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Take that!

20-2+18

Cruising along ,boys! Let me join in the cruise.

24-2+22

Go to 'New Topic'.

Pulling another one back for boys.There you go.....

74-2=72

Dear Priyatamau

About six years back, my elder brother made exactly the same mistake while filling up the I.I.T-J.E.E application form. He did exactly the same thing as your distressed friend has done.And you know what?His form was not rejected and he did take the examination!
Well, I do not know what may happen, for it happened six years back and I certainly am no expert to comment upon these things, but I just thought that sharing the example of my brother may prove to be of some help.Most probably, your friend will also have the same fortune as my brother had. Wish him luck from my side.

Goodbye!

I am glad to discover that so many people have expressed their views on this subject.It was most considerate of the Administration to have considered the expressed grievences and come up with a seemingly cure-all solution in the form of 'Nickle-Trickle'. I thank you all for giving a voice to the cause that was, to its credit,were listened to by the Administration.

Thanks.

What about the other people (31 uptil now) who have viewed this post? I request all those people who can, in some way, relate to the issue raised here to chip in with their views before the imminent relegation of this topic is to the next page.I wish to know if the concern raised by me and supported by nunoxic has any relevance for the roughly one thausand to two thausand users who log in each day,or is it just a few hapless, self-proclaimed introverts like me who are facing this problem.

I wonder if all this 'the nickel is the currency' thing is proving to be helpful for the students.I mean, introverts like me do not really know many people around and most of my classmates were members before I joined.Consequently, I have not earned a single nickel since joining.I suggest that nickels may be awarded to those who get a 'salute' for their answer, or some other system may be put in place which does not undermine the very reason why the nickel exists i.e. to get more people to join.What does the Administration think about this idea?Please, help me and many others like me by considering something on these lines.Thanks.

Hi

You have quite a luxury that you have to go to schoo only 3-4 days a week.Make use of that, buddy. My school compells me to attend school eveyday and that's nothing short of hell.

I wonder if all this 'the nickel is the currency' thing is proving to be helpful for the students.I mean, introverts like me do not really know many people around and most of my classmates were members before I joined.Consequently, I have not earned a single nickel since joining.I suggest that nickels may be awarded to those who get a 'salute' for their answer, or some other system may be put in place which does not undermine the very reason why the nickel exists i.e. to get more people to join.What does the Administration think about this idea?Please, help me and many others like me by considering something on these lines.Thanks.

Thanks to both of you, Nunoxic and Mr Puneet, for answering my query.

With regards,
Ashish

Dear Viewer

I am a student of Class11.I wanted to know if I should continue to use R D Sharma Mathematics for learning the basics of a topic.If not, then what books will you suggest at the basic level.

With regards

Ashish

Thanks to both Saurabh hr and Ms Rashmi for giving the answers.

With regards,
Ashish

Dear Saket

We know that

5.2=10
5.2.5.2=100

You would notice that whenever we multiply 5 with 2, we get a zero in the product and that the number of zeros we get is equal to the number of 5&2 pairs we have in the factors of the number.

For further example

5000=5.2.5.2.5.2.5

Three 5&2 pairs, three zeros, right!

Extending the concept further, whenever we multiply 5 with an even number, we get a zero in the product because an even number can always be expressed as a multiple of 2.

e.g. 5.12=5.2.2.3=60

Consider 5! ,

5! = 1.2.3.4.5 = 120

4 being an even number is multiplied with 5 to give us the lone zero in 120.

Moving onto 10! ,

10! = 5! .6.7.8.9.10 =120.6.7.8.9.5.2 = 3628800

Now, in 10! we have two 5&2 pairs (do not forget the one in 5!), and hence, two zeroes in the product.

So, it seems that if we could work out the number of 5s in n! , we would know the number of zeroes in the product.

Note that the number of 2s is not important because, the number itself may have 2 as its factor, otherwise, there are way too many even numbers surrounding each multiple of 5,eager to provide it with the 2 required to get a zero in the product.

Take the case of 15!
15! = ???????13.14.15 = ???????..13 . 7.2 . 5.3

So, the 2 belonging to 14 is used.

Now, let us get down to 100!

As explained above, we need to find out how many 5s one encounters in 100! i.e. the count of the multiples of 5 from 1 to 100.

Easy, divide 100 by 5. The answer is 20.

The numbers come out to be,

5 = 5.1
10 = 5.2
15 = 5.3
20 = 2.2.5
25 = 5.5

Oops! we forgot to consider the extra 5 of 25.

Similar is the case with 50, 75,100.

Add these four extra 5s to the tally of 20.

So, there are 24 5s in total.

Hence, there are 24 zeroes in 100! .

Notice that 100/25 = 4. [This may be used to find out the number of ?extra 5s?.]

Now, kindly find out the number of zeroes in 200! on your own. Be careful.

Good bye!

Well, my plans are also quite similar and I am quite confident that they will work.However, I would also like a few more people to air their opinions on this.