2015/052) Find four numbers in A.P. whose sum is 20 and sum of whose squares is 120?

without loss of generality we can choose numbers to be a - 3d, a - d, a + d, a-3d and d > 0

now sum = $(a-3d) + (a-d) + (a+d) + (a+3d) = 4a = 20$ or $a=5$

sum of squares = $(a-3d)^2 + (a- d)^2 + (a+d)^2 + (a+3d)^2 = 4a^2 + 20d^2 = 120$

so putting value of a we get d = 1 so numbers are 2,4,6,8
I found the problem at https://in.answers.yahoo.com/question/index?qid=20150528071206AAbn3Cz

2015/051)If $m\tan(a-30^\circ)=n\tan(a+120^\circ)$ show that $\cos2a=\dfrac{m+n}{2(m-n)}$

we have $\tan (a+ 120^\circ) = - \cot(a + 30^\circ)$ using $\tan (x+90^\circ) = - cot\, x$
so $mtan(a-30^\circ)= -n \cot ( a+ 30^\circ)$
so $\tan (a+30^\circ) tan (a-30^\circ) = - \dfrac{n}{m}$
$\dfrac{\tan\, a + \tan\, 30^\circ}{1- \tan\, a \tan\, 30^\circ}\dfrac{tan\, a -\tan\, 30^\circ}{1 + \tan\, a \tan\, 30^\circ} = -\dfrac{n}{m}$
or $\dfrac{\tan ^2 a - \tan ^2 30^\circ}{1- \tan ^2 a \tan ^2 30^\circ} = -\dfrac{n}{m}$

or $\dfrac{\tan ^2 30^\circ-\tan ^2 a}{1- \tan ^2 a \tan ^2 30^\circ} = \dfrac{n}{m}$

or $\dfrac{\frac{1}{3}-\tan ^2 a}{1- \frac{1}{3}\tan ^2 a} = \dfrac{n}{m}$

 or $\dfrac{1-3 \tan ^2 a}{3- \tan ^2 a} = \dfrac{n}{m}$

  using componendo dividendo to get

$\dfrac{4 - 4 \tan ^2 a}{- 2 - 2 \tan ^2 a} = \dfrac{n+m}{n-m}$

or 

$2 \dfrac{1 - 1 \tan ^2 a}{1 + \tan ^2 a} = \dfrac{n+m}{m-n}$

or 

$2 \dfrac{1 - \tan ^2 a}{sec ^2 a} = \dfrac{n+m}{m-n}$

or

$2 (1 - \tan ^2 a)(cos ^2 a) = \dfrac{n+m}{m-n}$

 or $(\cos^2 a-\sin ^2 a) = \dfrac{m+n}{2(m-n)}$
or $\cos 2a = \dfrac{m+n}{2(m-n)}$

This is solved at https://uk.answers.yahoo.com/question/index?qid=20120601212526AAguVnm

2015/050) If $ax + y + 1= 0, x +by+1=0,$ $x+y+c=0$

are concurrent then prove that $\dfrac{1}{1-a} + \dfrac{1}{1-b} + \dfrac{1}{1-c} = 1$

proof

We have
$a = - \dfrac{y+1}{x}$
or $1- a = \dfrac{x+y+1}{x}$
or $\dfrac{1}{1-a} = \dfrac{x}{x+y + 1} \cdots 1$

similarly

$x +by+1=0$
=> $x+1 = - by$
or $b = -\dfrac{x+1}{y}$
or $1-b= \dfrac{x+y+1}{y}$
or $\dfrac{1}{1-b} = \dfrac{y}{x+y+1}\cdots (2)$

and $x+y+c=0$
=> $-c = x + y$
=> $1-c = x+y+1$
or $\dfrac{1}{1-c} = \dfrac{1}{x+y+1}\cdots(3)$

adding all 3 we get the
$\dfrac{1}{1-a} + \dfrac{1}{1-b} + \dfrac{1}{1-c} = \dfrac{x+ y + 1}{x+y+1} =1$

