Q13/130) For real numbers a,b what is max(a,b)+min(a,b)

It is a + b

To prove there are 3 cases

1) a >b

In this case max(a,b) = a and min (a,b) = b and so sum = a +b

2) a= b

In this case max(a,b) = min (a,b) = a and so sum = 2a and a+ b= 2a

3) a < b

In this case max(a,b) = b and min (a,b) = a and so sum = a +b

so in call cases sum = a + b

Q13/129) What is the value of x in the following equation : (ab)^2 = (bc)^4 = (ca)^x = abc

From (ab)^2= abc we have ab = (abc)^(1/2)

Similarly (bc) = (abc)^(1/4)

ca = (abc)^(1/x)

multiplying all 3 we get (abc)^2 = (abc)^(1/2+1/4 + 1/x)

or 2 = ½ + ¼ + 1/x or x = 4/5

Q13/128) Consider n^2 + 20n + k = m^2 where k is an integer. As k varies, you can get cases of 0 or more solutions. What values of k causes 0 solutions?

if k is 2 mod 4 then there is no solution

Reason:
(n^2+20n + k) = (n+10)^2 + (k-100)

Now if n is odd then (n+10)^2 + (k-100) mod 4 = 1 + 2 = 3 so cannot be perfect square
if n is even then (n+10)^2 + (k-100) mod 4 = 2 so cannot be perfect square

so n is neither odd nor even .

so no solution

Q13/127) find integers in positive solutions to xyz + 20xy + yz + 5zx + 100x + 20y + 5z = 1913

Because we have xyz, xy, yz, xz, x, y, z terms in the left we should add some d and factor if possible

(x+a)(y+b)(z+c) – d

= xyz + ayz + bzx + cxy + bcx + abz + acy + abc – d

Comparing with given expression we get a = 1,  b = 5, c = 20

So we get (x+1)(y+5)(z+20) = xyz+ yz +20xy + 5xyz + 100 x + 5z + 20y + 100

Comparing with coefficient we get d = 100

So (x+1)(y+5)(z+20) = 1913 + 100 = 2013 = 3 * 11 * 61

So x = 2, y = 6, z= 41

Q13/126) If one of the roots of the equation 2x^2 - x -2 = 0 is a, prove that the other root is 4a^3 - 6a - 3/2

One solution is a so

2a^2 –a -2 = 0

Let other root be p = 4a^3 -6a – 3/2

So 4a^3 – 4 a = 2a^2

So 4a^3 -6a – 3/2 = 2a^2  – 2a – 3/2 = (2a^2 – a – 2) – a + ½ = ½ -- a

Now sum of roots = ½

Product of roots =  a(1/2- a) = ½a(1-2a) = 1/a – a^2 = - ½(2a^2- a)  = -1

So p is another root

Q13/125) For how many positive integer values of n does the equation 2x^2+689x+n=0 have an integer solution?

let (2x+ a)(x+b) = 2x^2+689x+n

as n is +ve both a and b positive or -ve, 

or 2x^2 + (a + 2b) = 2x^2+689x+n

a and b cannot - ve as sum is positive

0 < 2b < 689 and hence 0 < b < 344.5

so there are 344 values of b and also n

Q13/124) Let P(x) = x^3 - a x^2 + x - b Prove that there is a root between a and b (inclusive)

P(a) = a^3 - a^3 + a - b = a - b
P(b) = b^3 - a b^2 + b - b = b^2 (b - a)

If a = b then P(a) = P(b) = 0 so a and b are roots

If a and b are not same then P(a) and P(b) are of opposite signs so according to the intermediate value theorem, there exists c between a and b, such that

P(c) = 0.

Q13/123) Are there any repeating digit whole numbers that are perfect squares?

that is

Is there any number N that satisfies all of the following conditions?

(A) N is a whole number greater than 9.
(B) N is made of a single digit that repeats (1,111 or 777,777, e.g.)
(C) N is a perfect square.

Solution

The squares mod 10 can be ending with digits 1,4,9,6,5,0
so the numbers with all 2,3,7,8 cannot be perfect square

11 mod 4 , 55 mod 4, 99 mod 4 = 3, 66 mod 4 = 2

so the numbers with all 1,5,9,6

as numbers ending with all 4s ( all 11’s * 4) cannot be a perfect square

so there is no N.

Q13/122) If k is a positive integer and x is an integer such that x - 2 is a multiple of 7 and x^6 - 1 is a multiple of 7^k, show that (x + 1)^6 - 1 is also a multiple of 7^k.

We have 

(a+b)^6 = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + y^6

= a(a^5 + 6a^4 b + 15a^3b^2 + 20a^2b^3 + 15ab^2 + 6b^5) + y^ 6

If x -2 is divisible by 7 then taking a = 7k and b = 2

=> x = 7k+ 2 , b^6 mod 7 = 1

(a^5 + 6a^4 b + 15a^3b^2 + 20a^2b^3 + 15ab^2 + 6b^5) mod 7 = 6b^5 mod 7 not divisible by 7

now if a is divisible by 7^m and so (a^5 + 6a^4 b + 15a^3b^2 + 20a^2b^3 + 15ab^2 + 6b^5) is not divsible by 7

so x^6 – 1 mod 7^m is zero but x^6 – 1 mod 7^(m+1) is not zero

(x+1)^6 – 3^ 6 (as x+1 = 7k + 3) mod 7^(m+1) is is not zero

so if x^6 – 1 is divisible by 7^m <=> (x+1)^6 -1 is is divisible by 7^m

  I have proved that 
x^6 - 1 is a multiple of 7^k, iff (x + 1)^6 - 1 is also a multiple of 7^k.

  

Q13/121) Out of ( 2n + 1 ) tickets, consecutively numbered, 3 are drawn at random. Find the chance that numbers on them are in A.P.

The total number of cases (2n+1C3)

The lowest number can be 1 to 2n-1
if the lowest number is odd then there are even number of numbers and 2^nd number shall be in 1^st half.

So if lowest number id 2k +1 there are (n-k) choices for 2^nd and 3^rd number pairs

if the lowest number is even  then there are odd number of numbers and 2^nd number shall be in 1^st half and middle number not counting.

So if lowest number id 2k there are (n-k) choices for 2^nd and 3^rd number pairs

If lowest number is 1 total number of cases n( lower of the rest 2 from 2 to n+ 1
If lowest number is 2 total number of cases n-1
If lowest number is 3 total number of cases n-1
If lowest number is (2n-2) total number of cases 1( that is 2n-1 is the lower of the 2)
If lowest number is (2n-1) total number of cases 1( that is 2n is the lower of the 2)
So total number of cases = n(n-1)/2 *2 + n = n^2
So probability or chances= n^2/(2n+1C3)