Q2/121) The set of all integers n for which sqrt(n^2 + n) is an integer is.

a) the set {0,-1};
b) a finite set with at least three elements;
c) an infinite set;
d) none of these sets;

Proof:
sqrt(n^2 + n) = sqrt(n(n+1)) is inetger

n and n+1 are coprimes so either n= 0 or n = - 1 or n and n+1 both squares

n = x^2 and n+1 = y^2 => 1 = (x+y)(y-x) => x+y = 1 and y-x = 1 => x = 0 y =1 => n = 0

or x+y = -1 and y-x = -1 => x = 0 y =-1 => n = 0

so only solution 0 and -1 so ans is a)

Q12/120) Let a,b and c be the sides of a right angled triangle. Let theta be the smallest angle of this triangle.?

f 1/a, 1/b, 1/c are also the sides of a right angled triangle then show that Sin(theta) = (sqrt(5) - 1)/2;

Proof:
Let a < b < c


And theta (say t opposite to smaller side)

Sin t = a/c and cos t = b/c

Now 1/a > 1/b > 1/c and as it is right angled triangle we have

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

Or

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

(1/sin ^2 t) = (1/ cos^2 t ) +1

Or cos^2 t = sin ^2 t + sin ^2 t cos^2 t

Or 1- sin ^2t = sin ^2 t + sin ^2 t (1- sin ^2 t)
= 2 sin ^2 t – sin ^4 t
Or sin ^4 t – 3 sin ^2 t + 1 = 0

Sin^2 t = (3 +/- sqrt(9-4))/2 = (3 – sqrt(5))/2 , + cannot be taken as it shall be >1 not possible


So sin t = sqrt((3 – sqrt(5))/2) say sqrt(x) – sqrt(y) as there is sqrt(5) in square

So squaring we get

x+y – 2sqrt(ab) = (3 –sqrt(5))/2

so x+y = 3/2 and sqrt(xy) = sqrt(5)/4) or xy = 5/16

we can solve it as x = 5/4 and y= 1/4 or y = 1/4 and x = 5/4

so sin t = +/-(sqrt(5/4) - sqrt(1/4))

positive value to be taken as sint > 0 so sin t = sqrt(5/4) – sqrt(1/4) = (sqrt(5)-1)/2
 

Q12/119) Let a < b < c < d be four real numbers, such that all six pairwise sums ...? are distinct. The values of the smallest four pairwise sums are 1, 2, 3, and 4 respectively. What are the possible values of d?

Then the six pairwise sums

a+b, a+c, a+d, b+c, b+d, c+d

are all distinct.
there are 2 possibilities
case 1:
a+d < b+ c

Now, a+b = 1, a+c = 2, a+d = 3, b+c = 4.

∴ (a+c) - (a+b) = 2-1

∴ c-b = 1

∴ (b+c) + (c-b) = 4+1 ∴ 2c = 5 ∴ c = 5/2

∴ a+c = 2 gives a = 2-c = 2-(5/2) = -1/2

∴ a+d = 3 gives d = 3-a = 3-(-1/2) = 7/2

∴ d = 7/2. ........ Ans for case 1.
 case2 : b+ c< a + d
then we get a+b = 1, a+c = 2, b+c =3 a +d = 4 => 2a + b + c = 3 => a = 0 b = 1 c = 2 and d = 4

so d =4 is another solution for case 2

so d can be 7/2 or 4

Q12/118) Given 4 positive integers a,b,c and d such that a^5=b^4, c^3=d^2 and c−a=19, what is d−b?

as c^3 = d^2 so c will be a square let c = x^2

as a^5 = b^4 so a = y^4 for some y

now
c-a = 19
=> x^2 - y^4 = 19
=> (x-y^2)(x+y^2) = 19
hence x - y^2 =1 and x+y^2 = 19 as 19 is a prime
so x = 10 and y = 3

so a = y^4 or a^5 = y^20 = b^ 4 or b= y^5 = 243

c= x^2 => c= 100 and hence d^2 = 10^6 and so d = 1000

d-b = 1000 - 243 = 757

Q12/117 )100 positive integers are written in a row. The average of the first and second numbers is 1. The average of the second and third numbers is 2. The average of the third and fourth numbers is 3. This pattern continues, and the average of the 99th and 100th number is 99. What is the 100th number?

