2026/067) Given that $ \sum_{k=1}^{35}\sin\,5k =\tan(\frac{a}{b}) $ where angles are measured in degrees and a and b are relatively prime positive integers that satisfy $\frac{a}{b} < 90$ evaluate a + b

We shall use the formula for sum of sin of angles in AP

We have

$ \sum_{k=1}^{n}\sin\,kx = \dfrac{\sin\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin \frac{x}{2}}$

Put n = 35 and $ x = 5^\circ$

To get 

 $ \sum_{k=1}^{35}\sin\,5k^\circ = \frac{\sin\dfrac{175}{2}^\circ\sin\,90^\circ}{\sin \dfrac{5}{2}^\circ}$

$ = \dfrac{\sin\dfrac{175}{2}^\circ}{\cos \dfrac{175}{2}^\circ}=\tan  \dfrac{175}{2}^\circ$

From given condition $a=175,b=2,\dfrac{a}{b} = 87.5 <90$ 

So $a+b = 177$

2026/066) Determine all triples (a,b,c) of positive integers with $a^{b^c} = (a^b)^c$

 We are given $a^{b^c} = a^{bc}$

Or $a^{b^c-bc} = 1$
 

This has got 2 sets of solutions
 

1) a = 1 and b and c can be any positive integer
 

2) a = any positive integer other than 1 and
 

$b^c - bc = 0$
 

Or $b(b^{c-1}) = c$
 

As b is not zero so we have $ b^{c-1} = c $
 

if c = 1 we get b any number
 

Or $b= \sqrt[c-1]c$ $c>=2$
 

The root can be taken to be integer if c-1 = 1 or c =2 giving b = 2
 

So solution set $(1,b,c)$ or $(a,2,2)$ or $(a,b,1)$ where a,b,c are any integers

2026/065) If x and y are positive real numbers that satisfy the equation $x+4\sqrt{xy} -2 \sqrt{x} - 4 \sqrt{y} + 4y =3$ evaluate $\frac{\sqrt{x} + 2\sqrt{y} + 2014}{ 4- \sqrt{x} - 2\sqrt{y}}$

Solution 

We are given

$x+4\sqrt{xy} -2 \sqrt{x} - 4 \sqrt{y} + 4y =3$ 

Adding 1 to both sides

$x+4\sqrt{xy} -2 \sqrt{x} - 4 \sqrt{y} + 4y + 1=4$

Or $ (\sqrt{x} +  2\sqrt{y} -1)^2 = 4$ or $ (\sqrt{x} +  2\sqrt{y} -1) = 2$ as both square roots are positive

So  $ (\sqrt{x} +  2\sqrt{y}) = 3$

So  $\frac{\sqrt{x} + 2\sqrt{y} + 2014}{ 4- \sqrt{x} -  2\sqrt{y}}=  \frac{3 + 2014}{ 4- 3} = 2017$
 

2026/064) Given a,b,c,d are roots of the equation $x^4−7x^3+3x^2−21x+1=0$ Evaluate $(a+b+c)(b+c+d)(c+d+a)(d+a+b)$

Because a,b ,c ,d are roots of the equation $x^4−7x^3+3x^2−21x+1=0$

We have by vieta's formula
 

$a+b+c+d = 7\cdots(1)$
 

Let
 

$f(x) = x^4−7x^3+3x^2−21x+1=(x-a)(x-b)(x-c)(x-d)\cdots(2)$
 

From (1) we have
 

$a+b+c = 7-d\cdots(3)$
 

$b+c+d  = 7 -a\cdots(4)$
 

$c+d+a = 7 -b\cdots(5)$
 

 $d+a +b = 7-c\cdots(6)$
 

From (3), (4), (5),(6) we have
 

$(a+b+c)(b+c+d)(c+d+a)(d+a+b)= (7-d)(7-a)(7- b)(7-c) = f(7)\cdots(7)$ from (2)
 

As $f(x) = x^4−7x^3+3x^2−21x+1$
 

So $f(7) = 7^4 - 7 * 7^3 +3 *7^2 - 21 * 7 + 1 = 1$
 

From (7) and above we have
 

$(a+b+c)(b+c+d)(c+d+a)(d+a+b) = 1$

 

2026/063) Factor $n^4+6n^3+11n^2+6n+1$

Let $f(n) =  n^4+6n^3+11n^2+6n+1$

This does not have a change of sign so there is no positive root

So it can have -ve root if it has real root it has to be -1

Checking using the rational root theorem $(f(-1) = 1 - 6 + 11 -6 +1 = 1$

So -1  is not a root

So it has to be product of 2 quadratic polynomials 

This is a quartic polynomal. And the coefficients are symmetric.

So if n is a root then $\frac{1}{n}$ is a root. So the coeffcient of$x^2$ and constant should same in both polynomials. However I shall continue as below

$f(n) =  n^4+6n^3+11n^2+6n+1$

$=  (n^4+1) +11n^2+6(n^3+n)$ reordering the terms

$= (n^2+1)^2 -2n^2 +11n^2+6n(n^2+1)$ putting $n^4+1$ in terms of $(n^2+1)$

$=(n^2+1)^2 + 6n(n^2+1) + 9n^2$

$= (n^2+1)^2+ 2(3n)(n^2+1) + (3n)^2$ geting expressing in $a^2+2ab+b^2$ form

$=(n^2+1+3n)^2$

$= (n^2+3n+1)^2$ putting in standard form  

 

2026/062) Define $f(n)=LCM(1,2,\cdots\,n)$. Determine the smallest positive integer a such that $f(a)=f(a+2)$

This shall require that $a+1$ and $a+2$ do not increase the LCM.

This shall require both $a+1$ and $a +2 $to be composite and neither is a perfect power of a prime. 

Let us explain it for $a+1$ and the same logic holds for $s+ 2$

If $a+1$ is prime then we have not encountered the same in any of the numbers and this shall contribute to LCM and it shall increase. LCM should be multiplied by $a+ 1$ 

If  $a+1 = b^k$ where b is a prime then it should be multiplied by $b$ as $b^(k-1)$ must have come previously but not $b^k$

if $a+1=b^kc^m$ that is product of powers of 2 primes then $b^k$ and $c^m$ has already come and hence it shall not contribute to  a higher LCM

Same for $a+2$     

 So we require smallest n such that n+1, n+ 2 are composite and producut of power of at least 2 primes 

Looking at $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

$14 = 2 * 7 $so LCM does not increase

$15 = 3 * 5$ so LCM does not increase

For $(8,9)$ both power of primes ,

 For $(9,10)$, $9$ is power of prime

 So $a = 13$ 

Next $a = 19$ as  $20=2 ^2 * 5, 21 = 3 *7$ do not increase meets criteria  

  

2026/061) Five distinct 2-digit numbers are in a geometric progression. Find the middle term.

Without loss of generality we can assume that nmbers are in increasing sequence 

We have 3 powers of 2 as 2 digit numbers . They are 16,32,64. 

If we start at 10 we get 4 numbers in geometric progression when ratio is 2. 

So we need some ratio less than 2 . Common ratio can be fraction as long as we do not get a fraction after multiplying by common ratio . 

Because we need 5 numbers we need to multiply 4 times . So we start with a 4th power of 2 that is 16 and common ratio $\frac{3}{2}$ 16 giving 16, 24,36,54,81 and middle term is 36.

Basically  5 numbers $a^4,a^3b,a^2b^2,ab^3,b^4$ where a is starting number and $\frac{b}{a}$ as common ratio. form  GP.