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

Q3/017) If 30 and a are co primes then show that 60 divides a^2+ 59

30 and a are co primes so 2 , 3  5 none of these divide a

Now a^4-1 = (a+1)(a-1)(a^2 +1)

As a is odd so a+1 and a-1 are even and product is divisible by 4

As a is not divisible by 3 so (a+1) or (a-1) is divisible by 3

Further  a^4-1 mod 5 = (a+1)(a-1)(a^2 +1) mod 5

= (a+1)(a-1)(a^2-4) mod 5

= (a+1)(a-1)(a+2)(a-2) mod 5

Now (a-2), (a-1), a , (a+1), (a+2) being 5 consecutive numbers so one of them is divisible by 5 and a is not so one of the rest must be divisible

So a^4-1 is divisible by 5.

So   a^4-1 is divisible by 4 ,3 and 5 and hence product 60

So a^4-1 + 60 or a^4+ 59 is divisible by 60

Q3/016) Given that kx^3+2x^2+2x+3 and kx^2-2x+9 have a common factor, what are the?

GCD( kx^3+2x^2+2x+3,kx^2-2x+9) is not 1

GCD( kx^3+2x^2+2x+3,kx^2-2x+9)
= GCD(kx^2 - 2x + 9, kx^3+2x^2+2x+3 - x(kx^2-2x+9))
= GCD((kx^2 - 2x + 9, 4x^2 - 7x + 3)

say f(x) = kx^2 -2x + 9

4x^2 - 7x + 3 = (x-1)(4x-3)

x =1 => f(x) = k - 2 + 9 = 0 or k = -7

If (4x - 3) is a common factor, then f(3/4)= 0.

f(3/4) = 0
=>9k/16 + 15/2 = 0
=>k = -40/3

So  k = -40/3 or - 7

Q3/015) n^2+19n+130=f(n)

Find the sum of all the value of n for which f(n) is whole square.?

LHS = (n + 19/2)^2 - 361/4 + 130
= (n + 19/2)^2 + 159/4 = m^2

mulitply by 4 to get

(2n + 19)^2 + 159 = 4m^2 or 159 = (2m)^2 - (2n+19)^2 = (2m + 2n + 19)(2m-2n - 19)

now factors of 159 = 159 * 1, 53 * 3

taking 159 *1 we have 2m + 2n + 19 = 159 and 2m-2n - 19 = 1

or m+n = 70 and m-n = 10 => m = 40, n= 30

taking 53 * 3 we have 2m + 2n + 19 = 53 and 2m - 2n - 19 = 3
=> m+ n = 17 and m-n = 11 => m = 14, n= 3
  so n = 3 or 30

so sum= 33

 

Q3/014) The ratio of L.C.M & H.C.F of two numbers is 6:1 and the smallest number is 12, then find the larger number?

let LCM = a and HCF = b

for 2 numbers product of number = product of LCM and HCF

let larger number be l

12l = ab
and a = 6b or 6b^2 = 12l or b^2 = 2l

b is a factor of 12 but not 12 ( it is 1 or 2 or 3 or 4 or 6) and l >12 => b^2 > = 24 so b = 6 and l = 18

Q13/013) Alice and Bob were asked to solve a quadratic equation. While solving, Alice made a mistake in the constant term and got the roots as 8 and 2, while Bob made a mistake in coefficient of X and obtained the roots as -9 and -1.

Alice got 8 and 2 so equation

x^2-10x + 16 = 0 constant 16 is wrong

and bob got -9 and - 1 so

x^2 + 10 x + 9 = 0

as 10 is wrong (bob's mistake) and hence original equation

x^2-10x + 9 = 0 or (x-9)(x-1) = 0 so roots are 9 and 1

Q13/012) Find a four-digit natural number n, such that the last four digits of n2 are same as n.

This can be extended to

Find a k digit natural number n, such that the last k digits of n² are same as n.( k = 4)

Now n^2 –n = 0 mod 10^k

Or n(n-1)  = 0 mod 10^k

Now as n and n-1 consecutive so one of them is divisible by 2^k( any multiple)  and another by 5^k ( odd multiple)

For k = 1 we get n = 5 or 6

For k= 2 we have 5^2 = 25  and 24 is divisible by  2^2 so n = 25

                           Also 3 * 5^2 + 1 = 76 is divisible by 2^2 so n = 76

For k= 3 we have 5^3 = 125  and we need a multiple of 125 +/- 1 divisible by 8 that is (375+ 1) and (625-1) So Ans are 376 and 625

For k= 4 we have 5^4 = 625 so we need multiple of 625 +/- 1 divisible by 16 they are 624( 625-1) and 625 * 15 + 1 = 9376 but 625 is rejected as it is 3 digit number

So for 4 it is 9376

refer to

http://puzzle-a-day.blogspot.in/2011/07/four-digit-number.html

for another soultion

 

Q3/011) Rational or Irrational

Let a_n be defined as follows for all natural numbers n:
a_n = 0 if the number of divisors of n (including 1 and n) is odd
a_n = 1 otherwise.
Now consider the fraction 0.a₁a₂a₃....
Is this fraction rational or irrational? Explain.

Solution

The number of factors is odd for square number and it is even for non square numbers.
so a_n = 0 for no square number and = 1 for no square number
so in the decimal at each square place it is 1 and the gaps keeps on increasing .
so the digits do not recur and hence it is irrational