QUANTOMANIA

Last Non Zero Digit Of a factorial

2011-08-02T12:16:00.000-07:00

This is an oft repeated concept in different examinations but if you can grasp the following algorithm solving problems is a one min task Concept Lets say D(N) denotes the last non zero digit of factorial, then the algo saysD(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 #Problem

After One Year

2011-07-27T22:03:00.000-07:00

#Problem 17 If a,b,c,d are positive real numbers then find a possible value of a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)(a)11/5 (b)17/3 (c)2 (d)11/9 Solution Scheme And Approach: a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)>a/(a+b+d+c)+ b/(b+c+a+d) + c/(b+c+d+a) + d/(a+c+d+b)=1a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)<a/(a+b)+ b/(a+b)+ c/(c+d)+ d/(c+d) =2Hence the range of the expression

Digital root

2010-04-15T01:19:00.001-07:00

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Breaking a long overdue

2010-04-14T12:19:00.000-07:00

#Problem 15 Find the total number of solutions of | [x] - 2x |=4. (a)1,(b)2,(c)4,(d)6,(e)None of these [Where [x] is Greatest Int Func] Solution scheme and approach: Case 1 :( x is a whole number) so [x]=x =>|-x|=4 =>x=4 or x=-4 Case 2: ( x is a fraction) so x=a+f (f is the fractional portion) then the equation gets converted into |a-(2a+2f)|=4 =>|-a-2f|=4 =>a=4-2f =>a=3 (when f=1/2) =>x=3.5 or a

Polynomial

2010-02-27T10:30:00.000-08:00

#Problem 14 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 g(k)=kf(k)-1 As f(k) is a polynomial of deg 98 g(k) should be a polynomial of deg 99. =>g(k)=C*(k-1)*(k-2)*(k-3)*...*(k-99) =>kf(k)-1=C*(k-1)*(k-2)*(k-3)*...*(k-99)............(1) by putting k=0

Number Of Solutions

2010-02-11T04:12:00.000-08:00

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. #Problem 13.1 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

Powered Modulo

2010-02-09T05:33:00.000-08:00

When we need to calculate the modulus (or remainder) of a nth powered number, we have to find out out the pattern. #Problem 12.1 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

CAT CLASSIC : Number Theory

2010-02-01T08:17:00.000-08:00

Today I talked to one of my Pagalguy friend. He was right in saying that my blog is not good for all. I agree. What I try to discuss, here, is obviously not for all and sometimes the concept becomes too tough to understand. And there are a lot of people who are unable to attend coaching classes due to different reasons. So being a pagalguy member I should help them. And if this section can help

Concept 3: CATLAN numbers

2010-01-31T22:20:00.000-08:00

Can you tell me what is the next number of this series? 1,2,5,14,42,? You may face trouble in finding though. But here is the formula Tn=1/(n+1)*2nCn=(2n)!/[(n+1)!*n!] [For n>=1] So T5=132. The number of this series is popularly known as CATLAN numbers. And in different counting problems. I will try to explore some of them. (A)Mountain Ranges and Diagonal Avoiding Paths (i)How many mountain

Again a perfect square

2010-01-15T09:53:00.000-08:00

#Problem11 49=7^24489=67^2444889=667^2............................Now prove that if we insert 48 in the middle of the previous term of this series, like this way, in each case it will produce a perfect square. Solution scheme and approach Number is in this format [4444...4](n+1 numbers)[8888...8](n numbers)9 So we can write the number in the following way [4*10^(2n+1)+4*10^2n+4*10^(2n-1)+....4

CAT SHOT-1

2010-01-15T09:26:00.001-08:00

This section is dedicated to CAT. Here we deal with the problems which often appears in mocks and actual cat as well.Fasten your belt. Here we go. (1)2^y+2*(3^y)>3*(4^y) and y=3x^2+2x-2 Which of the following is a possible value of x? (a)-1.5 (b)-2.5 (c)-0.5 (d)0.7 (e)1.2 (2)For N values of t, where t<=250, the highest power of t in t! is t^5. Then which of the following is true? (a)N<=4 (b)N=6

Concept 2:Diophantine Equations

2010-01-14T16:06:00.000-08:00

Diophantine equation is an equation of polynomials where variables are integers only. Linear Diophantine Equation from the name it's clear that it's an equation of degree 1. ax+by=g.c.d.(a,b)=d ...(I) is also an linear Diophantine equation.According to Bézout's identity if equation number (I) holds true.i.e. gcd(a,b)=d satisfies then, there should be an integer solution of this equation.It's

Infinite primes

2010-01-14T11:07:00.000-08:00

#Problem 10 Prove that infinitely many prime numbers are possible. Solution Scheme and approach Let if possible there are a1,a2...an number of finite primes. Consider another number N=(a1*a2*a3*...*an)+1 Now we can say, N is such a number which can be divide by all primes but always gives a remainder 1. Hence, contradiction. Because any number>1 should be divisible by at least one prime number.