Average if 1st and 2nd number is 1 so sum is 2 so both positive integers have to be 1

so a1= 1
a2 = 1

now average of 2nd and 3rd is 2 so sum = 4 so a3= 3

average of 3rd and 4th is 3 so sum = 6 so a4= 3

average of 4th and 5th is 4 so sum = 6 so a5= 5

so we have nth term is n when n is odd and n-1 when n is even

this can be proved as below

for n odd 2 numbers are n and n and average = n
for n even the number is n-1 and next number is n+1 and average is n

so 100th number = 99

The integer 2006 is the product of three prime numbers p, q, and r. If p + q = r^2 (p + r), what is the value of q?

2006 = 2•17•59 (prime factorization)
so p q rare permutions of 2,17,59
now r^2 divides p+q so 17 and 59 are too large for r so r= 2
so p+q = 4 p + 8
or 3p = q - 8
so p < q-8 => so p=17 and 3p = 51 and q = 59 satisfying the condition
hence q = 59

What is the nth term that goes 1,2,2,3,3,3,4,4,4,4, etc

as n occurs n times

the last position of n = n(n+1)/2

last position of (n-1) = n(n-1)/2

so the position n(n-1)/2 + 1 to n(n+1)/2 contains n

kth position contains n and

n(n-1)/2 < k <= n(n+1)/2

or n(n-1) < 2k <= n(n+1)

n^2 - n < 2k <= n^2 + n

multiply by 4

4n^2 - 4n < 8k <= 4n^2 + 4n
(2n-1) ^2 - 1 < 8k <= (2n +1)^2 - 1
(2n-1) ^2 < 8k +1 <= (2n +1)^2

so n = (ceil(sqrt(8k+1) -1)/2

find last 3 digit of sum (9^k) k = 1 to 400

the sum is S = [1<= n <= 400] 9^n

we have 1000 = 8 * 125 so to compute S mod 8 and S Mod 125

now 9 = 1 mod 8 so 9^n = 1 mod 8

there are 400 numbers each is 1 mod 8 so sum = 0 mod 8

sum = 9/8(9^400-1)

as 9 and 125 are coprime as per http://www.cut-the-knot.org/blue/Euler.s…

9^φ(125) mod 125 = 1

now 125 = 5^3 so φ(125) = 125(1-1/5) = 100

so 9^(100) = 1 mod 125
or 9^400 = 1 mod 125
so 9/8(9^400-1) = 0 mod 125

now S = 0 mod 8 and S = 0 mod 125 and

hence S = 0 mod 1000 so last 3 digits are zero

Prove that average of the numbers : 2 sin2° , 4 sin4° , 6 sin6° ..... , 180 sin 180° = cot 1°

we have cos (n-1) - cos (n+1) = 2 sin n sin 1 ,,,1

now 180 sin 180 = 180 sin 0
178 sin 178 = 180 sin 2

so total sum = 180 ( sin 0 + sin 2 + sin 4 + ... + sin 88) + 90 sin 90
= 90 ( 2 sin 0 + 2 sin 2 + 2 sin 4 + 2 sin 88) + 90 sin 90

as there are 90 numbers average
= 2 sin 2 + 2 sin 4 + 2 sin 88 + sin 90 as
= ( cos 1 - cos 3)/ sin 1 + (cos 3 - cos 5)/ sin 1 + ( cos 87- cos 89 / sin 1 + 1
= ( cos 1 - cos 89)/ sin 1 + 1
= cot 1 - sin 1/sin 1 + 1 as cos 89 = sin 1
= cot 1 - 1 + 1
= cot 1

Given log(base a)x*log(base b)x + log(base b)x*log(base c)x +log(base c)x*log(base a)x = log(base a)x*log(base b)x*log(base c)x. prove that x = abc

devide both sides by

log(base a)x*log(base b)x*log(base c)x.
to get 1/ log(base c)x + 1/log(base a)x + log(base b)x = 1

now 1/ log(base c)x = log (base x) c

so we get log (base x) c + log (base x) a + log (base x) c = 1

or aply the product rule to get

log (base x) abc = 1

so abc = x^1 = x

proved