We have as 31 prime using Wilsons theorem
$30! + 1 \equiv 0 \pmod {31}$
or $30 * 29! + 1 \equiv 0 \pmod {31}$
as $30 \equiv - 1 \pmod {31}$
so we have $-1 * 29! + 1 \equiv 0 \pmod {31}$
Or $ 29! -1 \equiv 0 \pmod {31}$
Or $ 4 * 29! - 4 \equiv 0 \pmod {31}$
adding 124 which is multiple of 31 we get
$ 4 * 29! + 120 \equiv 0 \pmod {31}$
or $ 4 * 29! + 5! \equiv 0 \pmod {31}$
Proved
We have
$a+b+c=3\cdots(1)$
$a^2+b^2+c^2=3\cdots(2)$
Squaring (1) we get
$a^2+b^2 + c^2 + 2ab + 2bc + 2ca = 9$
Putting the value of $a^2+b^2+c^2$ from (2) we get
$ab+bc+ca = 3$
subracting above from (2) we get
$a^2+b^2+c^2 - ab - bc - ca = 0$
or $2a^2+2b^2+2c^2 - 2ab - 2bc - 2ca = 0$
or $(a^2-2ab + b^2) + (b^2-2bc+c^2) + (c^-2ca +a^2) = 0$
or $(a-b)^2 + (b-c)^2 + (c-a)^2= 0$
above is possible only if a=b= c
putting in (1) we get a = b = c = 1
We have $n^4-1 = (n^2+1)(n^2-1)$ difference of 2 squares
$= (n^2+ 1)(n+1)(n-1)$ difference of 2 squares
$= (n^2-4 + 5) (n+1)(n-1) as $n^2+ 1=n^2-4 + 5$
$= (n^2-4)(n+1)(n-1)+ 5(n+1)(n-1)$
$= (n-2)(n+2)(n+1)(n-1)+ 5(n+1)(n-1)$
$\frac{(n-2)(n-1)n(n+1)(n+2)}{n} + 5(n+1))(n-1)$
$5(n+1))(n-1)$ is divisible by 5
$(n-2)(n-1)n(n+1)(n+2)$ being product of 5 consecutive numbers is divisible by 5
As n is not dvisible by 5 so $\frac{(n-2)(n-1)n(n+1)(n+2)}{n}$ is divisible by 5
Hence the sum is divisible by 5 and so the given expression
Note: this can be done faster using mod 5 arithmetic but I have chosen a different way.
Aditionaly this is direct from fermats little theorem
Both a nd b have to be even or add as RHS is even as it has to be > 1.
Because if a is odd and b even or vice versa then LHS is odd.
Now let us consider 1st case that is a and bare odd
So we get
$a^3+b^3 = (a+b)(a^2-ab+b^2)$
a+b is even and $a^2-ab+b^2= a(a-b) + b^2$ is odd and it has to be 1
Or $a(a-b) + b^2=1$
The above is possble only if a-b = 0(else or a = b and we get a = b = 1
Now of the even case
Let $a=p.2^q$ and b= $m.2^r$ and without loss of generality let$ q >= r$ and p and m are odd.
