ISI - Difficult Maths Questions!!

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ISI - Difficult Maths Questions!!

Ridhika
Q1) When 3^2002 + 7^2002 + 2002 is divided by 29 the remainder is ?

Q2) The highest power of 18 contained in 50C25 is ?

Q3) The last digit of 2137^754 is?

Q4) The number of divisors of 6000 where 1 and 6000 are also considered as divisors of 6000 are?

Q5) The coefficient of x^2 in the binomial expansion of (1 + x + x^2)^10 is ?

Q6) The number of integer solutions of xy - 6(x+y) = 0 with x =<y is:
a) 5
b)10
c) 12
D) 9

Q7) The remainder when 3^37 is divided by 79 is?
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Re: ISI - Difficult Maths Questions!!

Dreyfus
For q3 I am getting last digit as 9
And for q4...there r 40 divisors including 6000 and 1( I used prime factorisation)
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Re: ISI - Difficult Maths Questions!!

Ridhika
So after prime factorization how do you count all the combinations of prime factors to count the # of non prime factors ?

How did you approach q3?



----- Reply message -----
From: "Vaibhav Garg [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: ISI - Difficult Maths Questions!!
Date: Mon, May 5, 2014 12:55 PM

For q3 I am getting last digit as 9
And for q4...there r 40 divisors including 6000 and 1( I used prime factorisation)



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Re: ISI - Difficult Maths Questions!!

don123
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Re: ISI - Difficult Maths Questions!!

Dreyfus
This post was updated on .
In reply to this post by Ridhika
Let's say for q4 instead of 6000 we have no.5 with prime factors being 5 now 5 has exactly two divisors ie 1 & 5. Now take another no. 10 now it has 2 prime factors 2 and 5 with their power being 1nd 10 has 4 divisors ie 1,2,5,10 ...
In both of the no.s their is similar pattern for counting no.of divisors that is for 5 ...the power of prime factor is 1 andwhen u add1 to the power u get 2.
Similarly for 10 u hv two prime factors each with powers 1 only! So no. Of divisors for 10 would be (1+power of prim factor a)*(1+power of prime factor b) so for 10, 2*2=4 are the no. of divisors including 1 nd 10
Extend this case with 150 also...u will get 12 divisors
Now 6000 can be written as 2^4*3*5^3
So total no of divisors= (4+1)*(1+1)*(3+1)=40
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Re: ISI - Difficult Maths Questions!!

Ridhika
Thanks a lot guys !

@ Khiladi - I dont know what you mean for the 2nd answer..
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Re: ISI - Difficult Maths Questions!!

don123
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Re: ISI - Difficult Maths Questions!!

Ridhika
Thanks a lot ! Please also look at my maths doubts posted here.. http://economicsentrance.weebly.com/forum.html

Please help with these isi questions also!
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Re: ISI - Difficult Maths Questions!!

Ridhika
link isnt working.. ISI 2007 ME-I Doubts.. Thanks :)