2021/078) What are two numbers whose HCF is 20 and LCM is 300?

Because HCF is 20 so the 2 numbers are 20m,20n where m or n = 1 or they are co-primes. without loss of generality let us assume that that m < n.

So we have LCM = 20mn = 300 or mn= 15

as mn = 15 and m and n are co-primes so m= 1 , n= 15 or m =3 nd n = 5 giving the 2 numbers (20,300) , (60,100) 

 

2021/077) Let $a>b>c$ be the real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $b(a+c)$.

To avoid radicals let $\sqrt{2014}=p$

So we get $px^3-(2p^2+1)x^2 +2 = 0$

Or factoring we get $(px-1)(x^2-2px-2)$ = 0

So one root is $x= \frac{1}{p}$ and  other two roots are roots of $x^2-2px-2=0$

For the equation $x^2-2px-2=0$ sum of the roots is 2p and product is -2. so one root has to be -ve and

The positive root shall be above 2p

So $b=\frac{1}{p}\cdots(1)$

And c is the -ve root and $a> 2p$

a,c are roots of $x^2-2px-2=0$ so $a+c = 2p\cdots(2)$

Hence $b(a+c) = 2$ using (1) and (2) 

2021/076) Evaluate $\sum^{89}_{n=0}\frac{1}{1+\tan^3(k^\circ)}$

We have $\sum^{89}_{n=0}\frac{1}{1+\tan^3(k^\circ)}$
 $= \sum^{89}_{n=0}\frac{\cos^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{\cos^3(0^\circ)}{\cos^3(0^\circ)+\sin ^3(0^\circ)} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{\sin^3(45^\circ)}{\cos^3(45^\circ)+\sin ^3(45^\circ)}+ \sum^{89}_{n=46}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ splitting into 4 parts
$=\frac{1}{1+0} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{\sin^3(45^\circ)}{\sin^3(45^\circ)+\sin ^3(45^\circ)}+ \sum^{89}_{n=46}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ as $\sin\,45^\circ = \cos \, 45^\circ$
$=\frac{1}{1+0} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{1}{2}+ \sum^{89}_{n=46}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ putting values and using  $\sin\,x^\circ = \cos \,(90^\circ-x)$
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} +  \sum^{89}_{n=46}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} +  \sum^{46}_{n=89}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ reversing order of sum of cos terms
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} +  \sum^{44}_{n=1}\frac{\cos^3( k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)+\cos^3( k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{3}{2} + \sum^{44}_{n=1} 1= 1.5 + 44 = 45.5$

2021/075) If $x^2-x-1$ divides $ax^{17}+bx^{16} + 1$ find a-b

If $x^2-x-1$ divides $ax^{17}+bx^{16} + 1$ then $x^2=x+1$ =>  $ax^{17}+bx^{16} + 1 = 0$
The above is so because x =t which is root of $x^2= x+1$ is also a root of $ax^{17}+bx^{16} + 1= 0$

We have $x^2=x+1$
Putting $y=\frac{1}{x}$ we get $\frac{1}{y^2} =  \frac{1}{y} +1$
or $y^2 = 1 - y$
 $=>y^4 = (1-y)^2 = 1-2y +y^2 = (1-2y) + (1-y) = 2-3y$
$=>y^8 = (2-3y)^2 = 4-12y +9y^2 = (4-12y) + 9(1-y) =13-21y$
$=>y^{16} = (13-21y)^2 = 169-546y +441y^2 = (169-546y) + 441(1-y) = 610-987y$
$=>y^{17} = 610y-987y^2 = 610y - 987(1-y) = 1597y - 987$
Putting back $x=\frac{1}{y}$ we get
$\frac{1}{x^{17}} =  \frac{1597}{x} - 987$
Or $987x^{17} - 1597x^{16} +1$ = 0

comparing with above we get a = 987, b = -  1597 so a - b = 2584

2021/074) A number n has sum of digits 100, while 44n has sum of digits 800. Find the sum of the digits of 3n.

The number can contain only the digits 1,2 besides 0. 1 * 44 = 44 and there is no overflow( if in the number the sum is one then it becomes 8, and if it is 2 the the sum of digits is 16) so the sum of digits is 8 times. if any digit is 3 to 9 then sum of digits less than 8 times so this shall give a lesser sum

Again 1 may be preceded/succeeded  by 0 or 1 as 11 * 44 = 484 . but 2 has to be    preceded/succeeded  by 0 as 21 * 44 = 924 and 12 * 44 = 538 and the sum of digits become less

So the number shall have 0 1 and 2 meeting above conditions so that sum of digits 100 and when we multiply by 3 (that is 3n) the digits shall be 0,3,6 and there is no overflow and sum of digits 300.

2021/073)Consider polynomials P(x) of degree at most 3 each of whose coefficients is an element of { 0,1,2,3,4,5,6,7,8,9}. How many such polynomials satisfy P(-1) = 9.

Because the polynomial is of degrees at most 3 with above coefficients we have 

$P(x) = ax^3+bx^2 + cx +d$ where a, b ,c, d are  element of { 0,1,2,3,4,5,6,7,8,9}.

now $P(-1) = -a + b -c + d = (b+d) - (a+c)$

now maximum sum of b+d can be 18 and as $b+d - (a+c) = 9$ so we have

$a+c =k$ and $b+d = 9+k$ and hence $0 <=<=9$

as the digits are from 0 to 9 for the sum to be k <= 9 one of the numbers can be x = 0 to k and another number can be k -x so there are k+1 possibilities

for the sum to be  $k >=9$ one of the numbers a should be $k-9$ to $9$ and another number c shall be $k-9$ and there are $(9+1) - (k-9)$ or $19-k$ choices.

for the sum to be  $k >=9= 9 + m$ there are  $19-(9+m)$ or $10-m$ choices.