So we get $p^32^{3q} + m^32^{3r} = 2^n$
Devideing both sides by $2^{3r}$ we get
$p^32^{3q-3r} + m^3 = 2^{n-3r}$
RHS is even as it is minimum 2
LHS is even if 3q=3r or q = r
Puttting it we get $p^3+q^3 = 2^{n-3r}$
As p and q are odd so both are same and hence $a= b = 1$ for odd and $a=b=2^x$ for even
We have $\frac{2+3z+4z^2}{2-3z+4z^2}$ real
Both numerator and denominator are expression with 3 terms
Subtracting 1 fron the expression we shall have it real and numerator is simpler
Or $\frac{2+3z+4z^2}{2-3z+4z^2}-1$ is real
Or$\frac{-6z}{2-3z+4z^2}$ is real
As imaglinary part of x is not zero so x is not zero so inverting
$\frac{2-3z+4z^2}{-6z}$ is real
Or $\frac{2-3z+4z^2}{z}$ is real
Or $\frac{2}{z}-3+4z$ is real
And adding 3 we get $\frac{2}{z}+4z$ is real
Or $\frac{1}{z}+2z$ is real
Now let $z= x+ iy$
So $\frac{1}{x+iy}+2(x+iy)$ is real
Or $frac{x-iy}{x^+y^2} + 2(x+iy)|$ is real
Or $\frac{-y}{x^+y^2} + 2y)=0$
as y is non zero $\frac{1}{x^2+y^2} -2=0$ or $|z|^= \frac{1}{2}$
As we see above power of 2 and 3 ( 9 is $3^2$) and 6 are invloved
Let $3^x = a$ and $2^x=b$
We get $ab + a^2 = 2 b^2$
Or $a^2 + ab - 2b^2 = 0$
or $(a-b)(a+2b) = 0$
$a=b$ or $a+2b=0$
as a and b are positive s a = b (a+2b=0 is inadmissible)
or $3^x = 2^x$ or x = 0
We have $x^2+xy+ y^2= 0\cdots(1)$
if x is zero the y is zero then x+y = 0 which is not possible as x+y is in denominator of the resultnat expression
Let $y=x\omega$
putting in (1) we get
$\omega ^2 + \omega + 1=0\cdots(2)$
Hence $\omega^3=1\cdots(2)$
Now $x+y = x + x\omega = x(1+\omega) = x(-\omega^2)= - x\omega^2$ using(1)
hence $\frac{x}{x+y} = \frac{x}{- x\omega^2} = - \omega\cdots(3)$
also $\frac{y}{x+y} = \frac{x\omega}{- x\omega^2} = - \frac{1}{\omega} = \omega^2\cdots(4)$
hence $(\frac{x}{x+y})^{2023} + (\frac{y}{x+y})^{2023}) = \omega^{2023} + (\omega^2)^{2023}$
= $- \omega^{2023} - \omega^{4036}$
= $- \omega^{3 * 677 + 1} - \omega^{3 * 1354 + 2 }$
= $- \omega - \omega^2 $ using (3)
= 1 using (2)
ew have 5n-13- (3n -10) = 2n -3 which is odd. So all 3 cannot be odd so one of them has to be 2 for all to prime.
and n has to be greater than 3 for all numbers to be positive
3n-10 = 2 gives n =4 and numebr are 2, 11, 7 all are prime
$6n - 13 \ge 11$ and $5n - 13 'ge 7$ as $n > 3$ so other numbers cannot be 2
so only solution n = 4
For the above to hold we must have $x= na^2$ and $x+60= nb^2$ where n,a,b are integers
So we get $n(b^2 - a^2) = 60$
or$n(b+a)(b-a) = 60$
now for a anb b to be integer we must have b+a and b-a to be both even or odd
we need to find 60 as product of 3 numbers with b+ a and b- a to be different
this gives us following cases
$n=1, b + a= 30, b- a = 2$ giving $n=1,b= 16, a = 14$ giving $x= 196, m = 900$
$n=3, b + a= 10, b- a = 2$ giving $n=3,b= 6, a = 4$ giving $x= 48, m =300$
$n=4, b + a= 5, b- a = 3$ giving $n=4,b= 4, a = 1$ giving $x= 4, m = 100$
$n=4, b + a= 15, b- a = 1$ giving $n=4,b= 8, a = 7$ giving $x= 196, m = 900$ (solution repeats)
