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Tuesday, August 2, 2011. Last Non Zero Digit Of a factorial. This is an oft repeated concept in different examinations but if you can grasp the following algorithm solving problems is a one min task. Lets say D(N) denotes the last non zero digit of factorial, then the algo says. D(N)=4*D[N/5]*D(Unit digit of N)[If tens digit of N is odd]. D(N)=6*D[N/5]*D(Unit digit of N)[If tens digit of N is even]; Where [N/5] is greatest Integer Function. Find the last non zero digit of 26! Solution Scheme and Approach.

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Tuesday, August 2, 2011. Last Non Zero Digit Of a factorial. This is an oft repeated concept in different examinations but if you can grasp the following algorithm solving problems is a one min task. Lets say D(N) denotes the last non zero digit of factorial, then the algo says. D(N)=4*D[N/5]*D(Unit digit of N)[If tens digit of N is odd]. D(N)=6*D[N/5]*D(Unit digit of N)[If tens digit of N is even]; Where [N/5] is greatest Integer Function. Find the last non zero digit of 26! Solution Scheme and Approach.

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QUANTOMANIA | quantomania.blogspot.com Reviews

https://quantomania.blogspot.com

Tuesday, August 2, 2011. Last Non Zero Digit Of a factorial. This is an oft repeated concept in different examinations but if you can grasp the following algorithm solving problems is a one min task. Lets say D(N) denotes the last non zero digit of factorial, then the algo says. D(N)=4*D[N/5]*D(Unit digit of N)[If tens digit of N is odd]. D(N)=6*D[N/5]*D(Unit digit of N)[If tens digit of N is even]; Where [N/5] is greatest Integer Function. Find the last non zero digit of 26! Solution Scheme and Approach.

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quantomania.blogspot.com

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QUANTOMANIA: e is irrational

http://quantomania.blogspot.com/2010/01/e-is-irrational.html

Monday, January 11, 2010. Prove that e is irrational. Solution scheme and approach. Let e is rational so e=p/q. Multiplying equation number (i) by q! Is an integer and q! E is also an integer. So we can say other terms should be an integer. Let I signifies the rest terms whose summation is an integer. 1/(q 1) 1/(q 1)*(q 2) 1/(q 1)*(q 2)*(q 3) . 1/(q 1) 1/(q 1) 2 1/(q 1) 3 .=1/(q 1)/[1-1/(q 1)]=1/q 1. So, e should be an irrational number. Q.E.D. This solution has been proposed by Fourier].

2

QUANTOMANIA: PROBLEM OF THE WEEK-1

http://quantomania.blogspot.com/2010/01/problem-of-week-1.html

Monday, January 11, 2010. PROBLEM OF THE WEEK-1. Here I start a new section called problem of the week. Solution of this question will be posted in the coming week. Till then have fun. Alphonse and Beryl are back! They are playing a two person game with the following. 8226; Initially there is a pile of N stones, with N = 2. 8226; The players alternate turns, with Alphonse going first. On his first. Turn, Alphonse must remove at least 1 and at most N −1 stones from. Strategy.Where N =100 [Difficult].

3

QUANTOMANIA: Powered Modulo

http://quantomania.blogspot.com/2010/02/powered-modulo.html

Tuesday, February 9, 2010. When we need to calculate the modulus (or remainder) of a nth powered number, we have to find out out the pattern. If a 2 b 2 is a multiple of 7 2009 then prove that ab is a multiple of 7 2010. [USA Math Talent Search]. Solution scheme and approach. A 2 b 2=7 2009*k. As right hand side is divisible by 7 left hand should be divisible by 7. Now we can deduce the the following table. Number- - -Square- - - -remainder (modulo 7). 0- - - - - - -0- - - - - - -0. A 2 b 2=7 2009*k.

4

QUANTOMANIA: Number Of Solutions

http://quantomania.blogspot.com/2010/02/number-of-solutions.html

Thursday, February 11, 2010. This is a typical pattern. If you can remember different formats of this specific problem, cracking questions is just a matter of few seconds. Let's have a look. Let n ≥ 1 be an integer and let t denote the number of positive integer divisors of n 2. Show. That the number of positive integer solutions (a, b) of the equation 1/a−1/b = 1/n is precisely. Equal to (t − 1)/2. Solution scheme and approach:. From the above equation it's clear that 1/a 1/n = n a. Equal to t /2. A)6 2...

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QUANTOMANIA: Polynomial

http://quantomania.blogspot.com/2010/02/polynomial.html

Saturday, February 27, 2010. Let f be a polynomial of degree 98, such that f(k)=1/k. [k=1,2,3,4.99]. Find the value of f(100). [USAMath Talent Search 2009]. Solution scheme and approach. Consider a function g(k) such that. As f(k) is a polynomial of deg 98 g(k) should be a polynomial of deg 99. Subscribe to: Post Comments (Atom).

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The Unfinished Evening | THE GOD OF SMALL THINGS | Page 2

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THE GOD OF SMALL THINGS. Number Theory’s Concept. December 13, 2010. This too shall pass only the history shall remain.Whether history will select you or throw you at the dustbin depends on your choice.So time to finish THE UNFINISHED EVENING. November 18, 2010. November 14, 2010. 8220;Hope is a good thing,may be the best of the things and no good thing ever dies.”. Now time I get back to my desk to finish all those futile elements before the morning starts. September 26, 2010. 8220;Friends and liars.

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QUANTOMANIA

Tuesday, August 2, 2011. Last Non Zero Digit Of a factorial. This is an oft repeated concept in different examinations but if you can grasp the following algorithm solving problems is a one min task. Lets say D(N) denotes the last non zero digit of factorial, then the algo says. D(N)=4*D[N/5]*D(Unit digit of N)[If tens digit of N is odd]. D(N)=6*D[N/5]*D(Unit digit of N)[If tens digit of N is even]; Where [N/5] is greatest Integer Function. Find the last non zero digit of 26! Solution Scheme and Approach.

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