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}