$n=5, b + a= 6, b- a = 2$ giving $n=5,b= 4, a = 2$ giving $x= 20, m = 180$
n= 15 gives a+ b and b-a to be same so no further solution
So solution set $(196,900),(48,300), (4,100),(20,180)$
We shall prove it by taking the GCD
$GCD((n+1)!+1, n!+1)$
$= GCD((n+1)!+1- (n!+1), n!+1)$ using GCD(a,b) = GCD(a-mb,b) for any integer m
$= GCD((n+1)!- n!, n!+1)$
$= GCD((n!(n+1-1), n!+1)$
$= GCD(n!.n, n!+1)$
$= GCD(n!, n!+1)$ we can devide 1st term by n and GCD shall not change and 2nd term is not divisible by n
$= GCD( n!, n!+1-n)$ using GCD(a,b) = GCD(a-mb,b) for any integer m
$=GCD(n!,1)= 1$
So these are relativvly primes
We have $a+b+c = 0$
Hence $ a+ b = -c $
Squaring both sides $a^2+2ab + b^2 = c^2$
Adding $a^2+b^2$ on both sides we get
$2(a^2+ab+b^2) = a^2 + b^2 + c^2$
Or $a^2+ab + b^2 = \frac{1}{2}(a^2+b^2+c^2)$
So $\frac{ab}{a^2+ab + b^2} = \frac{2ab}{a^2+b^2+c^2}\cdots(1)$
Similarly we have $\frac{bc}{b^2+bc + c^2} = \frac{2bc}{a^2+b^2+c^2}\cdots(2)$
And $\frac{ca}{c^2+ca + a^2} = \frac{2ca}{a^2+b^2+c^2}\cdots(3)$
Adding (1) (2) and (3) we get
$\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}= \frac{2ab+2bc+2ca}{a^2+b^2+c^2}\cdots(4)$
Now staring with $a+b+c=0$ squaring both sides we get
$a^2+b^2+c^2 + 2ab + 2bc+2ca= 0$
Or $a^2+b^2 + c^2 = - (2ab+2bc+2ca)$
Or $\frac{2ab+2bc+2ca}{a^2 +b^2+ c^2} = -1\cdots(5)$
Form (4) and (5) we get $\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}= - 1$
First we note that $gcd(x,x+1) = 1$
$gcd(x, x^2 + x + 1) = GCD(x, x(x+1)) = 1$
and $gcd(x+1, x^2 + x + 1) = GCD(x+1, x(x+1)) = 1$
So $x,x+1,x^2+x+1$ are pairwise co-prime and to prove that the given expression is divisible by $x(x+1)(x^2+x+1)$ we need to show that it is divisible by $x$, $x+1$ and $x^2+x+1$
Now Let $P(x) = (x+1)^7−x^7−1$
To show that it is dvisible by x we need to show P(0) = 0
$P(0) = 1^7 - 0^7 -1 = 0$ so it is divisibe by x
To show that it is dvisible by x+1 we need to show P(-1) = 0
$P(-1) = (-1+1)^7 - (-1)7 -1 = 0$ so it is divisibe by x+1
Now we need to show that it is divisible by $x^2+x+1$
That is if $x^2+x+1=0$ then $P(x) = 0$
Because
$x^2+x+1=0\cdots(1)$ we get
$x+1 = - x^2\cdots(2)$
And multiplying by $x-1$ we have $x^3-1 = 0$
Or $x^3 = 1\cdots(3)$
Now $P(x) = (x+1)^7−x^7−1$
$= (-x^2)^7 - x^7 -1$ using (1)
$=x^{14} - x^7 -1$
$= -(x^3)^4 * x^2 - (x^3)^2 * x - 1$ using (2)
$= -x^2 - x - 1$
$= - (x^2+x+1)=0$ using (1)
So P(x) =0 if $x^2+x+1=0$
Hence $x^2+x+1$ is a factor
As $x,x+1,x^2+x+1$ are factors and are pairwise co-prime so $(x+1)^7−x^7−1$ is divisible by $x(x+1)(x^2+x+1)$
We have $n^4 -1 = (n^2+1)(n^2–1) = (n^2–4 + 5)(n^2–1) = (n^2–4)(n^2–1) + 5(n^2–1)$
$= (n+2)(n-2)(n+1)(n-1) + 5(n^2–1)$
now the 2nd term is multiple of 5 and 1st term is product of 4 numbers which along with are 5 consecutive numbers. so one of them has to be divisible by 5. but as n is not divisible by 5 so one of the 4 other numbers is a multiple of 5 and hence the product. as given number is sum of 2 numbers each multiple of 5 and hence the given expression.
