We are given
$ab+bc+ca = abc$
Hence $bc(a-1) = abc - bc = ab + ac = a(b+c)$
Hence
$\frac{(b+c)}{bc(a-1)} = \frac{(b+c)}{a(b+c)} = \frac{1}{a}$
or
$\frac{(b+c)}{bc(a-1)} = \frac{1}{a}\cdots(1)$
similarly
$\frac{c+a)}{ca(b-1)} = \frac{1}{b}\cdots(2)$
and
$\frac{(a+b)}{ab(c-1)} = \frac{1}{c}\cdots(3)$
Adding (1),(2),(3) we get
$\frac{(b+c)}{bc(a-1)} + \frac{(c+a)}{ca(b-1)} + \frac{(a+b)}{ab(c-1)}$
$=\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
$=\frac{bc+ca+ab}{abc} = \frac{abc}{abc}=1$
We are given $x^3+3367=2^n$
We know $3367=7 * 13 &*37
So let is work $x^3=2^n \pmod 7$
Working in mod 7 we have $x^3 \in \{1,-1,0\}$ and $2^n \in \{1,2,4\}$ so we get 1 as common and for that n has to be multiple of 3.
So we get
$x^3= 2^{3k} \pmod 7$
Going back to the original equation we get
$x^3 + 3367 = 2^{3k}$
Or $(2^k)^3 - x^3 = 3367$
Or $y^3 -x^3 = 3367$ where $y = 2^k$
As $(y-x) | y^3-x^3$ so $y-x | 3367\cdots(1)$
Further
As we know $(y-x)^3 = y^3-3y^2x + 3yx^2 - x^3 = (y^3-x^3) - 3yx(y-x) \lt y^3-x^3$ when $y-x \gt 0$
So $y-x\lt 15\cdots(2)$
So $y-x \in \{1,7\}\cdots(2)$ from (1) and (2)
$y-x =1$
gives $x^3 + 3367 = (x+1)^3 = x^3 + 3x^2 + 3x + 1$
Or $3367 = 3x^2+ 3x + 1$
Or $3x^2+3x = 3366$
Or $x^2 + x = 1122$
x = 33 and y = 34 and y is not power of 2 so not a solution
$y-x=7$
gives
$x^3 + 3367 = x^3 + 21x^2 + 147 x + 343$
or $3024 = 21x^2+ 147x
or $x^2+ 7x = 144 $
or $x = 9$ or $x=-16$
only positive value admissible giving x = 9 and y = 16
$2^n = y^3$ giving $2^n = 2^{12}$ or n = 12
Hence x=3 , n = 12
We have
$1+2^x+ 2^{2x+1} = y^2$
Or $2^x+ 2^{2x+1} = y^2-1$
Let us check one case x = 0 for which $2^x$ is odd that is 1 and we get $1 + 1+ 2 = 4 = y^2$ or
(0, 2), (0, -2) are 2 solutions.
$2^x(1+2^{x+1}) = y^2-1 = (y+1)(y-1)\cdots(1)$ in this x is positive
As LHS is even number so RHS is even.
Now y+1 and y-1 both must be even and one has to be an odd multiple of power of 2 and other one odd multiple of 2.
So $y = 2^{x-1}p\pm 1$ where p is odd.
Or $y = 2^{x-1}p +q$ where p is odd and q is $\pm 1$
Putting in the equation (1) we get
$2^x(1+2^{x+1}) = y^2-1 = (2^{x-1}p +q)^2-1 = 2^{2x-2}p^2 + 2^x pq$
Or $1+2^{x+1} = 2^{x-2} p^2 + pq$
Or $1-pq = 2^{x-2}(p^2-8)$
q =1 gives $1-p = 2^{x-1}(p^2-8)=>(p^2-8)<0$ or $p=1$ which does not satisfy the condtion.
If q = -1 then we have
$1+p = 2^{x-2}(p^2-8)$ giving $1 + p > p^2 -8$
Or $p^2-p -7 <=1$ this has $p <= 3$ as p = 3 giving x = 4 and $y = \pm 23$
So 4 solutions are $(0,2),(0,-2),(4,23), (4,-23)$
We have
$a+b+c = 1$
So $1+a + b+c = 2$
Or $1+a = 2 - b-c = (1-b)+(1-c)$
because (1-b) and (1-c) are positive using AM GM inequality we have
$(1-b)(a-c) >= 2\sqrt{(1-b)(1-c)}$
or $(1+a) >= 2\sqrt{(1-b)(1-c)}\cdots(1)$
similarly
$(1+b) >= 2\sqrt{(1-c)(1-a)}\cdots(2)$
and
$(1+c) >= 2\sqrt{(1-a)(1-b)}\cdots(3)$
Multiply above 3 equations we get
$(1+a)(1+b)(1+c) >= 8(1-a)(1-b)(1-c)$
Let there be finite number of prime numbers of the form 4n+ 3 the number of numbers by k with $p_k = 4 q_k +3$
Now consider the number $4p_1p_2\cdots p_k -1 $
The number above is of the form 4n+3.
