2010/065) an example of chinese remainder theorem

What would be the least total no. which give the remainder 1,2 and 3 when divided by 7,9 and 11?

1,2 and 3 when divided by 7,9 and 11?

hence
x= 1 mod 7
x= 2 mod 9
x =3 mod 11

now 7 ,9 and 11 are pair wise coprime

x = 1 a1b1 + 2 a2b2 + 3 a3b3 mod 7*9*11(or 693)


where a1 = 9*11 = 99
a2 = 7 * 11 = 77
a3 = 7 * 9 = 63

and
a1b1 = 1 mod 7
a2b2 = 1 mod 9
a3b3 = 1 mod 11

a1 = 99 so b1 = 99 mod 7 or 1 mod 7 so b1 = 1

77b2 mod 9 = 1 so 5b2 mod 9 = 1 so b2= 2 (as 5*2 = 10 mod 9)

63b3 = 1 mod 11 or 8b3 =1 mod 11 so b3 = 7 as 7 * 8 = 56 = 1 mod 11

So x = 1 * 99 * 1 + 2 * 77 * 2 + 3 * 63 * 7 mod 693 = 1730 mod 693

= 344 mod 693

So x if of the form 693n + 344 or lowest x = 344

2010/064) if a+b+c=1, ab+bc+ca=2, abc=3, then

find the value of 1/(a+bc)+1/(b+ca)+1/(c+ab).


we have

a+b+c=1 ...1

ab+bc+ca=2 .. 2

abc=3 ...3

so a b c are roots of equation

x^3-x^2 + 2x - 3 = 0

further

as a+ b+ c = 1

so a = 1- (b+c)
so a + bc = 1-(b+c) + bc = (1-b)(1-c)

similarly b+ ca = (1-c)(1-a)
c + ba = (1-b)(1-a)

we need to find 1/(a+bc)+1/(b+ca)+1/(c+ab) = 1/ ((1-b)(1-c)) + 1/ ((1-c)(1-a)) + 1/ ((1-b)(1-a))

= (1-a) + (1-b) + (1-c)/(1-a)(1-b)(1-c) = ((a-1) + (b-1) + (c-1))/(a-1)(b-1)(c-1)



so we need to form an equation whose roots are a-1 , b- 1 and c- 1

as a b and c are roots of f(x) = x^3-x^2 + 2x - 3

so a-1 b- 1 and c- 1 are roots of

f(x+1) = (x+1)^3 - (x+1)^2 + 2(x+1) - 3 = x^3 + 2 x^2 + 3x -1

so (a-1) + (b-1) + (c-1) = - 2

(a-1)(b-1)(c-1) = 1

so given expression = ((a-1) + (b-1) + (c-1))/((a-1)(b-1)(c-1)) = - 2

2010/063) prove the inequality a+b/c+b+c/a+c+a/b>6 for a,b,c > 0 .?

this can be proved in at least 2 ways


method 1
we need to prove

(a+b)/c+(b+c)/a+(c+a)/b>6

or (a+b+c)/c + (a+b+c)/a + (a+b+ c)/b > 6

or (a+b+c) ( 1/a + 1/b+ 1/c) > 9

we know by AM GM inequality (a+b + c)/ 3 >= (abc)^(1/3)

again (1/a+ 1/b+ 1/c)/3 > = (1/abc)^(1/3)

multiplying we get (a+b + c)/ 3 * (1/a+ 1/b+ 1/c)/3 > = 1

or (a+b + c) * (1/a+ 1/b+ 1/c) >= 9

it is equal when a = b = c

proved


method 2
x+ 1/x >= 2 as (x+1/x) = (sqrt(x) -1/sqrt(x))^2 + 2

puttiing x = a/b . b/c and c/a in 3 times we get 3 eqautions

a/b + b/a >= 2
a/c+ c/a > = 2
b/c + c/b > = 2
adding above 3 and rearranging the terms we get the result.

2010/062) What are all the rational numbers x such that x + 1/x is an integer?

x + 1/x = (x^2 + 1)/x

This will be an integer if x^2 + 1 = kx, for some integer k. Equivalently,

x^2 - kx + 1 = 0

==> x = [ k +/- sqrt(k^2 - 4) ] / 2


for this to be an integer sqrt(k^2 - 4) has to be an integer say h( same parity as k that is both odd or both even)

k^2-4 = h^2

k^2 - h^2 = 4

so (k-h)(k+h) = 4 = 1 * 4 and 2* 2 and -1 * -4 and -2 * - 2

so k+ h = k-h = 2 or k-h = k+ h = - 2

k+ h = k-h = 2 => k = 2 giving x = 1

k+ h = k-h = 2 => k = -2 giving x = -1

so x = 1 or - 1

2010/061) prove the equality 2 + tan1° tan2° = cot1° tan2°

We know

Tan 2° = tan (1°+1°) = 2 tan 1°/(1- tan ^2 1°)
So tan 2° – tan 2° tan ^2 1° = 2 tan 1°

dividing by tan 1° we get

So tan 2° / tan 1° – tan 2° tan 1° = 2
Or
tan 2° cot 1° – tan 2° tan 1° = 2

So

tan 2° cot 1° = 2 + tan 2° tan 1°

2010/060) Proof of Bezout Identity

Bezout Identity states that

If a and b are integers (not both zero) then there exists integers u and v such that

Gcd(a,b) = au + bv

(Note: u and v are not unique

For example

au + bv = a(u-b) + b(v+a))

this can be used by backtracking the euclid’s equation to find Gcd(a,b) in terms of a b. this can be found in a number of books and is the standard process.

However this can be proved using pigeon hole principle also as below

We know that au is divisible by Gcd(a,b) so au mod b is divisible by Gcd(a,b)

Let gcd(a,b ) = k and
b / gcd(a,b) = t

Now taking an mod b( n from 1 to b/gcd(a,b)-1 that is t -1 ) there are

t-1 remainders

they are kn1, kn2, kn3, so on

there are t-1 remainders and all are divisible by k

no remainder can be zero

because na (n from 1 to t-1) cannot be a product of b.

no 2 remainder can be same if they are then difference is a multiple of b

so there has to a n such that one of the remainder is k

as all t- 1 remainders has to be different and values from 1 to n-1 so one remainder has to be 1

an = k mod b or an + bm = k

(u = n, v= m satisfy the condition)

2010/059) find sqrt((48*49*50*51)+1)

we know

x(x+1)(x+2)(x+3) + 1
= (x(x+3))((x+1)(x+2)) +1
= (x^2+3x)(x^2+3x+2) + 1

put x^2+3x = t

so we get = t(t+2)+1) = t^2 + 2t + 1 = (t+1)^2
= (x^2+3x+1)^2

so sqrt(x(x+1)(x+2)(x+3)+1) = x^2+3x + 1

putting x= 48 we get
sqrt(48 * 49 * 50* 51+1) = 48^2 + 3 * 48 + 1 = 48 * 51 + 1 = 2449

