Q13/031) solve this greatest integer equation? [x+4]=3[x]-8

[x+4] = 3 [x] - 8
because 4 is integer so [x+4] = [x] + 4
so
[x] + 4 = 3[x] - 8

or 2[x] = 12
or [x] = 6

or 6 <= x < 7 or in set notation [6, 7 }

Q13/030) For any integer n=>1, prove that 1/6{n(n+1)(2n+1)} is an integer.?

that is n(n+1)(2n+1) is divisible by 6

n(n+1)(2n+1) = n(n+1)(n-1 + n + 2)
= n(n+1)(n-1) + n(n+1)(n+2)

1st number product of 3 consecutive numbers is divisible by 6 and 2nd number for the same reason

hence {n(n+1)(2n+1)} is divisible by 6 so 1/6{n(n+1)(2n+1)} is an integer.

Q13/029) Prove that 21n^22+22n^26+34n^32≡0(mod 77)

we have 77 = 7 * 11
as 7 annd 11 are co-primes
so a mod 77= 0 if a mod 7 = 0 and a mod 11 = 0

21 mod 7 = 0 , 22 mod 7 = 1, 34 mod 7 = -1

so 21n^22+22n^26+34n^32 = n^ 26 - n^ 32 = n^26(1 - n^6)

if n = 0 mod 7 the n^26 = 0 mod 7 else n is co-prime to 7

so n^6 =1 mod 7 ( by Fremat's Little theorem)
so
21n^22+22n^26+34n^32≡0 mod 7 ..1

now for mod 11
21 = -1 mod 11 , 34 = 1 mod 11, and 22 = 0 mod 11

so
21n^22+22n^26+34n^32
= - n^22 + n^ 32 = n^22 (n^10-1)

if n = 0 mod 11 then n^22 = 0 mod 11

or n is corpime to 11 so n^ 10 -1 = 0 mod 11

so 21n^22+22n^26+34n^32 = 0 mod 11...2

so 21n^22+22n^26+34n^32≡0(mod 77)  (from 1 and 2)

Q13/028) Rationalise the denominator 1/(√2 +√3 +√5)?

This type is done one at a time
(√2 +√3 +√5) = (√2 +√3) +√5)

so multiply both numerator and denominator by (√2 +√3) -√5)

to get ((√2 +√3) -√5))/(( (√2 +√3)^2 -√5) ((√2 +√3) +√5))
= ((√2 +√3) -√5))/((√2 +√3)^2 -5)
= ((√2 +√3) -√5))/((2 +3+ 2√6 -5)
= ((√2 +√3) -√5))/(2√6)

now multiply both numerator and denominator by √6

to get √6((√2 +√3) -√5))/12

Q13/027) solve for ø sin^2ø+cos^2ø+tan^2ø+sec^2ø+ csc^2ø+cot^ø=7

sin^2ø+cos^2ø+tan^2ø+sec^2ø+ csc^2ø+cot^ø
= 1 + tan^2ø + 1 + tan^2ø + 1 + cot^ø + cot^ø = 7 ( as sin^2ø+cos^2ø = 1, sec^2ø = tan ^2ø + 1 , csc^2ø = cot ^2ø + 1)
= 3 + 2 ( tan ^2 ø + cot ^2ø) = 7
or ( tan ^2 ø + cot ^2ø) = 2
or ( tan ^2 ø + cot ^2ø) - 2 tan ø cot2ø = 0
or ( tan ø - cot ø)^ 2 = 0

or ( tan ø = cot ø) = 1 or -1 so ø = pi/4 is a solution other solutions are 3pi/4, 5pi/4, 7pi/4

Q13/026) Two roots of the polynomial x^3 + ax^2 + 15x -7 = 0 are equal and rational. Find "a"

if 2 roots are rational then 3rd must be rational.
possible roots are -7 , -1, 1, 7
now the double root shall be +1 or - 1 as 7 or -7 shall give product module 49 or more.

so 3rd root has to be 7 ( as product has to be +ve)

so f(x) = x^3 + ax^2 + 15x -7 = 0 = f(7)

or 343 + 49 a + 105 - 7 = 0

or a = 9

check:
x^3 - 9x^2 + 15x - 7 = (x-1)^2(x-7)

alternatively we have(x-1)^2(x-7) and multiplying we get  x^3 - 9x^2 + 15x - 7 or a = 9

Q13/025) Prove that sin(1degree) is not a rational number?

we have

cos 2 t = 1 - 2 sin ^2 t

so if sin 1 is rational cos 2 is rational

now cos 0 =1 is rational

cos (n-2 ) + cos (n+2) = 2 cos n cos 2

so cos (n+2) = - cos (n-2) + 2 cos n cos 2

if cos n and cos n- 2 are rational then by strong induction cos n+2 is rational

hence proceeding we get cos 30 = sqrt(3)/2 is rational which is contradiction

hence cos 2 and then sin 1 are not rational

Q3/024) If a=(4√6)/{(√2)+(√3)} then the value of {(a+2√2)/(a-2√2)}+{(a+2√3)/(a-2√3)… is

to keep in simple for without radicals

let x= 2√2 and y = 2√3

a=(4√6)/{(√2)+(√3)} = 2(xy)/(x + y)

so
a/x = 2y/(x+y)

using componedo dividendo (a+x)/(a-x) = ( 3y + x)/(y-x)

siminalrly (a+y)/(a-y) = ( 3x + y)/(x-y)

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

hence {(a+2√2)/(a-2√2)}+{(a+2√3)/(a-2√3)} = 2

Q13/023) Given the following 2^p=3^q=12^r,? Find r in terms of p and q

from above

2 = 12^(r/p)
3 = 12^(r/q)

now

12 = 2 ^2 * 3 = 12^(^2r/p) 12^(r/q)

or 1 = 2r/p + \r/q

or 1/r = (2/p+ 1/q) = (p+2q) / (pq) or r = pq/(p + 2q)

Q13/022) How many ordered pairs of integers (a,b) are there such that 1/a+1/b=1/200?

1/a+ 1/b= 1/200

or 200 b + 200 a = ab

or ab - 200 a - 200 b = 0
or (a-200) (b-200) = 40000 = 5^4 * 2 ^ 6
it has got (4+1) * (6+1) or 35 ordered solution for (a b) and 35 for (b a) so 70 solutions as 40000 can be factored in 35 ways( positive solutions)

Q13/021)What is the smallest integer k such that 1^2+2^2+3^2+…+k^2 is divisible by 100?

we have ^2+2^2+3^2+…+k^2 = k(k+1)(2k+1)/6 divisible by 100

so k(k+1)(2k+1) divisible by 600

as k(k+1)(2k+1) is always divisible by 3 so

k(k+1)(2k+1) divisible by 200 = 5^2 * 2^8

now 2k + 1 is odd so k or k+1 should be divisible by 8

that leads to 4 cases
 k  is divisible by 8 and k+1 or 2k+1 by 25
k+ 1 is divisible by 8 and k+1 or 2k+1 by 25

so we look for k mod 8 = 0, k+1 mod 25 = 0 => k = 24

k mod 8 = 0 2k + 1 mod 25 = 0 => k = 112

  k+1 mod 8 = 0 k mod 25 = 0 shall have a larger k = 175
k+1 mod 8 = 0 2k +1 mod 25 = 0 gives k = 87

clearly smallest k = 24