Maximum value of a variable

2010-01-13T13:01:00.000-08:00

#Problem 9 a+b+c+d+e=8 a^2+b^2+c^2+d^2+e^2=16. [a,b,c,d,e are real numbers] Find the maximum value of e. Solution Scheme and approach: This problem can easily be done with Cauchy-Schwartz inequality. But will explain this in my next topic. Let's do it without Cauchy. (a-r)^2+(b-r)^2+(c-r)^2+(d-r)^2+(e-r)^2 =(a^2+b^2+c^2+d^2+e^2)-2r(a+b+c+d+e)+5r^2 =16-16r+5r^2[ By putting the values] we can

PROBLEM OF THE WEEK-1

2010-01-11T07:53:00.000-08:00

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 rules: • Initially there is a pile of N stones, with N >= 2. • The players alternate turns, with Alphonse going first. On his first turn, Alphonse must remove at least 1 and at

e is irrational

2010-01-11T06:33:00.000-08:00

#Problem 8 Prove that e is irrational Solution scheme and approach e=1+1/1!+2/2!+3/3!+.....(i) Let e is rational so e=p/q multiplying equation number (i) by q! we get q!*e=q!+q!/1!+q!/2!+q!/3!+....+q/q!+ other terms Now as q!+q!/1!+...+q!/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. I=q![

Number's application

2010-01-07T09:34:00.000-08:00

#Problem 7 Show that there is no integer a such that a^2-3a-19 is divisible by 289.[R.M.O. 2009] Solution Scheme and Approach a^2-3a-19 =a^2-3a-70+51 =(a+7)(a-10)+51 ...(i) Now as (i) divides by 289 it should be divisible by 17 as well. So, it is clear at least one of (a+7) and (a-10) should be divisible by 17. Let, a-10=17k =>a=10+17k therefore a+7=17(k+1) Hence we can write a^2-3a-19=289k(k+1

Cubic Equation

2010-01-02T07:26:00.000-08:00

#Problem 6 The cubic equation x^3+2x-1 = 0 has exactly one real root r. Note that 0.4 < r < 0.5. (a)Prove that an increasing sequence of positive integers a1,a2,a3,.....exists, such that 1/2=r^a1+r^a2+r^a3+.... is possible (b) Prove that the sequence that you found in part (a) is the unique increasing sequence with the above property. Solution Scheme and approach (a) As r is a root then we may

Perfect square

2010-01-02T02:11:00.000-08:00

This is one of the important topics which often comes in various competitive exams. #Problem 5(A) Find the total number of values of n for which n^2 + 24n + 21 is a perfect square.Where n is natural number. Solution scheme and approach Let n^2+24n+21 is a perfect square.Hence we can write n^2+24n+21=k^2 =>n^2+2*n*12+12^2-12^2+21=k^2 =>(n+12)^12-k^2=123 Case I (n+12+k)(n+12-k)=3*41=(-3)*(-41) So,

Concept 1 :KiSsInG cIrClEs

2010-01-01T00:51:00.000-08:00

Descartes' theorem (1964) If four mutually tangent circles have curvature ki (for i = 1,...,4), (ki=1/ri, Where ri is radius) (1) When trying to find the radius of a fourth circle tangent to three given kissing circles, the equation is best rewritten as: (2) The ± sign reflects the fact that there are in general two solutions. Ignoring the degenerate case of a

How many primes

2009-12-31T11:17:00.000-08:00

#Problem 4 How many primes among the positive integers, written as usual in base 10, are such that their digits are altenating 1’s and 0’s, beginning and ending with 1? [Putnam] Solution scheme and approach Let, X=1010101…1=1+10^2+10^4+…+10^2n=(10^2n-1)/99=(10^n+1)*(10^n-1)/99 For n=2 X=101 which is a prime number For n>=3 If n is even { As (10^n+1)>99 (10^n-1)=(9+1)^n -1=0(mod 9) So,(10^n-

Ultimate Divisibility

2009-12-31T01:07:00.000-08:00

This is one of the famous puzzles. Try it. #Problem 3 Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements: a. The number should be divisible by 9. b. If the most right digit is removed, the remaining number should be divisible by 8. c. If then again the most right digit is removed, the

Infinite descent

2009-12-30T16:46:00.000-08:00

If x, y, z and n are integers, there are no solutions of x^n+y^n=z^n with n>2 and x,y,z>0.This is known as Fermat's last theorem. This problem can be an extension of Fermat's Infinite Descent #Problem2 Find all solutions to c2 + 1 = (a2 − 1)(b2 − 1), in integers a, b, and c. Solution Scheme and approach Clearly a = b = c = 0 is one integer solution. Also, if c = 0, a^2 − 1 = ±1, and so

Number of solutions

2009-12-30T14:23:00.001-08:00

Here I start my blog with a very interesting problem. #Problem 1 Find the value of x and y. x^3+y^3=7163.[Where x and y are positive integers] Solution scheme and approach 1 First of all one number should be even another is odd. Let, x=(2n+1) and y=2m So, x^3+y^3=7163 =>(2n+1)^3+(2m)^3=7163.....(i) =>8n^3+12n^2+6n+1+(2m)^3=7163 =>2n(4n^2+6n+3)+8m^3=7162 =>n(4n^2+6n+3)+4m^3=3581 As RHS is odd