2010/058) factorize: 9x^2 - 9(a + b )x + (2a^2 + 5ab + 2b^2) = 0?

we know

2a^2 + 5ab + 2b^2 = (2a+b)(2b+a)

let s = 2a+b and t = 2b + a

s + t = 3a + 3b

9x^2 - 9(a + b )x + (2a^2 + 5ab + 2b^2)

= 9x^2 - 3(s+t)x + st
= 9x^2 - 3sx - 3tx + st
= 3x(3x-s) - t( 3x -s)
=(3x-s)(3x-t)
= (3x - 2a - b)(3x - a - 2b)

2010/057) For what positive integers n is the polynomial x^2n + x^n + 1 irreducible?

let f(x) = x^3-1 = (x-1)(x^2+x+1)

let g(x) = x^2+ x + 1

if w is complex cube root of 1 then f(w) = 0 so g(w) = 0 and g(w^2) = 0

now consider p(x,n) = x^2n + x^n + 1

n cannot be of the form 3k+1 or 3k+2 as

p(x=w , 3k+1) = w^2(3k+1) + w^(3k+1) +1 = w^2 + w +1 = 0 as w^3k = 1

p(x=w^2, 3k+1) = w^4(3k+1) + w^2(3k+1) +1 = w + w^2 +1 = 0 as w^3 = 1

so p(x,3k+1) is divisible by x^2+ x + 1

similarly p(x,3k+2) is divisible by x^2+ x + 1

so n cannot have a facor 3k+1 or 3k+ 2

so h has to of the form 3^k ( k >= 1)

replacing x by x+1 and applying Eisenstein crieteria works and shows that x^2n + x^n + 1 is irreducible for all n of the form 3^m.

as a special case n = 1 means x^2 + x + 1 is irreducible

so n = 1 or e^k for k k >= 1 of 3^k ( k >=0)

2010/056) Show that if n is a positive odd integer then 9^(n+3)+4^n is divisible by 65

Show that if n is a positive odd integer then 9^(n+3)+4^n is divisible by 65

65 = 5 * 13 prime factors and we need to show that it is divisible by 5 and 13

9 = -1 mod 5 and 4 = -1 mod 5


we have 9^(n+3)+4^n mod 5

= (-1)^(n+3)+ (-1)^n mod 5
= (-1)^n((-1)^3+ 1)) mod 5 = 0

so divisible by 5

now for 13
9 = -4 mod 13 so 9^(n+3) mod 13 = (-4)^(n+3) mod 13 = (-4)^n . (-64) mod 13 = (-4)^n mod 13

so 9^(n+3)+4^n mod 13 = (-4)^n + 4^n = 0 mod 13 as n is odd

divisible by 13 and 5 so 65

2010/055) Prove ABC+AB'C+ABC'+A'BC=AB+BC+CA by using boolean laws only.

ABC+AB'C+ABC'+A'BC = ABC+AB'C+ABC'+A'BC + ABC + ABC

as ABC + ABC = ABC so we can add 2 copies of ABC

= ABC+AB'C+ABC'+ ABC + A'BC + ABC ( rearranging)
= AC(B+B') + AB(C'+C) + BC(A' + A) ) grouping)
= AC + AB +CA as B+B' = 1 A+A'=1 and C+C'= 1

2010/054) Show that in the expansion of the (a+b)^n

Show that in the expansion of the (a+b)^n, the sum of binomial coefficients of all odd terms is equal to the sum of he binomial coefficients of all even terms.

proof: this can be done algebraically
putting
a= 1, b= - 1

we see the sum = (1-1)^n = 0

all the odd terms(power of n) shall be -ve and all the even terms positive and hence both sums must be same for the total to be zero.

but this is not ineresting.

this can be done using method of combinotrics
now the coefficient of a^kb^n-k is the number of ways we can pick k objects out of n

now there are 2 cases

1) n is odd

we break n into two 2 parts k and n-k

if k is odd then n-k is even(number of ways k element can be selected is same as number of ways

so the number of ways odd numbers can be selected is same as the number of number of ways even number can be selected

so they are same.

2) n is even

for this case we pick one element and keep it aside. now n-1 is odd and the number of ways we pick odd elements is the number of ways we can pick even elements. now adding that to even element we do not change the number of ways but make odd elements and adding to odd elements we do not change the number of ways but make even elements

so by adding (n+1)st element number of odd selections is same as number of even selections and not adding the numbers are same.

hence number of odd element selection is same as number of even elements selection

hence proved

2010/053) If r and s are the roots of x^2 + x + 7 = 0, compute 2r^2 + rs + s^2 + r + 7

2r^2 + rs + s^2 + r + 7 =
r^2+rs + s^2 + (r^2+r+7)
= r^2+rs + s^2 as r^2+r+ 7 = 0 as r is a root
= (r+s)^2-rs ..1

now as x^2+x+7 has root r and s so rs = 7 and r+s = - 1

so from (1) we get given expression = 1-7 = - 6

2010/052) For which real values of p and q are the roots of x^3-px^2+11x-q=0 three consecutive integers

the three consecutive integers can be taken to be

s = a-1, r = a, t = a +1

coefficient of x = 11 = rs + rt + sr = a(a-1) + a(a+1) + (a-1)(a+1) = 3a^2 - 1 = 11

so a = 2 or -2
a=2 gives
roots are 1,2,3

so (x-1)(x-2)(x-3) = x^3- 6x^2 + 11x -6 or p = q = 6

a = -2 gives
roots as -1 ,-2,-3

(x+1)(x+2)(x+3) = x^3+6x + 11+ 6 so p=q = - 6

so 2 solutions are

1) p= q = 6
2) p =q = - 6

2010/051) factor: x^2 - y^2 - z^2 -2yz + x + y + z

= x^2-(y^2+z^2+2yz) + (x+y+z)
= x^2-(y+z)^2 + (x+y+z)
= (x+y+z)(x-y-z) + (x+y+z)
= (x+y+z)(x-y-z+1)

2010/050) Prove that the last 6 digits of 7^10000 is 000001

we know 7^4 = 2401

so 7^10000 = (2401)^2500

now 2401^2500 = (2400+1)^2500

if we collect nth term it it (2500 C n) (2400)^n

for n > 2 2400^n is divisible by 10^6

so we need to look for n = 2 and n =1

n=0 gives 1 and 1-1 = 0

n = 2 => (2500C2)(2400)^2 = 2500*1200*2400 so 10^6 is a factor
n =1 =>(2500)(2400) = 6000000 so 10^6 is a factor

so all the elements except last that is 1 is divisible by 10^6 and last element is 1

so last 6 digits are 000001 or 7^10000 mod 10^6 = 1

2010/049) Let's say a number ABCDE is such that ABCDE*4=EDCBA. where A,B,C,D,E are no.s between 0 and 9. then ABCDE

