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

Q13/020) Prove that all integer positive powers of 3 have an even tens digit

proof:
we need to show that

3^ n mod 20 is one digit number

n = 1 => 3
n = 2 => 9
n= 3 => 27 = 7 mod 20
n =4 => 21 = 1 mod 20
after n= 4 it repeats and hence 10's digit is even

Q3/019) Find f(x) given that f(x+2)=x^2+7x+14

f(x+2)=x^2+7x+14

put x+ 2 = y => x = y- 2 to get
f(y) = (y-2)^2 + 7(y-2) + 14 = y^2 + 3y + 4 on expansion

put y =x to get

f(x) = x^2 + 3x + 4

Q3/018) If the hands of a clock are at 12 noon? how long is it before the hands cross again

let it cross after x minutes

in 60 minutes minute hand has covered 360 degree and so in x minutes

6x degree

in 60 minutes hr hand has covered 30 degrees so x/2 degrees in x minutes

now minute hand travels faster and 2 shall meet when difference is 360 ( as for one hr it has advanced 360)

or 6x - x/2 = 360 or 11/2x = 360 or x = 720/11 minute or 1 hr and 60/11 minutes