Proved 

2105/049) For what positive integral values of x is $3^x-x^2$ is divisible by 5

As 3 and 5 are coprimes

So $3^4 = 1$ mod 5
$3^{4k+1} = 3$ mod 5
$3^{4k+2}= 4$ mod 5
$3^{4k+3} = 2$ mod 5

Now $(5k+m)^2$ mod 5 = 0 if m= 0, 1 if m= 1 or 4, 4 if m = 2 or 3 mod 5

now $3^x= x^2$ mod 5 if

x = 4k that is 0 mod 4 and 1 mod 5 ( in both cases remainder 1)
or x = 4k and 4 mod 5( in both cases remainder 1)
or x = 4k+2 or 2 mod 4 and 2or 3 mod 5
x = 0 mod 4 and 1 mod 5 => x= 16 mod 20
or x = 4k and 4 mod 5 => x = 4 mod 20

or x =2 mod 5 and 2 mod 4 => x = 2 mod 20

or x = 2 mod 4 and 3 mod 5 => x= 18 mod 20

x = a mod 4 ,b mod 5 can be solved by Chinese remainder theorem
I have not detailed steps

2015/048) What could be possible value of integer a if $10^ {2n +1} + a.7^ {2n +1}$ is divisible by 51 exactly

we have $10^2 = 100 = -2$ mod 51

so $10^ {2n +1} = 10(-2)^n$

$7^2 = -2$ mod 51

so $a.7^ {2n +1} = 7a (-2)^n$ mod 51

$10^{2n +1} + a.7^ {2n +1}$ mod 51
= $10(-2)^n  + 7a (-2)^n = 0$ as divisible by 51

so $10 + 7a = 0$ mod 51
$7a = - 10$ mod 51 = 41 mod 51

$7a = 41$ mod 51

we need to find inverse of 7 mod 51( we can find by extended euclid algorithm as below)

$51 = 7 * 7 + 2$  or  $2 = 51- 7 * 7$

$7 = 2 * 3 + 1$ or $1= 7 - 2 * 3 = 7 - (51- 7 * 7) * 3 = 51 * 3 - 22 * 7$

so 22 = inverse of 7 mod 51

so $a = 41 * 22$ mod 51 or 35 mod 51 or 51k + 35

 

2015/047) Prove that if $x = log_a(bc),y = log_b(ca),z = log_c(ab)$ then the value of $ xyz - x - y - z $ is 2

we have

$x= log_a(bc)$

so $1 + x = 1 + log_a(bc) = log_a(a) + log_a(bc) = log_a(abc)$

or $\dfrac{1}{1+x} = log_{abc}(a) \cdots (1)$

similarly

$\dfrac{1}{1+y} = log_{abc}(b) \cdots (2)$

$\dfrac{1}{1+z} = log_{abc}(c) \cdots (3)$

 

 Adding (1), (2) and (3) we get
$\dfrac{1}{1+x}+ \dfrac{1}{1+y}+ \dfrac{1}{1+z} = log_{abc}(a) + log_{abc}(b) + log_{abc}(c) = log_{abc}(abc) = 1$

or
$(1+y)(1+z) + (1+z)(1+x) + (1+x)(1+y) = (1+x)(1+y)(1+ z)$

or

$1 + yz + y + z + 1 + xz + x + z + 1 + xy + x + z = 1 + x + y + z + xy + yz + zx + xyz$

or $2 + x + y+ z = xyz$

or $xyz – x – y -z = 2$

2015/046) Prove that for any natural number n, $11^{n + 2} + 12^{2n + 1}$ is divisible by 133

we have

$11^{n + 2} + 12^{2n + 1}$

= $11^2 * 11 ^n + 12 * 12^{2n}$

= $121 * 11^n + 12 * 144^n$

= $121 * 11^n + 12 * 11^n + 12 * 144^n – 12 * 11^n$

= $133 * 11^n + 12( 144^n – 11^n)$

the 1^st term is multiple of 133 and 2^nd term is divisible by 144-11 $( a^n-b^n)$ is divisible by a – b and hence the sum.