AB has to be < 25 because25* 4 = 100 so RHS shall be 6 digit number.
A cannot be 1 as from RHS A has to be even
B is Odd as only multple of 4 *n n < 10 ending with 2 is 12 or 32 which causes odd + 4* n
so AB has to be 21 or 23
if AB = 23 DE >= 92
do D = 9
which is not possible as 4*8 is 2 ending but 4*9 is not

AB = 21
So E = 8
DE *4 should end in 12 so D = 7 as
now number is
so the number = 4*(21078+100C) = 87012+100C
or 84312+400C = 87012+ 100C
300C = 2700 or C =9
so number = 21978 * 4 = 87912

2010/048) given rational numbers p, q , r satisfying

pq+qr+rp=1

prove that (psquare +1)(qsquare +1)(rsquare +1) is the square of a rational no.

proof:
pq+qr+rp = 1

so p = (1-qr)/(q+r)

if we chose q = tan A and r = tan B

we get

1/p = (q+r)/(1-qr) = (tan A + tan B)/(1- tan A tan B) = tan (A+B)

or p = cot (A+B)

we can chose q and r to be <1 -ve="" also="" br="" case="" else="" even="" in="" positive="" want="" we="">(p^2+1) (q^2+1)(r^2+ 1) = sec^2 A sec ^2 B cosec^2 (A+B)


= (sec^2 A sec^B)/ sin^2 (A+B)

this is square of reciprocal of sin (A+B) cos A cos B
sin (A+B) cos A cos B
=( sin A cos B + cos A sin B)cos A cos B
= sin A cos A cos ^2B + cos^2 A sin B cos B
= tan A cos ^2 A cos ^2 B + tan B cos ^2 A cos ^2 B
= (tan A + tan B)/(sec^2 A sec ^2B)
= (tan A + tan B)/(1+ tan ^2 A)(1+ tan ^2B)

so (p^2+1) (q^2+1)(r^2+ 1) = ((1+ tan ^2 A )(1+tan ^2B)/(tan A + tan B))^2

if tan A and tan B that is q and r are rational then

((1+ tan ^2 A )(1+tan ^2B)/(tan A + tan B)) is rational and so (psquare +1)(qsquare +1)(rsquare +1) is the square of a rational no.

2010/047) prove that if gcd(a,b)=1 and gcd(a,c)=1 then gcd(a,bc)=1 using Bézout's identity

as GCD(a,b)= 1

so 1 = ax + by

so c = acx + bcy ...1

now GCD(a,c) = 1

so 1 = ma+ nc = ma + n(acx+ bcy) ftom 1
= a ( m + ncx) + bc(ny)

as we can put 1 as linear combination of a and bc so Gcd(a,bc) = 1

2010/046) If 2cos(a)=x+1/x and 2cos(b)=y+1/y.so then what is the value of [x/y+y/x]

x^2 - 2x cos a + 1 = 0

so x = (2 cos a +/- sqrt(4 cos^2 a - 4))/2 using discriminant
= cos a +/- sqrt(cos^2 a - 1)
= cos a +/- sqrt(- sin ^2a)
= cos a +/- i sin a = e^ia or e^-ia

taking e^ia

similarly y = e^ib

x/y = e^i(a-b) = cos(a-b) + i sin (a-b)

y/x = e^-i(a-b) = cos(a-b) - i sin (a-b)

adding we get x/y + y/x = 2 cos (a-b)

2010/045) If x + y = 2 and x^3+y^3 = 38,

then what does xy equal?

we know

(x+y)^3 = (x^3+y^3) + 3xy(x+y)

putting value from given conditions


2^3 = 38 + 3xy * 2 or 8 = 38 + 6xy => xy = - 5

2010/044) Determine five infinite sets whose union is the natural numbers and whose mutual intersections are all empty.?

Determine five infinite sets whose union is the natural numbers and whose mutual intersections are all empty.?

S1 = { 5k : k >1}
S2 = { 5k+1 : k>=0}
S3 = { 5k+2 : k>=0}
S4 = { 5k +3: k>=0}
S5 = { 5k+4 : k>=0}

2010/043) Given that 5cos x + 12 cos y = 13 ? maximum value of 5 sin x + 12sin y

What is the maximum value of 5 sin x + 12sin y , given that 5cos x + 12 cos y = 13 ?

we start with

(5 sin x + 12sin y)^2 + (5cos x + 12 cos y)^2 = 169 + 120 cos(x-y)

or (5 sin x + 12 sin y)^2 + 169 = 169 + 120 cos(x-y)

or 5 sin x + 12 sin y = 120 cos(x-y)

5 sin x + 12 sin y = sqrt( 120 cos(x-y))

this is maximum when cos(x-y) is maximum

theoritically cos(x-y) is maximum = 1 when x = y

but is is possible under given case that is x =y satisfies


then we get

5 cos x + 12 cos x = 13 so cos x = 13/17

it is possible

so maximum value of 5 sin x + 12sin y = sqrt(120)

2010/042) Prove that if a < b and c < d then ad + bc < ac + bd

a < b

so (b-a) > 0

also (d-c) > 0
as both are positive hence

(b-a)(d-c) > 0

or bd - bc -ad + ac > 0

or (bd+ac) - (bc+ad) > 0

or (bd + ac) > (bc + ad) or (bc+ad) < (bd+ac) or ad + bc < ac + bd

2010/041) If (x-k)^2 is a factor of x^3 + 3px + q

prove that 4p^3 + q^2 =0
proof:
(x-k)^2 is a factor of x^3+3px + q so x-k is factor of this as well as derivative of this

so (x-k) is a factor so k^3 + 3pk + q = 0

and (x-k) is a factor of d/dx (x^3+3px + q) or 3x^2 + 3p or x^2+p =0

so k^2 + p = 0 ... 1 or p = -k^2

sum of zeros zero so 3rd factor = -2k

product = -2k^3 = q ...2

from 1 p = - k^2 or p^3 = -k^6 or 4p = -4k^6
and q^2 = 4k^6by adding we get 4p^3+q^2 = 0

2010/040) Let f(x) be a polynomial of degree>2

if f(x) is divided by (x-1) and(x-2)...?
it leaves a remainder 0,2 respectively.what is the remainder when f(x) is divided by (x-1)(x-2).

if f(x) is divided by (x-1)(x-2) remainder shall be a polynomial of degree 1

f(x) = p(x)(x-1)(x-2) + a x +b

f(1) = a + b = 0 so a = - b is the remainder when divided by (x-1)

f(2) = 2a + b = 2 so -2b +b = 2 or b - 2 and a = 2

so remainder = 2x - 2

2010/039) A problem in fibonacci sequence

show that
Σ (n = 1 to ∞) F(n)/x^n = x/(x^2-x-1) where F(n) is n th Fibonacci number

Where "x" is any integer >= 2

proof:
this is based on Geometric series and not on relationship in Fibonacci sequence which can be found at http://en.wikipedia.org/wiki/Fibonacci_number#Power_series

We know

F(n) = (p^n – (1-p)^n)/ sqrt(5) where p is phi (value = (1+ sqrt(5))/2)

So sqrt(5) F(n)/x^n = (p/x)^n – (1-p)/x)^n

So Σ (n = 1 to ∞) sqrt(5) F(n)/x^n = Σ (n = 1 to ∞)
(p/x)^n – Σ (n = 1 to ∞) (1-p)/x)^n

As x is positive
The sum converges when (p/x) < 1 and (1-p)/x < 1

Or x > p that is x > the golden ratio and need not be integer.