We have $\frac{n^7}{7} + \frac{n^{13}}{13} + \frac{71n}{91}$
$=\frac{n^7-n +n }{7} + \frac{n^{13}-n + n}{13} + \frac{71n}{91}$
$=\frac{n^7-n} {7} + \frac{n^{13}-n}{13} + \frac{n} {7}+ \frac{n}{13} + \frac{71n}{91}$
$=\frac{n(n^6-1)}{7} + \frac{n(n^{12}-n) }{13} + n$
If n is divisble by 7 1st term is integer and if n is not divisibl by 7 $n^6-1$ is divsible by 7 as per Fermats Little Theorem is 1st term is integer similarly the 2nd term and 3rd term is integer for any integer n and hence the given expression
Let us have a look at the $t^{th}$ term.
Let us look at difference for a couple of terms
$t_2 = t_1 = 5- 1 = 4$
$t_3-t_2 = 10 -5 = 5$
$t_4-t_3 = 16 -10 = 6$
From the above we see that $t_k = t_{k-1} + k + 2$
From this let us find $t_n$ in terms of n
We have $t_n = 1 + \sum_{k=2}^{n} (k + 2)$
$= 1 + \frac{n(n+1)}{2} - 1 + 2(n-1)$
$= \frac{n^2}{2} + \frac{5n }{2} - 2$
So sum $S_n = \sum_{k=1}^n (\frac{k^2}{2} + \frac{5k }{2} - 2)$
$= \sum_{k=1}^n (\frac{k^2}{2}) + \sum_{k=1}^n\frac{5k }{2} - \sum_{k=1}^n2$
$=\frac{1}{2}(\frac{n(n+1)(2n+1)}{6} + \frac{5}{2} \frac{n(n+1)}{2} - 2n $
$= \frac{2n^3 + 3n^2 +n }{12} + \frac{5n^2 +5n}{4} - 2n$
$= \frac{2n^3 + 3n^2 + n + 15n^2 + 15n -24n}{12}$
$= \frac{2n^3 + 18n^2 -8n}{12}$
$= \frac{n^3 + 9n^2 -4n}{6}$
We have
$\sqrt{(x^2 + 4x + 13)}$ Completing the square we get
$= \sqrt{x^2+ 4x+ 4 + 9} = \sqrt{(x+2)^2 + 3^2}$
The above in a plane is the distance from $(-2,-3)$ to $(x,0)$
$\sqrt{(x^2 - 8x + 41)}$ Completing the square we get
$= \sqrt{x^2 - 8x+ 16 + 25} = \sqrt{(x-4)^2 + 5^2}$
The above in a plane is the distance from $(4,-5)$ to $(x,0)$ also from (4,5) to $(x,0)$ and because of symmetry as the line has to be to $(x,0)$ we can choose $(4,-5)$
The sum of the 2 expression is the distance from $(-2,3)$ to $(x,0)$ and from $(x,0)$ to $(4,5)$ or $(4,-5)$ as the 2 distances are same. it is better to chose a point on the other side of x axis so (4,5).
