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8 1 4

8 1 4

4 8 1

This satisfies the condition that Sushil got higest in the last round. So in the first round Vijendar 4 got the middle number.

Yes the answer is 1.

Sagar I have a different and simple approach.

Here comes our next basic funda.

All of us know that the remainder obtained when a number is divided by10 is nothing but the unit's digit of that no.Had u ever thought of the remainder that is obtained when divided by 9.It is nothing but the single digit sum of that number.

For example the single digit sum of 4658 = 4 + 6 + 5 + 8 = 23 = 2 + 3 =5.

Now when 4658 is divided by 9 the remainder is nothing but the remainder obtained when sum of the digits of 4658 i.e 23 is divided by 9 which in turn is nothing but the remainder obtained when 2+3 = 5 is divided by 9.

So that's the funda here folks.

Finally the remainder obtained when 4444^3333 is divided by 9 is nothing but the single digit sum.4444^3333 % 9

= 7^ 3333 % 9

= 343^ 111 % 9

= 1.

Since the remainder obtained when N is divided by 9 is 1 the single digit sum is 1.Remember if the remainder obtained is 0 then the single digit sum is not 0 but ofcourse it is 9

Please explain the part in blue. How is it true?

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Can you please tell the solution to this question

Set A is formed by selecting some of the numbers from the first 100 natural numbers such that the HCF of any two numbers in the set is the same.

Q: If every pair of numbers of set A has to be relatively prime and set A has the maximum number of elements possible, then in how ways can the set A be selected?

64

96

72

108

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Time occured: Mon, 20 Nov 2006 06:00:08 +0000

One of my friends, Venkat Naidu.. a member of this forum suggested me to have a look at this

Hi I'm Ananth. I am working in CTS presently. I have taken CAT once and got just 75 percentile.. I wish to crack CAT this time.. and well thats why I'm here.

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