The sum = 1/(1-(p/x)) – 1/(1-(1-p)/x)
= x/(x-p) – x/(x-1+p)
= x((1/(x-p) – 1/(x-1+p))
= x(x-1+p) –(x-p))/)(x^2- x –(p(p-1))

As p(p-1) = 1 this is x^2-x-1
we get
So Σ (n = 1 to ∞) sqrt(5) F(n)/x^n = x(2p-1)/ (x^2-x-1)

As p = (1+ sqrt(5))/2) so (2p-1) = sqrt(5)

So Σ (n = 1 to ∞) sqrt(5) F(n)/x^n = x sqrt(5)/ (x^2-x-1)

Or Σ (n = 1 to ∞) F(n)/x^n = x/ (x^2-x-1)
Proved

2010/038) For a natural number n, let An = n^2 + 20

If Dn denotes the greatest common divisor of An and An+1,?
then show that Dn divides 81.

proof:
GCD(An , An+1)as we need to get a constant
= GCD(n^2+20, (n+1)^2 + 20) (one of 1st or 2nd term need to be constant by proper transform)
= GCD(n^2+20, n^2+2n + 21)
= GCD(n^2+20, 2n+1) as GCD(a,b) = GCD(a,b-a)
[now we need to eliminate n^2 from 1st term so we double the 1st term as 2nd term is odd so GCD(a,b) = GCD(2a,b) when b is odd]
= GCD(2n^2+40, 2n+1)
= GCD(2n^2+40-n(2n+1),2n+21) as GCD(a,b) = GCD(a-kb,b) for any k
= GCD(40-n, 2n+1)
= GCD(80-2n, 2n+1) [ same argument above as 2n+1 is odd double the 1st term
= GCD(81, 2n+1) as GCD(a,b) = GCD(a+b.b)

now 81 is constant and GCD must devide 81

2010/037) Sum of six consecutive whole squares can never be a whole square

let the 1st number be n^2

sum of 6 consecutive squares is

n^2+(n+1)^2+(n+2)^2+(n+3)^2+(n+4)^2 + (n+5)^2

= 6n^2+ 30n + 55
= 6n(n+5) + 55

1st part 6n(n+5)is divsible by 4 as either n or n+ 5 is even

so 6n(n+5) + 55 mod 4 = 55 mod 4 = 3

it cannot be a perfect square as perfect square mod 4 = 0 or 1

2010/036) if n is an integer give an example of n consecutive composite integers

in the 1st attempt the solution appears to be

(n+1)! + 2 to (n+1)!+n+1 the kth term is divisible k+1

but it can be bettered as

LCM(2,3,... n+1) + 2 to
LCM(2,3,... n+1) + n+ 1

and LCM(2,3,... n+1) is the lower of the two

It can still be bettered to the product of all primes (<= n+1)say P(n) + 2 to that + n+ 1

as the number from 2 to n+1 if a prime the product has that prime as a factor

and if not a prime then P(n) and and k have a common factor so sum is not a prime

(one of my friends informed me)

2010/035) Show that

27 x (23^n) + 17 x (10^(2n)) is divisible by 11 for all positive integers n

proof:
23 = 1 mod 11
so 23^n = 1 mod 11

10^2 = 100 = 1 mod 11

so 10^2n =1 mod 11

so 27 x (23^n) + 17 x (10^(2n)) = 27 * 1 + 17 * 1 = 44 = 0 mod 11

so divisible by 11

2010/034) Factorise x^4+ x^2+1

x^3 and x are missing so we can try to do it by difference of squares adding and subtracting x^2 that is middle term

= (x^4+2x^2+1)- x^2
= (x^2+1)^2 - x^2
= (x^2+x+1)(x^2-x+1)

2010/033) Let S be the sum of all the real coefficients of the expansion of (1 + ix)^2009. what is log_2(S)?

Let S be the sum of all the real coefficients of the expansion of (1 + ix)^2009. what is log_2(S)?

the sum of all the real coefficients of the expansion of (1 + ix)^2009 is real part of (1+i)^2009 by putting x = 1

now (1+i) = sqrt(2) cis(pi/4) where cis(x) = cos x + i sin x

so (1+i)^2009 = sqrt(2)^2009(cis (2009pi/4))

so real part = sqrt(2)^2009 cos(2009pi/4) = sqrt(2)^2009 * cos(pi/4) = sqrt(2)^2008 = 2^1004

hence log_2(s)= 1004

2010/032) If p and q are distinct primes, show that p^q + q^p ≅ (p + q) mod pq

If p and q are distinct primes, show that p^q + q^p ≅ (p + q) mod pq.
as

q is prime

so p^(q-1) mod q = 1 (as per fermats little theorem)


so p^(q-1) = mq + 1

multiply by p on both sides p^q = mpq + p = p mod pq

similarly as q is prime q^p = p mod pq

adding we get (p^q + q^p) mod pq = (p+q)

proved

2010/031) 5-digit number whose half is a perfect cube and one-third is a perfect square?

5-digit number whose half is a perfect cube and one-third is a perfect square?

solution
we have
2n^3 = 3p^2
so n has to be multiple of 3 say 3x and p multiple of 2 say 2a
2(3x)^3 = 3(2a)^2
or 54x^3 = 12a^2
or 9x^3 = 2a^2
so a has to be divisible by 3 (say a = 3y) (sqrt of 9) and x by 2 say x = 2b
so n= 6y and p = 6b

and 9(2y)^3 = 2(3b)^2

or 4y^3 = b^2

so b has to be divisible by 2 say b = 2c that is p = 12c

then 4y^3 = 4c^2 or y^3 = c^2 so y has to be a prfect square and c a perfect cube that it

y= m^2 and c= m^3

so n = 6m^2 and p = 12m^3

now it shall not be difficult to find 2n^3 or 2(6m^2)^3 which is a 5 digit number

by checkin we find that it is m = 2 so n = 24 and ans = 2 * 24^3 or 27648

2010/030) factor a^2(b+c) + b^2(a+c) + c^2(a+b) + 2abc

a^2(b+c) + b^2(a+c) + c^2(a+b) + 2abc
= a^2(b+c) + (b^2a + b^2 c + ac^2 + bc^2 + 2abc)
= a^2(b+c) + a(b^2+c^2 + 2ab) + (b^2 c + bc^2)
= a^2(b+c) + a(b+c)^2 + bc(b+c)
= (b+c)(a^2+ ab + ac + bc)
= (b+c)(a+b)(a+c)

2010/029) Find the maximum value of, (√p + √q + √r ). (plz see details. . . )?