The distance ins minimum if $(-2,3),(x,0)$ and $(4,5)$ are in same line and so minimum distance is distance from $(-2,3)$ to $(4,5)$ $= \sqrt{(4+2)^2 + (5+3)^2} = 10$
Because it is a degree 6 polynomial so there are six roots let them be $x_1,x_2,x_3,x_4,x_5,x_6$
Because coefficient of $x^5$ is zero so sum of roots is zero
So $\sum_{i=1}^{6}x_i = 0\cdots(1)$
Because coefficient of $x^45$ is zero so double sum of roots is zero
So $\sum_{i=1}^{6}\sum_{j=1. j\ne i}^{6}x_i x_j = 0\cdots(2)$
Hence from (1) and (2) we have $\sum_{i=1}^{6}x_i^2 = 0$
If roots are real all $x_i$ are zero which is impossible if a,b,c,d are non zero
Hence there is a contradiction and all roots cannot be real
An odd number is of the form 2n+ 1, so let 3 odd numbers be 2a+1, 2b+1,2c+ 1
Now $(2a+1)^2 = 4a^2 + 4a + 1 = 4(a^2+ a) + 1$
$(2b+1)^2 = 4b^2 + 4b + 1 = 4(b^2+ b) + 1$
$(2c+1)^2 = 4c^2 + 4c + 1 = 4(c^2+ c) + 1$
We observe that square of an odd is of the from 4n + 1
Now $(2a+1)^2 + (2b+1)^2 + (2c + 1)^4 = 4(a^2 + a + b^2 + b + c^2 + c) + 3$
Above is an odd number and of the form 4n + 3 so cannot be a perfect square
So the answer is No
$log(10+3\sqrt{10}) + log(10+\sqrt{90+\sqrt{90}})+ log(10- \sqrt{90+\sqrt{90}})$
Solution:
Add the 2nd and 3rd expression and knowing $log\, a + log\, b = log\, ab$
$log(10+\sqrt{90+\sqrt{90}})+ log(10- \sqrt{90+\sqrt{90}})$
$= log((10+\sqrt{90+\sqrt{90}})(10-\sqrt{90+\sqrt{90}}))$
$= log(10^2 - (\sqrt{90 +\sqrt{90}})^2$ using $(a+b)(a-b) = a^ 2- b^2$
$=log( 100 - (90 + \sqrt{90}))$
$= log(100 - 90 - \sqrt{90})$
$= log (10 - \sqrt{3^2 * 10})$ factoring 90 to product of a square and no square
$= log( 10 - 3 \sqrt{10})$
So given expression = $log(10+3\sqrt{10}) + log( 10 - 3 \sqrt{10})$
$= log ( (10+3\sqrt{10})(10- 3\sqrt{10}))$ knowing $log\, a + log\, b = log\, ab$
$=log (10^2 - (3\sqrt{10})$
$= log (100 - 9 * 10)$
$=log\, 10 = 1$
We shall prove the same by principle of mathematical induction.
Let $P(n) = 6^{n+2}+7^{2n+1}$
We shall show that base step and induction step are true
Base step
Here we show that P(1) is divisible by 43
$P(1) = 6^ 3 + 7^3 = 216 + 343 = 559 = 43 * 13$ is divisible by 43
So base step is true
Now for induction step
Induction step
Let it to true k that is P(k) is divisible of 43
We shall show that P(k+1) is divisible by 43
P(k) is divisible by 43 so there exists an integer b such that
$P(k) = 6^{k+2}+7^{2k+1}= 43b$
We have $P(k+1) = 6^{(k+1)+2}+7^{2(k+1)+1}$
$= 6(6^{k+1}) + 49 * 7^{2k+1} $
$= 6(6^{k+1}) + (6+43) * 7^{2k+1} $
$= 6(6^{k+1}) + 6 * 7^{2k+1} + 43 * 7^{2k+1}$
$= 6(6^{k+1} + 7^{2k+1}) + 43 * 7^{2k+1}$
$= 6 * 43 b + 43 * 7^{2k+1}$
$= 43( 6b + 7^{2k+1})$
This is multiple of 43
As we have proved both the base step and induction step hence this is true hence proved
Let the 2 numbers be ma and mb where GCD(a,b) = 1 and hence m is GCD of the 2 numbers and a > b
Sum = m(a+b) = 253 and LCM = mab = 644
As GCD(a,b) = 1 so gcd(a+b, ab ) =1 so m is gcd of 253 and 644
253 = 23 * 11
644 = 23 * 28
So the GCD = 23
The numbers are 23a and 23b
$a+ b = 11\cdots(1)$
$ab = 28\cdots(2)$
So $(a-b)^2 = (a+b)^2 - 4ab = 11^2 - 4 * 28 = 9$
Or $a-b= 3\cdots(3$
Adding (1) and (3) we get $2a =14$ or $a= 7$ and putting in 1) we get b = 3
Hence 2 numbers are 23 * 4 = 92 and 23 * 7 = 161
