I must say the concept that you discuss are not found in any of the books, thanks for posting such concepts.
I have a doubt to rectify, would you please share your mail id or drop me a mail at email@example.com
This is exactly what I was searching for. I have come across similar questions and was not sure how to solve them. Thanks a lot and keep up the good work 🙂
When cubes there are only 5 possible remainders… How did you find this out ? Is there a method to find this because this could be very helpful during exams. Or is it just trial and error ?
Vaibhav, try numbers from 13k, 13k+1, 13k+2….13k+12 and check the remainders of their cubes with 13. You will find there are only the above possibilities.
In natural number question we have to find out max number of set of number that should be 29+3=31 instead of 29.
i mean 32
Is there any thumb-rule for identifying pigeons and pigeonholes in a problem?
Not really….you have to use common sense, which is what makes it so difficult for a lot of people 😛
Hi J! I found your post really helpful.. but I am not getting the solution of q1. If it was asked to have the minimum number of boxes with the same number of oranges, then the answer would had been 4 , am I right? Your post probably has an error in the final answer part.. 🙂
No, the minimum value would have been 0. There could be 0 boxes having a particular number! But this is a different question – it is asking for a minimax, that is, the minimum value of the maximum possible number. Read the earlier 3 posts again.
Let me phrase it differently. Suppose I say, you want to find the mode of the distribution, what could be the least possible value for the mode? You will find that it could be 5 but not less.
Sir but the minimum possible value for the mode can be 4 right? As 5×31=155. which implies that some weights must have no. of boxes less than 5. Please correct me if I’m wrong.
Remember that we are asked the minimum possible value of the maximum. As you said, Some may be less than 5, but do note that all cannot be less than 5. So the maximum is at least 5.
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