Let α, β, γ > 0 and (α + β + γ) = π/2. If p = (tan α tan β) + 5, q = (tan β tan γ ) + 5
and r = (tan γ tan α) + 5 ,
find the maximum value of (√p + √q + √r ).

tan γ = tan [ (π/2) - (α+ß) ] = cot(α+ß) = 1 / tan(α+ß)

tan γ = (1 - tan α tan ß) / ( tan α + tan ß )

tan γ tan α + tan ß tan γ = 1 - tan α tan ß

(r-5) + (q-5) = 1 - (p-5)

p + q + r = 16 ................. (1)

Let p = a^, q= b^2 and r = c^2

We are given p+q+r = 16 => a^2+b^2+c^2 = 16

We need to maximize a + b + c

We know
(a+b+c)^2 = a^2+b^2 +c^2 + 2ab + 2bc + 2ca
(a-b)^2 = a^2 + b^2 – 2ab
(b-c)^2 = b^2 + c^2 – 2bc
(c-a)^2 = c^2 + a^2 – 2ac

Adding we get (a+b+c)^2 +(a-b)^2 + (b-c)^2 + (c-a)^2 = 2 (a^2+b^2 + c^2) = 32
Clearly a+b+c is maximum when a=b=c because as (a-b)^2 + (b-c)^2 + (c-a)^2 can not be lower than zero and is zero when a= b= c

So 2 (3a^2) = 32 or a^2 = p = 16/3

so maximum value of (√p + √q + √r ) = 4 √3

2010/028) If x + y + z = xyz

Prove that x(1-y^2)(1-z^2) + y(1-z^2)(1-x^2) + z(1-x^2)(1-y^2) = 4xyz?

proof

If x + y + z = xyz,

putting x = tan A, y = tan B, z = tan C,

tan A + tan B + tan C = tan A tan B tan C, i.e., S₁ = S₃

∴ tan ( A+B+C ) = ( S₁ - S₃ ) / ( 1 - S₂ ) = 0

∴ A+B+C = nπ for integral n.

2A + 2B +2C = 2nπ

∴ tan ( 2A+ 2B+ 2C ) = tan 2nπ = 0

∴ tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

∴ 2x / (1-x²) + 2y / (1-y²) + 2z / (1-z²) = [ 2x / (1-x²) ]•[ 2y / (1-y²) ]•[ 2z / (1-z²) ]

∴ x(1-y²)(1-z²) + y(1-z²)(1-x²) + z(1-x²)(1-y²) = 4xyz ............

Hence proved

2010/027) What is the square root of -i

THis can be done both with and with out De Moivre's formula as mentioned in

http://en.wikipedia.org/wiki/De_Moivre%2…

1st with it

-i = 0 + (-1) i = r cos t + r sin t

r^2 = 1

cos t = 0 and sin t = - 1 so t = 3pi/2

so -i = e^(3pi/2) i

so sqrt(i) = e^(3pi/4) i or e^5pi/4 taking (3pi/2 + 2npi/2) n = 0 and 1 n =2 3 gives 1st value

= cos 3pi/4 + i sin 3 pi/4 or cos 5pi/4 + i sin 5 pi/4

= - 1/sqrt(2) + 1/sqrt(2) i or 1/sqrt(2) - 1/sqrt(2) i

in case you are not familiar with De Moivre's formula then say

sqrt(-i) = (a+ib)

sqare both sides

-i = a ^2-b^2 + 2abi

so a^2-b^2 -= 0
and 2ab = -1
a^2-b^2 = 0 => a = +/-b so ab = -1/2 and we get a = 1/sqrt(2) b= - 1/sqrt(2)

or a = -1/sqrt(2) b= + 1/sqrt(2)

giving same results

2010/026) Strong induction problem

Suppose x is a real number, x does NOT equal 0, and (x + 1/x) is an integer

Prove for all n ≥ 1, x^n + 1/(x^n) is an integer.
x+ 1/x is in integer = n (given)

so (x+ 1/x)^2 = n^2

x^2 + 1/x^2 = n^2-2 is an integer
so true for 1 => true for 2

let it be true for upto k
now x^k+ 1/x^k

(x^k+ 1/x^k)(x+ 1/x)

= x^(k+1) + 1/x^(k-1) + x^(k-1) + 1/x^(k+1)

so x^(k+1) + 1/x^(k+1) = (x^k+1/x^k)(x+1/x) - (x^(k-1) + 1/x^(k-1))

if it is true for all n upto k so RHS is integer then LHS is integer so for k+ 1 so the induction step is proved

hence proved

2010/025) For which positive integer values of n occur EXACTLY two 0's at the end of 1^n + 2^n + 3^n + 4^n (base 10)

As it can have maximum 2 zeroes
Let us first find the period at which 1^n+2^n+3^n+ 4^n mod 100 repeats.

The period has to be < 100 as there are maximum number if 100 remainders



1^n mod 100 =1

As 3 is coprime to 100

3^40 = 1 because 40 is for n = 100 as per http://en.wikipedia.org/wiki/Euler%27s_t…

Because 2 and 4 are factors of 100 we need to find period n for which 2^n repeats to base 25. The for 25 is 20

That is 2^20 mod 25 = 1 and 4^20 mod 25 = 1

3 has a period factor of 40( 40 is maximum) and 2 and 4 have a period factor of 20 that is after that \it repeats

So 1^n,2^n, 3^n, 4^n has a period of 40 that is it repeats every 40 times for mod 100

For 1 to 40 we need to find for which the remainder 1^n+2^n+3^n+4^n is 0 mod 1

By a computer program it is found to be 3,5,7,11,15,19,23,25,27,31,35,39.

