times....+1)+81
=For 11..80 times+11...79 times+.....+1, the divisibility by 9 cann be checked by finding the sum of them=80+79+78+...1= 81.80/2
So we have a factor of 81 in the the number represented by 81 unities. Hence it is divisible by 81.
But i dont think it is a general case. Will have to find a big number where the sum of digits is equal to 81 but the number is not divisible by 81.

[vineet.nitd]

Originally Posted by amar_kashyap

1111...81 times= 10^80+10^79+10^78+............1
= (10^80-1)+(10^79-1)+(10^78-1)+......(10^1-1)+ 81
=9(11..80times+11..79 times+11..78 times....+1)+81
=For 11..80 times+11...79 times+.....+1, the divisibility by 9 cann be checked by finding the sum of them=80+79+78+...1= 81.80/2
So we have a factor of 81 in the the number represented by 81 unities. Hence it is divisible by 81.
But i dont think it is a general case. Will have to find a big number where the sum of digits is equal to 81 but the number is not divisible by 81.

Wotever number i have taken whose sum is 81 , is coming out to be a perfect multiple ...
can u give me a NUMBER whose digit sum is 81 and not divisible by 81..???

[amar_kashyap]

Originally Posted by vineet.nitd

Wotever number i have taken whose sum is 81 , is coming out to be a perfect multiple ...
can u give me a NUMBER whose digit sum is 81 and not divisible by 81..???

88888888881..try dividing it by 81...

[vineet.nitd]

Originally Posted by amar_kashyap

88888888881..try dividing it by 81...

hmmmmm

[varun nakra1]

Originally Posted by amar_kashyap

In between if there are other methods of doing problem#1...Please post

Another method is write 32^32 =(30+2)^32 Now opening the binomial expression gives us the last two terms as 30*32*2^31 whch gives (....80) as a number and the last term as 2^32..Now 2^32 =[(1024)^3]*4..then 24*24 gives 76 as the last two digits then 76*24 again gives us 24 and finally 24*4 gives 96 as the last two digits (...96)..Thus the binonial expression gives us all zeroes at the unit's place and a 6 at the unit's place coz of the last term..and it gives all zeroes at the tenth place and one 9 and one 8 at the tenth place..i.e. .........00+......00+........+.......80+........96 =............76 as the answer

But i still feel modulo by 100 is a better method

[amar_kashyap]

problem#3
A number N has 3 prime factors and 27 factors which are perfect cubes. If 125 of the factors of N are perfect squares , how many factors does N have...

A.648 B. 729 C. 900 D. 1000.