We have
$x^3+x = y^3 + y$
$=>x^3-y^3 + x - y= 0$
$=>(x-y)(x^2+xy+ y^2) + (x - y)= 0$
$=>(x-y)(x^2+xy+ y^2+1) = 0$
$=>$ $x-y = 0$ $=>x = y$
or $x^2+xy+y^2 + 1 = 0$
but $x^2+xy+y^2 +1 = (x-\frac{y}{2})^2 + \frac{3}{4}y^2 + 1 > 1$ which does not have real solution
so solution x = y
Let us consider $2^{1987}+3^{1987}+\cdots+(n)^{1987} $
We have $k^{1987} + (n+2-k) ^{1987 }$ is divisible by n+ 2
If n is odd taking k from 2 to $\frac{n-1}{2}$ we have |frac{n-1}{2}$ pairs and eac pair is divisible by n+2 and adding 1 does not divide by 1
If n is even taking k from 2 to $\frac{n}{2}$ we have $\frac{n}{2}$ pairs and each pair is divisible by n+2 and middle number is $(\frac{n+2}{2})^{1987} $
If n+ 2 is a multiple of 4 this is even and adding 1 makes it odd and hence the sum is not divisible by n+ 2
If n+2 is of the form 4m+ 2 and in this case divisible by 2m+ 1so adding 1 does not make it divisible by 2m+ 1 so not divisible by n+ 2 proved
We shall prove it by contradiction.
Let the smallest solution of the same be $(a,b,c)$ such that $a^2 + b^2 = 3c^2\cdots(1)$
Working in mod 3 we have $x^2 \equiv 0 \pmod 3\cdots(2) $ or $x^2 \equiv 1 \pmod 3\cdots(3)$
in (1) RHS is divisible by 3 so LHS is also divisible by 3
For this to be true using (2) and (3) we must have
$a^2 \equiv 0 \pmod 3$
and $b^2 \equiv 0 \pmod 3$
So a and b are multiples of 3 that is $a=3p$ and $b= 3q$ for some p and q
Putting in (1)
$3p^2 + 3q^2 = c^2$ or $c^2$ is multiple of 3 or c is multiple of 3
So hence $\frac{a}{3}, \frac{p}{3}, \frac{c}{3}$ is a smaller solution
Which is a contradiction
So equation does not have integer solution
This method is known as proof by infinite descent
Without loss of generality assume $ a \ge b \ge c$ so $a+b+c \le 3a$
We have $abc \le 3a$
Or $bc \le 3$
That is c = 1 and b =2 giving a = 3
Or c = 1 and b = 3 giving a = 2 but this violates the condition
So solution is $(3,2,1)$ or any permutation because of symmetry
We have
$2x = 3y-5$
Cubing both sides as in the required condition we have $8x^3$ and $27y^3$
We get $8x^3 = (3y-5)^3 = (3y)^3) - 5^3 -3(3y)5(3y-5)$ (using $(a -b)^3 = a^3- b^3 -3ab(a-b)$
or $8x^3 = (3y-5)^3 = (3y)^3 - 5^3 -3(3y)*5 * 2x$ as $3y-5 = 2x$
or $8x^3 = 27y^3- 125 - 90xy$
or $8x^3-27y^3 + 90xy +125 = 0$
We have $\lfloor x \rfloor \le x$ and equal ony of x is integer
so $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor + \lfloor \frac{n}{6} \rfloor \le \frac{n}{2} + \frac{n}{3}+ \frac{n}{6} = n$
so $\lfloor \frac{n}{2} \rfloor = \frac{n}{2}$
$\lfloor \frac{n}{3} \rfloor = \frac{n}{3}$
$\lfloor \frac{n}{6} \rfloor = \frac{n}{6}$
the above is so because if any expression above is not true LHS is less that RHS in any of them shall make the sum < n
so $\frac{n}{2}, \frac{n}{3} ,\frac{n}{6}$ all are ntegers of n is a multiple of 2,3 and 6 that is multiple of LCM(2,3,6) that s 6. so n is of the form 6k
We have $(n-1)! = n^2-1 = (n+1)(n-1)$
So n-1 =0 which gives LHS = 2 RHS = 1 which is contradiction
Or $(n-2)! = n + 1$
Put n- 2 = k giving $k! =k + 3$
As LHS is divisible by so is RHS so k is a factor of 3 k = 1 or 3
k =1 gives n = 3 which is not a solution as it does not satisfy the criteria
k =3 gives n= 5 and as $4! + 1 = 25 = 5^2$ so n = 5 is a solution
n = 0 is not a solution so $n > 0$.