The values are 40k + 3,40k + 5, 40k + 7, 40k + 11, 40k + 15, 40k + 19, 40k + 23, 40k + 25, 40k + 27, 40k + 31, 40k + 35, 40k + 39 where k is an integer >= 0

2010/024) If the roots of [b-c]x*x+[c-a]x+[a-b]=0are equal then prove that 2b=a+c

roots are equal so discriminant is zero

so (c-a)^2 - 4(b-c)(a-b) = 0

or c^2-2 ac + a^2 - 4(ab-ac- b^2 + bc) = 0

or a^2+4b^2+c^2+2ac - 4ab - 4bc = 0

or(a+c-2b)^2 = 0

or a+c-2b = 0 or a + c = 2b

proved

2010/023) prove that a^4+b^4+c^4> or =abc(a+b+c)

we know

(a^2-b^2) ^2 >= 0

or a^4+b^4 >= 2a^b^2

similarly

b^4 + c^4 >= 2 b^2 c^2

c^4+ a^4 >= 2 c^2 a^2
adding all 3 above and deviding by 2

a^4+b^4+c^4 >= (a^2b^2+b^2c^2 + c^2 a^2) ... 1


now a^2+c^2 >= 2ac

multiply by b^2 on both sides

b^2(a^2+c^2) >= 2b^2ac -- 2

similarly
a^2(b^2+c^2) >= 2a^2bc ... 3

and c^2(a^2+b^2) >= 2c^2ab .. 4

adding (2) (3) and (4) we get

2(a^2b^2+b^2c^2+c^2a^2) >= 2abc(b+a+c)

or (a^2b^2+b^2c^2+c^2a^2) >= abc(a+b+c) ...5

from 1 and 5 we get a^4+b^4+c^4>=abc(a+b+c)

proved

2010/022) What is the largest number of 0's that can occur at the end of 1^n+2^n +3^+4^n

for every positive integer n

1^n + 2^n + 3^n + 4^n

n = 1 gives 10 and hence one zero
n =2 gives 41 so zero
n =3 gives 100 so 2 zeros

and it cannot end with > 2 zeros as it can be proved in steps as below

for it to end with k zeroes it need to be divisible by 10^k that is 2^k and 5^k. If we can show that is it not divisible by 8 then we are through.

for n > = 3, 2^n and 4^n are divisible by 8

now 1^n =1 for all n
3^n = 1 mod 8 for even n and 3 mod 8 for odd n

So 1^n + 3^n = 2 mod 8 for n even and 4 mod 8 odd n

So not divisible by 8 for any n and hence it cannot have >2 zeros

so ans is 2(for n = 3)

2010/021) Prove that there is no a for which a^2-3a -19 is divisible by 289

Proof:
As a first step as we see that 289 = 17^2.

Now a^2-3a-19 = (a-10)(a+7) + 51

The 2nd term that is 51 is divisible by 17 and for the 1st term that is product to be divisible by 17 either (a-10) or (a+7) is divisible by 17. but if one of them is divisible by 17 then the 2nd one is divisible by 17.

So 1st term is divisible by 289 and 2nd one is not divisible by 289 so sum is not divisible by 289. Or the 2nd term is divisible by 17 and 1st term is not divisible by 17 so sum is not divisible by 17.

So the expression is not divisible by 289.

2010/020) Form a cubic equation with odd coefficients and rational root that fits x^3 + ax + b

because coefficient of x^3 is 1 so rational roots are integers say m, n , p

product is b odd so all roots (b= -mnp) are odd

coefficient of x^2 = -(m+n+p) = 0 which is not possible as m n and p are odd

so no solution

2010/019) Suppose two complex numbers z = a + ib and w = c + id

Suppose two complex numbers z = a + ib and w = c + id satisfy the equation ...?
(z + w)/z = w/(z + w).

Then,
a) both a and c are zero;
b) both b and d are zero;
c) both b and d must be non zero;
d) at least one of b and d is non-zero;
Kindly explain...

(z + w)/z = w/(z + w)
=> (z+w)^2 = wz
=> z^2+wz+w^2 = 0

let z/w = t so we get

t^2+t+1 = 0 and hence t is cube root of 1 that is cis 120 or cis 240

so z and w in complex plane and at angle 120

so at least one of b and d must be non zero because in case b and d both are zero then z/w is real

hence ans is d

2010/018) The set of all integers n for which sqrt(n^2 + n) is an integer is...?

The set of all integers n for which sqrt(n^2 + n) is an integer is...?
a) the set {0,-1};
b) a finite set with at least three elements;
c) an infinite set;
d) none of these sets;
Kindly explain...

Ans:
sqrt(n^2 + n) = sqrt(n(n+1)) is inetger

n and n+1 are coprimes so either n= 0 or n = - 1 or n and n+1 both squares

n = x^2 and n+1 = y^2 => 1 = (x+y)(y-x) => x+y = 1 and y-x = 1 => x = 0 y =1 => n = 0

or x+y = -1 and y-x = -1 => x = 0 y =-1 => n = 0

so only solution 0 and -1 so ans is a)

2010/017) Lim (n -> infinity) {(1 + 1/n)^n - (1 + 1/n)}^-n is

(a) 1; (b) 1/(e - 1); (c) 1 - e^-1; (d) 0;?

Ans:
take (1+1/n) out to get

(1+1/n)^-n((1+1/n)^(n-1)-1)^-n

as n->infinite (1+1/n)^-n = 1/e

((1+1/n)^(n-1)-1)^(-n) = (e-1)^(-n) = 0

so product = 0

hence d

2010/016) Let 1, a1, ... ,a6 denote the distinct roots of x^7 - 1.

Let 1, a1, ... ,a6 denote the distinct roots of x^7 - 1. Then the product (1 - a1)(1 - a2)(1 - a3) ...(1 - a6) is

is
a) 0;
b) 1;
c) 6;
d) 7;
Kindly explain...
x^7- 1 = (x-1)(x-a1)(x-a2)(x-a3)(x-a4)(x-a5)(x-a6… by definition of roots

dividing by (x-1) on both sides

x^6+x^5+x^4+x^3+x^2+x+1 = (x-a1)(x-a2)(x-a3)(x-a4)(x-a5)(x-a6)

putting x= 1 on both sides we get 7 = (1-a1)(1-a2)(1-a3)(1-a4)(1-a5)(1-a6)

hence d

2010/015) Suppose that one moves along the points (m,n) in the plane where m and n are integers in ...?

such a way that each move is a diagonal step, that is, consists of one unit to the right or left followed by one unit either up or down.
a) Which points (p,q) can be reached from the origin?
b) What is the minimum numbers of moves needed to reach such a point (p,q) ?

Ans:
) In each move x cordinate goes up/down by 1 and y cordinate goes up or down by 1 so x+y sum changes by an even number.

the starting point is (0,0) so sum of x and y is even so p+q should be even.

b) if p + q is odd then we cannot go

so let p+ q be even.

then number of steps required is maximum(|p|,|q|)

to explain say q and p positive and q> p

so we require q steps to wards y side up and out of it p + (q-p)/2 steps to the right and (q-p)/2 steps to left

note: if some term above is -ve then move that many positive steps in opposite direction

2010/014) Let a,b and c be the sides of a right angled triangle. Let theta be the smallest angle of this triangle.?

