Q2/131) Find 2^(1/2) in ratio of integer form(approximate taht is p/q where p and q are integers)

Solution      

We know 2^(1/2) is irrational and let us assume2^(1/2)  = x/y

So we get x^2 = 2y^2 or x^2-2y^2 = 0

This does not have integer solution.

But if we chose x^2-2y^2= 1 then it has  integer solution x= 3 , y=2 is one of them

for each of the equations we can generate solutions with higher x and y for solution of either one of the equation and as we get x and y higher we get x/y as close to 2^(1/2)

for that we need 2 solution ( or 1 solution can be used 2 times) ( assuming x/y and a/b)

x^2-2y^2= 1 ..1 => (x+ 2^(1/2)y)(x- 2^(1/2)y) = 1  …1

a^2-2b^2= 1 ..1 => (a+ 2^(1/2)y)(x- 2^(1/2)y) = 1  …2

multiply (1) by (2) to get (x+ 2^(1/2)y)( (a+ 2^(1/2)b) (x - 2^(1/2)y)(a -2^(1/2)b) = 1

or ( ax + 2by + 2^(1/2)(bx + ay))( ax + 2by- 2^(1/2)(bx + ay)) = 1 .. 3

so if (x,y) and (a,b) are solution then (ax+2by, bx+ay) is a solution

we can chose (x,y) and(a,b) as different of (x,y) = (a,b) also giving a solution

now taking (x,y) = (a,b) = (3,2) we get

(ax+2by, bx+ay) = (17,12) is one solution : note that 17^2 – 2 * 12^2 =1 confirms the same

Taking (x,y) = (3,2) and (a,b) = (17,12) we get

(ax+2by, bx+ay) = (99, 70) is next solution : note that 99^2 – 2* 70^2 =  1 confirms the same

We can proceed to get solutions as (3,2), (17,12), (99,70) and so on and 3/2(= 1.5 ) , 17/12 ( = 1.416 ) and 99/70 (=1.414285 = correct upto 4 decimal places) are approximately 2^(1/2) closer and closer . Few more iterations shall give the values much closer.

Q12/130) What is the sum of the squares of the roots of the equation x^2 - 7[x] + 5 = 0?

7[x]>  =  7x

hence any solution must satisfy the quadratic equation
x^2 - 7x + 5 <= 0
So x is between 0.8 and 6.2.(approximate)
It follows that the possible values of [x] are 0 to 6. So we need to check for of [x] from 0 to 6 and find x to be in proper range

There are altogether 4 distinct solutions.

They are sqrt(2), sqrt(23), sqrt(30), sqrt(37)

The sum of their squares is 2 + 23 + 30 + 37 = 32+60 = 92

Q2/129) For how many 2 digit numbers the sum of digits is greater than the product of digits?

Solution

(x+y) > xy
or  xy - x - y < 0
x(y-1) < y

  y = 0 for all x so all multiples of 10 is a set of solution 

x < y/(y-1)

or x < 1+ 1/(y-1)
so x has to be 1

so multiples of 10 or any one of digits should be 1

that 10, 20, 30, 40,50,60,70,.80,90, 11,12,13,14,15,16,17,18,19, 21,31,41,51,61,71,81,91

Q2/128) Solve in positive integers (1+1/x) (1+ 1/y))(1 +1/z) = 2

Without loss of generality we can choose x <=y <=z and answer  shall be a permutation of if

 

Now x < 4 as (5/4)^3 < 2 ( it is 125/64)

x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution

so x = 2 or 3

if x = 2 we get

3/2(y+1)(z+1) = 2yz

Or 3(y+1)(z+1) = 4 yz

Or yz – 3y – 3z = 3

(y-3)(z-3) = 12 by adding 9 on both sides

( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4

So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7) are the 3 sets of solutions

if x = 3 we get

4/3(y+1)(z+1) = 2yz

Or 2(y+1)(z+1) = 3 yz

Or yz – 2y – 2z = 2

(y-2)(z-2) = 6

( y-2)(z-2) = 1 * 6 or 2 * 3

So (x,y,z ) = (3,3,8) , ( 3,4,5)

Q2/127) If a, b, c are different and the equations ax+a^2y+(a^3+1)=0 bx+b^2y+(b^3+1)=0 cx+c^2y+(c^3+1)=0 are consistent, prove that abc+1=0...

Let us have a function f(t) = xt+yt^2+(t^3+1)

We have f(a) = 0 , f(b) = 0, f(c) = 0

And f(t) = t^3 + yt^2 + xt + 1 …1

So a,b, c are zeroes of f(t) which is a cubic polynomial

So f(t) = m(t-a)(t-b)(t-c) …2

Comparing coefficient of t^3 from (1) and (2) m = 1 

Now constant term = -mabc = 1 or –abc = 1 or abc + 1 = 0

Proved

Q2/126) When does a/b repeat when it terminates

Reduce a/b to the lowest form p/q such that p and q are co-primes.

Now q is of the form 2^m5^nc

We can convert  power of 2 and 5 to power of 10 by multiplying properly so the denominator can be converted to 10^r c

If c is 1  then it terminates.

If c is not 1 then c is coprime to 10 and the remainder and hence the quotient repeats. 

Q12/125) Assume x and y are integers, such that (x^2+1)=2y. Now prove that y is the sum of squares of two integers?

x has to be odd so let x= 2m + 1

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

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

proved

Q2/124) Show that in a sequences of 3 numbers one number is always divisible by 3

This can be proved by pigeon hole principle. As there are 3 numbers there are 3 remainders and the remainders can be 0 or 1 or 2 and no 2 remainder can be same if they where same then difference is divsible by 3 but difference cannot be be >2 so this is not possible.

So one of the remainders has to be zero.

Hence proved.

However this can be proved as below

Add 3 to all the terms starting from 1^st number and keep adding. Then we get all the terms to the right. Subtract 3 from 3^rd term and keep subtracting. We get all the numbers and none of them is divisible by 3. which is a contradiction. So one of the numbers has to be divisible by 3.

Q2/123) Find integer sided isosceles triangle whose area is integer

We have Pythagorean triplet that is integer length sides for a right angled triangle with hypotenuse c as

a= u^2-v^2
b= 2uv
and c = (u^2+v^2)

Now it may be noted that as b is even so ab/2 is integer.

Now if we double any of the base and hypotenuse as the two other sides we have area is integer

So the 3 sides are (u^2+v^2, u^2+v^2, 4uv) or (u^2+v^2, u^2+v^2, 2u^2-2v^2)

Q2/122) : find a , b, c,d such that 1/a = 1/b + 1/c + 1/d

We know that

1/n = 1/(n+1) + 1/(n(n+1))  .. (1)

Puting n = a we get

1/a = 1/(a+1) + 1/(a(a+1))

Now putting n = a(a+1)

We get

1/(a(a+1)) = 1/(a(a+1) + 1) + 1/((a(a+1))(a(a+1) + 1) =

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

Hence b= a+ 1

c= a^2 + a + 1 and d = abc is the solution

for example if a= 2 we have

1/2 = 1/3 + 1/6 ... using 1

1/6 = 1/7 + 1/42 using 1

hence 1/2 = 1/3 + 1/7+ 1/42 ( a= 2 , b= a+1 = 3, c= a^2 + a + 1 = 7 and d= abc = 42)

we can proceed indefinitely to any number of reciprocals. and for any starting reciprocal as well