We need to find $3^n+ 4^n \mod 5$
Let us take the $3^n \mod 5$ and see the behaviour As 5 is prime we have
$3^4 \equiv 1 \mod 5$
and also
$4^4 \equiv 1 \mod 5$
so $3^n + 4^n \mod 5 $ have period 4
Let us consider $f(n) = 3^n + 4^n \pmod 5$
We have
$f(0) = 2$
$f(1) = 2$
$f(2) = 0$
$f(3) = 1$
And $f(n) = f(4k+n)$
We see that $f(2)$ is zero so $f(4k+2)$ is zero
So we need to find the smallest k suck that $4k + 2 > 99$ and get $k = 25$ and smallest number $102$
Basis
To find the number of positive divisors of an integer, find its prime factorization, add one to each exponent, and multiply those sums together.
That is If $N=p_1^{q_1}p_2^{q_2}\cdots p_n^{q_n}$
it shall have $(q_1+1) (q_2+1)\cdots(q_n+1)$ factors
kindy note that $p_1,p_2\cdots p_n$ are relatively prime and not $q_1+1,q_2+1\cdots q_n+1$
Solution
Let is factor 24 in as many ways as we can and put N as product of power of primes and for the number to be lower higher power shall be with lower numer
24 = 24 this gives number $2^23$
24 = 8 * 3 giving $2^7 *3^2= 1152 $= this is smaller
can we make it smaller
24 = 4 * 2 *3 = 4 * 3 * 2 giving $ 2^3 *3^2 * 5 = 360$
if we try to make a smaller number we get 3 * 2 * 2 * 2 giving $2^2 * 3 * 5 * 7 = 420$ and it is larger
so the ans is $360$
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 $a+b = 177$
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
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 $\frac{\sqrt{x} + 2\sqrt{y} + 2014}{ 4- \sqrt{x} - 2\sqrt{y}}= \frac{3 + 2014}{ 4- 3} = 2017$
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
$(a+b+c)(b+c+d)(c+d+a)(d+a+b) = 1$
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