For the value  $b+d - (a+c) = 9$ a + c can be k and b+d can be k+9 so there are (k+1)(10-k) ways 

now the value k can be from 0 to 9 so we have sum

$\sum_{k=0}^9(k+1)(10-k) =\sum_{n=1}^{10}n(11-n) = 11\sum_{n=1}^{10}n - \sum_{n=1}^{10}n^2 = 11 * \frac{10*11}{2} - \frac{10 * 11 * 21}{6} = 605- 385 = 220$

So number of polynomials = 220  

2021/072) if in a triangle $a^2+b^2= \frac{19}{9}c^2$ evaluate $\frac{\cot\,C}{\cot\, A + \cot \, B}$

We are given
$a^2+b^2= \frac{19}{9}c^2$
Using the above and law of cosine we get
$2ab\cos\, C = a^2+b^2-c^2 = \frac{19}{9}c^2 - c^2 = \frac{10}{9}c^2$
or $ab\cos\, C = \frac{5c^2}{9}\cdots(1)$

Further
$\cot\, A +\cot\, B= \frac{\cos\, A}{\sin\, A} +  \frac{\cos\, B}{\sin\, B}$
$= \frac{\cos\, A\sin\, B +  \sin\,A\cos\, B}{\sin \, A\sin\, B}$
$= \frac{\sin(A+B)}{\sin \, A\sin\, B}$
$= \frac{\sin(\pi-C)}{\sin \, A\sin\, B}$ as $A+B+C=\pi$
$= \frac{\sin\,C}{\sin \, A\sin\, B}$

Hence $\frac{\cot\,C}{\cot\, A + \cot \, B} = \frac{\cos\, C \sin\, A\sin\, B}{\sin ^2C}$
$=\frac{ab\cos\,C}{c^2}$ (using law of sin)
$=\frac{5}{9}$ 

2021/071) For the triangle with angles A,B,C, the following trigonometric equality holds. $\sin^2B+\sin^2C−\sin^2A=\sin\,B\sin\,C$ Find the measure of the angle A.

Using law of sin's $\sin A = ka, \sin B= kb, \sin C = kc$

We get

$b^2+c^2 - a^2 = bc$

Or $a^2 = b^2 + c^2 + bc\cdots(1)$

By law of cos

$a^2 = b^2 + c^2 - 2bc \cos A \cdots(2)$

from (1) and (2)

$2 \cos A = - 1$ or $\cos A = \frac{-1}{2}$ or $A = 12^circ$

2021/070) Evaluate $\dfrac{1}{1-\cos \dfrac{\pi}{9}}+\dfrac{1}{1-\cos \dfrac{5\pi}{9}}+\dfrac{1}{1-\cos \dfrac{7\pi}{9}}$.

We know that $\cos \frac{\pi}{9},\cos \frac{5\pi}{9},\cos \frac{7\pi}{9}$ are different and they are

roots of equation $\cos 3x = \cos \frac{\pi}{3} = \frac{1}{2}$

or $4\cos^3 x - 3\cos\,x =\frac{1}{2}$

or

so $\cos\frac{\pi}{9}, \cos\frac{5\pi}{9}, \cos\frac{7\pi}{9}$ are roots of equation

$x^3 - \frac{3}{4}x - \frac{1}{8}= 0$

let $x_1= \cos\frac{\pi}{9}, x_2 = \cos\frac{5\pi}{9}, x_3=\cos\frac{7\pi}{9}$

Now $x_1,x_2,x_3$ are roots of equation

$f(x) = x^3 - \frac{3}{4}x - \frac{1}{8}= 0\cdots(1)$

by Vieta's formula we have

$x_1 + x_2 + x_3 = 0\cdots(2)$

$x_1 x_2 + x_2x_3 + x_3 x_1 = \frac{-3}{4}\cdots(3)$

Further $f(1) = (1-x_1)(1-x_2)(1-x_3) =  1- \frac{1}{4} - \frac{1}{8} = \frac{1}{8}\cdots(3)$

And we need to evaluate $\frac{1}{1-x_1 } + \frac{1}{1-x_2} + \frac{1}{1-x_3}$

Now

$\frac{1}{1-x_1 } + \frac{1}{1-x_2} + \frac{1}{1-x_3}$

$= \frac{(1-x_2)(1-x_3) + (1-x_1)(1-x_3) + (1-x_1)(1-x_2)}{(1-x_1)(1-x_2)(1-x_3)}$

$= \frac{1-x_2 - x_3 + x_2x_3 + 1-x_1 - x_3  + x_1x_3  + 1-x_1 - x_2 + x_1x_2}{(1-x_1)(1-x_2)(1-x_3)}$

$= \frac{3 - 2(x_1 + x_2 + x_3)  + (x_2x_3 + x_3x_1  + x_1x_2)}{(1-x_1)(1-x_2)(1-x_3)}$

$= \frac{3 - 2 * 0 + \frac{-3}{4}}{\frac{1}{8}}$ putting the values using (2) , (3) and (4)

$= 18$

Hence $\frac{1}{1-\cos \frac{\pi}{9}} + \frac{1}{1-\cos \frac{5\pi}{9}} + \frac{1}{1-\frac{7\pi}{9}}= 18$

 

2021/069) Prove that there are no integers $a,\,b,\,c$ and $d$ such that the polynomial $ax^3+bx^2+cx+d$ equals 1 at $x=19$ and 2 at $x=62$.

We have  $f(62)−f(19)=a(62^3−19^3)+b(62^2−19^2)+c(62−19)=1$

Or  $(62−19)(a(62^2+62∗19+19^2)+b(62+19)+c)=1$ 

LHS is a multiple of 43 and RHS is 1 so this does not have integer solution