No friend I dont think that this number is the right answer .... becoz it is very large number and we have to find the smallest number which are divisible by 3,5 and 7 as well as sum of the digit also divisi ble by the same...

I can give you a smaller number than this

999999999936105

Now this number adds up to 105 and is divisible by 3, 5 and 7 also.

But again this is also not the smallest number.... so I am still waiting for the right answer....

Lage raho......

OK friends still there is no answer for the above question... well i will wait for another few days....

IN THE MEANWHILE I will add a twist to this question ....

In the same question can u write down the smallest number such that the digits are also only 3,5and 7. So now there are two questions

1) FIND the smallest number such that the number is divisible by 3,5and 7 and digits of the number also adds upto a number which is divisible by 3,5 and 7.

2) Find the smallest number such that the number is divisible by 3,5and 7 and digits of the number also adds upto a number which is divisible by 3,5 and 7 such that the digits used is only 3, 5 and 7.

Waiting to hear the right answer soon...

Lagey raho...

You've missed the second condition

My guess is :

210210...35 times

I might be wrong Feel free to correct me

No friend I dont think that this number is the right answer .... becoz it is very large number and we have to find the smallest number which are divisible by 3,5 and 7 as well as sum of the digit also divisi ble by the same...

I can give you a smaller number than this

999999999936105

Now this number adds up to 105 and is divisible by 3, 5 and 7 also.

But again this is also not the smallest number.... so I am still waiting for the right answer....

Lage raho......

lehmanbrothershereicome Saysu take LCm of 15 and 7 and that is 105 which is the answer :D

You've missed the second condition

My guess is :

210210...35 times

I might be wrong Feel free to correct me

Here comes another interesting question.......

and its really an interesting and exciting one >>

Q Find the smallest number divisible by 3, 5 and 7 such that the sum of the digits of the number is also divisible by 3, 5 and 7.

u take LCm of 15 and 7 and that is 105 which is the answer 😁

Here comes another interesting question.......

and its really an interesting and exciting one >>

Q Find the smallest number divisible by 3, 5 and 7 such that the sum of the digits of the number is also divisible by 3, 5 and 7.

yes luvufunda1 you are right. whatever correction u made is centpercent right. I just forgot to add the sign of power. Good job all the answer are correct but try Q2 also. Its not tough to crack.Find some logic for the same.

Well others thread are already in advance stage so I think that for a beginner it will be of much help.

And the first set of quiestions are:

Q1. Factorise 1001.(well this question is very easy , i hope every one knows)

Q2. Find any factor of 1000100010001 except the number and the unity itself. (hope this will need some thinking )

Q3. All the divisors of 72 are multiplied. The product can be written in the form 2a3b. Then the value of a + b is

a. 28

b. 32

c. 5

d. 30

Q4.Find the smallest positive integer n such that if 10n = MN, where M and N are positive integers, then at least one of M and N must contain the digit 0.

For q1 yes it was very easy and i think u have given a clue to or against for the second Question.

Q1 >> 7*11*13

Q2 >> not gettin any clue ....

Q3 >> ans should be 30.(but i think.... it is 2^a*3^b )

Q4 >> ans should be 8. As it should be again 10^n = 2^n *5^n ... now min value of n =8 then N will have one zero because 5^8=390625.

I think it was a great effort to bring some easy and logical Question. Great work Gaurav .. keep it up and pour lot of Question without errata.

Well others thread are already in advance stage so I think that for a beginner it will be of much help.

And the first set of quiestions are:

Q1. Factorise 1001.(well this question is very easy , i hope every one knows)

Q2. Find any factor of 1000100010001 except the number and the unity itself. (hope this will need some thinking )

Q3. All the divisors of 72 are multiplied. The product can be written in the form 2a3b. Then the value of a + b is

a. 28

b. 32

c. 5

d. 30

Q4.Find the smallest positive integer n such that if 10n = MN, where M and N are positive integers, then at least one of M and N must contain the digit 0.

Where are the problems??

Anyways..right now we have too many threads on quant ..if you want to do some maths ..you are very much welcome there..

May be you can visit those threads and see for yourself.

Frankly you wont see much participation on this thread now...May be few months later...

Still, if you wish you can carry on..depending upon Moderators' decision.

All the best..:D

Hi Puys,

The sole purpose of this thread is to keep learning and solving new problems and concepts. Mathematics is beautiful as a lady and simply simple which is just opposite to the nature of a lady. Though this is very early to devote a thread to CAT'07 but I think its never early to build your platform for big show. My aim is to unconditionally help anyone who seeks knowledge because I believe that this is the right approach to keep increasing one's knowledge. So all the peers are invited to pour suggestions and bring innovative ideas to solve QUANT.We would start with very basic fundas and problems then we will go the zenith of problem solving techniques near CAT. Learning is an adventure and I hope this thread will help us all. So gear up and start cracking . Your alternate solution and concepts are welcome.

All the best.

dude there are many more threads available for this purpose use quant marathon or quantagious for ur discussion do not open threads for which there are already available ones.