deviding both sides by n we get
$n + 19 = (n-1)!$
putting n = x+ 1 we get $x+20 = x!$
for this to be valud x is a factor of 20
so we check with 1 LHS = 21 and RHS = 1
x = 2 LHS = 22 RHS = 2 no
x = 4 LHS = 24 LHS = 24 so x = 4 is a solution
x = 5 LHS = 25 RHS = 125 not a solution
if we take x larger RHS grows larger as compared to LHS so no solution
so solution x = 4 or n = 5.
If the umber is of the form $\prod_{n=1}^{k}p_n^{q_n}$ then the number of factors = $\prod_{n=1}^{k}(q_n+1)$
Because 5 is prime the number must be of the form $p^4$ where p is prime
We have the number of digits =2 and we should find p such that $ 9 \lt p^4 \lt 100$
The only number that satsfies the condition is n = 2 and n is a prime and $2^4= 16$
So 16 is the only 2 digit number having 5 factors
HCF is 9 so the 2 numbers are 9x, 9y where HCF(x,y) = 1 and x and y >0 and without loss of generality let x > y
product of the 2 numbers are 9xy = 270
or $xy = 30\cdots(1)$
sum of the 2 numbers are 9(x+y) = 99 or $x + y = 11\cdots(2)
we have $(x-y)^2 = (x+y)^2 - 4 * xy = 11^2 - 120 = 1$
so $x - y = 1\cdots(3)$
so we get x = 6 and y = 5 and numbers are $54, 45$
we have $4^x = 2^{2x}$
now working in mod 3 we get $1-5^y \equiv 0 \pmod 3$
or $5^y \equiv 1 \pmod 3$
as we know $5 \equiv -1 \pmod 3$ so y has to be even say 2m
now $4^x - 5^y = 2^{2x} - 5^{2m} = 39$
Farctoring we get $(2^x + 5^m)(2^x-5^m) = 39 = 39 *1 = 13 * 3$ (39 can be factored in 2 ways)
so we have 2 cases
$2^x+5^m= 39$ and $2^x - 5^m=1$ adding we get $2^x *2 = 40$ and this does not have integer
or
$2^x+5^m= 13$ and $2^x - 5^m=3$ adding we get $2^x *2 = 16$ or x = 3 and subtracting $2 * 5^m = 10$ and m = 1
so x = 3 and y = 2
Let the 2023 consecutive numbers be from n-1011 to n + 1011
For all to be natual numbers $n>=1012$
The sum of them = 2023n
$2023 =17 ^2 * 7$
So if we choose n to be of the form $7m^2$ then the sum becomes a perect square
Not $n >= 1012$ or $7m^2 >=1012$ of $m > 13$
So the 2023 number starting from $7m^2-1011$ where $m > =13$ satisfy the criteria
Let persons enter A instances and leave B instances
A+B = $3^{1999}\cdots(1)$
A-B = $3^{1000} + 2\cdots(2)$
Subtracting 2nd from the 1st we get
$3B = 3^{1999} - 3^{1000} - 2$
LHS is multiple of 3 but RHS is not so it is not possible
Dividing a polynomial f(x) by $(x-1)^2$ the remainder is $g(x) = x+1$-
so dividing by (x-1) the remainder is $g(1) = 1 + 1 = 2$
Dividing by $x^2(x-1) $ is $ax^2+bx+c$
so deviding $ax^2+bx+c$ by $x-1$ remainder must be 2
so $f(1) = a + b+ c = 2$
