thisisananth

@thisisananth

One solution for this is

A S V
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?
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
Hi PG, I tried subscribing to the quant a day mailing list and the following error occured Please resolve it ...


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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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