If 1/a, 1/b, 1/c are also the sides of a right angled triangle then show that Sin(theta) = (sqrt(5) - 1)/2;

proof
Let a < b < c


And theta (say t opposite to smaller side)

Sin t = a/c and cos t = b/c

Now 1/a > 1/b > 1/c and as it is right angled triangle we have

(1/a)^2 = (1/b)^2 + (1/c)^2

Or

(c/a)^2 = (c/b)^2 +1

(1/sin ^2 t) = (1/ cos^2 t ) +1

Or cos^2 t = sin ^2 t + sin ^2 t cos^2 t

Or 1- sin ^2t = sin ^2 t + sin ^2 t (1- sin ^2 t)
= 2 sin ^2 t – sin ^4 t
Or sin ^4 t – 3 sin ^2 t + 1 = 0

Sin^2 t = (3 +/- sqrt(9-4))/2 = (3 – sqrt(5))/2 , + cannot be taken as it shall be >1 not possible


So sin t = sqrt((3 – sqrt(5))/2) say sqrt(x) – sqrt(y) as there is sqrt(5) in square

So squaring we get

x+y – 2sqrt(ab) = (3 –sqrt(5))/2

so x+y = 3/2 and sqrt(xy) = sqrt(5)/4) or xy = 5/16

we can solve it as x = 5/4 and y= 1/4 or y = 1/4 and x = 5/4

so sin t = +/-(sqrt(5/4) - sqrt(1/4))

positive value to be taken as sin t > 0 so sin t = sqrt(5/4) – sqrt(1/4) = (sqrt(5)-1)/2

2010/013) Form a quadratic equation whose coefficients are integers and in AP and it has got rational roots

Let the quadratic equation be

ax^2+bx+c = 0


as the coefficients are in ap so let common difference be y so b-a = y c-b = y

or a = b-y and c = b+ y

for the equation to have rational roots we have

discriminant is a perfect square that is say n^2


b^2 – 4 ac = n^2

or b^2-4(b-y)(b+y) = n^2

or b^2-4(b^2 – y^2) = n^2

or 4y^2-3b^2= n^2

dividing by n^2 and letting b/n = t and y/n = s we ge

4s^2-3 t^2= 1 ..1

s and t are real numbers

And from inspection we have a root s = 1/2, t =0 is a rational root and to find another rational root we draw a straight line from (0,1/2)

We get s = mt+ 1/2 is a general root …(2)

So we put in 1 to get the intersection of the curve given in 1 with this straight line to get

(2mt+1)^2 - 3t^2 = 1

Or 4m^2t^2 +4mt+1 – 3t^2 = 1

Or (4m^2-3)t^2 + 4mt = 0

ignoring the starting solution t = 0 we get

t= - 4m/(4m^2-3)

so s= mt+1/2 we get s = -(4m^2-3)/(8m^2-6)

now putting n = 8m^2-6 to get rid of denominator we get

b = -8m and y = -4m^2+3

so we get

a = 4m^2 – 8m + 3
b = - 8m
c = -4m^2 -8m -3 as roots

and taking m as different integers we get different solutions in integers

for example
m = 1 gives (-1.-8, -15) multiply by one we get x^2+ 8x + 15 = 0 giving factor (x+3)(x+5) rational
m = 2 gives (3.-16, -35) we get 3x^2 -16x -35= 0 giving (3x+5)(x-7)
m =3 gives (15,-24,-63) divide by 3 to get 5x^2-8x-21 = (5x+7)x-3)

so on

2010/012) If 0 < x < 1, then the sum of the infinite series (1/2)X^2 + (2/3)X^3 + (3/4)X^4 + ... is ...

A) log[(1+x)/(1-x)];
B) x/(1-x) + log(1+x);
C) 1/(1-x) + log(1-x);
D) x/(1-x) + log(1-x);
Explain your answer...

et f(x) = (1/2)x^2 + (2/3)x^3 + (3/4)x^4 +

df/dx = x + 2 x^2 + 3x^3 + ... ....
= (x+x^2 + x^3...) + (x^2 + x^3 ...) + (x^3+x^4 + ....)
= x/(1-x) + x^2/(1-x) +
= x/(1-x)^2
= (1-x-1)/(1-x)^2
= 1/(1-x) - 1/(1-x)^2
integrating you get
f = log(1-x) + 1/(1-x) + constant of integration log(1-x) as 1- x > 0
this can be checked to be zero so ans is C

2010/011) If cos(x)/cos(y) = a/b, then a tan(x) + b tan(y) equals

A) (a+b)cot((x+y)/2); B) (a+b)tan((x+y)/2); ...?
C) (a+b)(tan(x/2) + tan(y/2));
D) (a+b)(cot(x/2) + cot(y/2));
Kindly explain your answer...

we have (a+b) on RHS of (a)(b)(c) and (d) so let us eliminate a and b from numerator of
( a tan(x) + b tan(y))/(a+b)

b cos x = a cos y
b / a = cos y/ cos x
(b+a)/a = (cos y + cos x)/cos x

a/(a+b) = cos x/(cos y+ cos x)
and b/(a+b) = cos y /(cos y + cos x)

(a tan x + b tan y)/(a+b)
= (a/(a+b)) tan x + (b/(a+b) tan y
= (cos x tan x + cos y tan y)/(cos y + cos x)
= (sin x + sin y)/(cos x + cos y)
= (2 sin (x+y)/2 cos (x-y)/2 )/(2 cos (x+y)/2 cos (x-y)/2 )
= tan (x+y)/2

hence ans is B

2010/010) Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number. Show that n? is a prime.

1 + x + x^2 + x^3 + ... + x^n-1 = (x^n-1)/(x-1)

now in case n is not a prime then say n= pq

now x^n-1 = (x^(pq)-1) = (y^q-1)/(x-1) where y = x^p
= (y-1)/(x-1)(y^(q-1) + y^(q-2) + ...1)
= (x^p-1)/(x-1) (y^(q-1) + y^(q-2) + ...1)
= (x^p-1 + +1)((y^(q-1) + y^(q-2) + ...1)
has got 2 factors and hence not a prime
so n must be prime

2010/009 Evaluate : Lim(n-> infinity) {(1 + 1/2n)(1 + 3/2n)(1 + 5/2n)...(1+(2n - 1)/2n)}^(1/2n)?

this can put as sum of

((1+1/2n)(1+2/2n) .....* * (1+(2n)/2n))^(1/2n)/
(1+2/2n)(1+4/2n) ........(1+ 2n/2n)^(1/2n)

basically multiply it by the missing terms and then devide back

now numerator say
N = ((1+1/2n)(1+2/2n) .....* * (1+(2n)/2n))^(1/2n)
take log to base e

log N = 1/2n (log (1+2/n) + log (1+ 2/2n) + .... + 2)

as n-> infinite this tehds to

1/2 int (log x) for x from 1 to 2

integral of log x = x log x - x
definite integral = (2 ln 2 -1) - (-1) =2 ln 2
so N = 1/2e^(2 ln 2) = e^2

now you can evaluate the denominator as
ln d = 1/2n (ln (1+2/2n ) + ln (1+4/2n ) + ...+ ln (2))
as n->infinite this is int log x

integral of log x = x log x - x
definite integral = (2 ln 2 -1) - (-1) =2 ln 2
so D = e^(2 ln 2) = 2e^2
so N/D = value = 1/2

2010/008) Show that there is exactly one value of x which satisfies the equation,?

2 cos^2 (x^3 + x) = 2^x + 2^(-x).

proof:
the maximum value of LHS = 2

as cos^2(x^3+x) <= 1

the minimum value of RHS

let 2^x = y

so y + 1/y = (sqrt(y) - 1/(sqrt(y))^2 + 2

so minum value of RHS = 2


so both sides are same when both are 2

RHS = 2 when y = 1 or x= 0

LHS = 2 when 2 cos^2(x^3+x) = 2

or cos^2(x^3+x) = 1

x = 0 satisies LHS

so x = 0 is the only value that satisfied equality

2010/007) prove 64 {Cos^8(x) + Sin^8(x)} = cos8 x + 28cos 4x + 35

We know

(a+b)^8 = a^8 + 8 a^7b + 28 a^6b^2 + 56 a^5b^3 + 70a^4b^4 + 56 a^3b^5 + 28 a^2b^6+8ab^7+b^8

And (a-b)^8 = = a^8 - 8 a^7b + 28 a^6b^2 - 56 a^5b^3 + 70a^4b^4 - 56 a^3b^5 + 28 a^2b^6 -8ab^7+b^8

So (a+b)^8 + (a-b)^8 = 2(a^8 + b^8) + 56 (a^6b^2+a^2b^6) + 140(a^4b^4)

Putting a = e^ix and b = e^-ix

LHS = (e^ix+e^-ix)^8 + (e^ix-e^-ix)^8 = 2(e^i8x + e^-i8x)+ 56(e^4ix + e^-4ix) + 140

Or (2 cos x)^8 + (2i sin x)^8 = 2 ( 2 * cos 8x) + 56 * 2 * cos 4x + 140

Or 2^8(cos ^8x + sin ^8x) = 4 * ( cos 8x + 28 cos 4x + 35)

Or 2^6(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)

Or 64(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)

2010/006) equation sin(w) = (a^2 + b^2 + c^2)/(ab+bc+ca), where a,b,c are fixed nozero real numbers

has a solution for w
A) Whatever be a,b,c;
B) iff a^2 + b^2 + c^2 < 1;
C) iff a,b and c all lie in the interval (-1,1);
D) iff a = b = c;
Explain your answer.

ans

we know by GM AM enaquality a^2+b^2 >= 2ab
a^2+c^2 >= 2ac
and b^2+c^2 >= 2bc

adding all 3 we get 2(a^2+b^2+c^2) >= 2(ab+bc+ca)

or (a^2+b^2+c^2)/(ab+bc+ca)>= 1 and equal only when a= b= c

sin (w) cannot be > 1 so ans is when (a^2+b^2+c^2)/(ab+bc+ca) = 1 or a= b =c

that is d

2010/005) If 0 < x < 1, then the sum of the infinite series (1/2)X^2 + (2/3)X^3 + (3/4)X^4 + ... is ...

let f(x) = (1/2)x^2 + (2/3)x^3 + (3/4)x^4 + ...

df/dx = x + 2 x^2 + 3x^3 + ... ....
= (x+x^2 + x^3...) + (x^2 + x^3 ...) + (x^3+x^4 + ....)
= x/(1-x) + x^2/(1-x) + ...
= x/(1-x)^2
= (1-x-1)/(1-x)^2
= 1/(1-x) - 1/(1-x)^2
integrating you get
f = log(1-x) + 1/(1-x) + constant of integration log(1-x) as 1- x > 0
this can be checked to be zero
so f(x) = 1/(1-x) + log(1-x)

2010/004) Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number.

Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number. Show that n?
is a prime.

Proof:
1 + x + x^2 + x^3 + ... + x^n-1 = (x^n-1)/(x-1)

now in case n is not a prime then say n= pq

now x^n-1 = (x^(pq)-1) = (y^q-1)/(x-1) where y = x^p
= (y-1)/(x-1)(y^(q-1) + y^(q-2) + ...1)
= (x^p-1)/(x-1) (y^(q-1) + y^(q-2) + ...1)
= (x^p-1 + +1)((y^(q-1) + y^(q-2) + ...1)
has got 2 factors and hence not a prime
so n must be prime

2010/003) The equation x^3 + 2x^ + 2x + 1 = 0 and x^200 + x^130 + 1 = 0 have ...?

exactly one common root;
b) no common root;
c) exactly three common roots;
d) exactly two common roots;
Kindly explain...

ans:
x^3 + 2x^2 + 2x + 1 = x(x+1)^2 + (x+1) = (x+1)(x^2 + x + 1) has zeros -1, w and w^2 where w is cube root of 1

-1 is not a zero of x^200 + x^130 + 1 since the expression equals 3 when x = -1.

So we are down to 2 possible roots in common.

let f(x) = x^200 + x^130 + 1

f(w) = w^200 + w^130 + 1= w^2+ w + 1 = 0 so w is a root
f(w^2) = w^400 + w^260 + 1 = w+w^2+1 = 0 so w^2 is a root

so there are exactly 2 common roots and hence d)

2010/002) The numbers 12n + 1 and 30n + 2 are relatively prime for ...?

any positive integer n;
b) infinitely many, but not all, integers n;
c) for infinitely many integers n;
d) none of the above;
Kindly explain...

Ans:
GCD(12n + 1 ,30n + 2)
= GCD(12n + 1 ,2(15n+1))
= GCD(12n + 1 ,15n+1) as 12n+1 is odd
= GCD(12n+1,(15n+1-(12n+1)) as GCD(a,b) = GCD(a,b-a)
= GCD(12n+1,3n)
= GCD(3n, 12n+1- 4*3n) as GCD(a,b) = GCD(b,a- n * b))
= GCD(3n,1)
= 1 for all n
hence ans is a

2010/001) Prove that (2222^5555)+(5555^2222) is divisible by 7

Prove that (2222^5555)+(5555^2222) is divisible by 7

Proof:
we know

2222 mod 7 = 3

and 5555 mod 7 = 4 or -3

so (2222^5555)+(5555^2222) mod 7

= 3^5555 + (-3)^ 2222 mod 7
= 3^5555+ 3^2222 mod 7
= 3^2222(3^3333 + 1) mod 7


as 3^2222 is not divisible by 7 so

we need to show that

3^3333 + 1 mod 7 = 0

now as 7 is prime so as per format's little theorem

3^6 mod 7 = 1

3^3333 mod 7 = 3^(3333 mod 6) mod 7
= 3^ 3 mod 7 = 27 mod 7

so 3^3333 + 1 mod 7 = 28 mod 7 = 0

hence proved