If the number is prime then we have found a prime number above the largest prime and we are done.
If it is not a prime number it has got some prime factors.
All the prime factors cannot be of the 4n+1 because in that case product shall be of the from 4n+ 1
So it has got a prime factor of the form 4n+3 above $p_k$. which is again a conradiction.
Hence there are infinite prime numbers of the form 4n+3
We have
${2p}\choose{p}$$=\frac{(2p)!}{p!p!}$ bu definition
$=\frac{p!\prod_{k=p+1}^{2p}k}{p!p!}$ bu expansion
$=\frac{\prod_{k=p+1}^{2p}k}{p!}$ cancelling p! from both numerator and denominator
$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!}$ by taking 2 p ouut
working in mod p we get $(p+n) \equiv n \pmod p$
So we get
$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!} \pmod p$
$=\frac{2p\prod_{k=1}^{p-1}k}{p!} \pmod p$
$=2\frac{p\prod_{k=1}^{p-1}k}{p!} \pmod p$
$=2\frac{p!}{p!} \pmod p$
$=2$
Proved
When we break a number into smaller parts the product become larger as number of parts becomes larger and each is smaller however no part can be 1 as it shall provide a smaller value. so we can break it into 1012 parts and each is 2 giving a product $2^{1012}$.
However there is exception to ir as $2^3 \lt 3^2$ so we can group terms of 3 getting 674 3 and left with 1 2 giving a product $3^{674} * 2$
Proof:
A cube is either even or odd
If it is odd we can write it as $2n + 1$ which can be written as $(n+1)^2 - n^2$
If it is even the the number is even say $2n$ so cube is $8k$ where $n= k^3$ can be written as $(2k+1)^2 -(2k-1)^2$
We are given $\frac{x}{y+7} + \frac{y}{x+7}= 1$
or $x(x+7) + y(y+7) = (x+7)(y+7)$
or $x^2+7x + y^2 +7y = xy + 7x + 7y + 49$
or $x^2 + y^2 -xy = 49$
now $x^2- xy$ we need to form in form of squares by adding some teerm
let us multiply by 4 to above to get
$4x^2 + - 4xy + 4y^2 = 196$
or $4(x^2-4xy + y^2) +3y ^2 = 196$
or $(2x-y)^2 + 3y^2 = 196$
as it is sum of squares we need to check a finite number of values(fron pair of values we consider only positive ones)
$y = 3$ gives $2x-y =13$ giving $x = 8, y = 3$
$y= 5$ gives $2x-y = 11$ giving$ x = 8, y = 5$
$y = 7$ gives $2x -y = 7$ giving $x = y = 7$
$y = 8$ gives $2x -y = 2$ giving $x = 5 y = 8$
or $2x - y = -2$ giving $x = 3 and y = 8$
so the solutions are $(x=8,y=3), (x=8,y=5), (x=7,y=7),(x=5,y=8), (x=3,y=8)$
https://www.quora.com/What-is-the-equation-of-a-circle-which-passes-through-three-points-0-0-a-0-and-0-b
Because one point $(0,0)$ is origin, one point $(a,0)$ lies on x-axis and another $(0,b)$ on y axis it is clear that is is a right angled triangle. so the circumcenter is the midpoint of the hypotenuse that is $(\frac{a}{2},\frac{b}{2})$
so the equation of the circle is
$(x-\frac{a}{2})^2 +(y-\frac{b}{2})^2=(0-\frac{a}{2})^2 + (0-\frac{b}{2})^2 $ the RHS is square of distance of origin to to the centre
simplifying we get
$x^2-ax+y^2-by=0$
We are given
$x^2 = 2^ y + 2023$
Now working mod 4 we have
$2^ y + 2023 \equiv 2^y + 3 \pmod 4$
y cannot be greater than 1 as $2^y + 3 \equiv 3 \pmod 4$
As a square cannot be $3 \pmod 4$ so only possible value is y = 1 giving x= 45 and $x+y=46$
We know for any integer x
$\lfloor x+y \rfloor = x + \lfloor y \rfloor$
So we have
$\lfloor n + k + \sqrt{n+k} \rfloor = n + k + \lfloor \sqrt{n+k} \rfloor$
From the given 2024 numbers by putting k = 0 t0 2023 we get
$\lfloor n + \sqrt{n} \rfloor = n + \lfloor \sqrt{n} \rfloor$
$\lfloor n + 1 + \sqrt{n+1} \rfloor = n + 1 + \lfloor \sqrt{n+1} \rfloor$
$\lfloor n + 2 + \sqrt{n+2} \rfloor = n + 2 + \lfloor \sqrt{n+2} \rfloor$
$\lfloor n + 2023 + \sqrt{n+2023} \rfloor = n + 2023 + \lfloor \sqrt{n+2023} \rfloor$
They are consecutive if we have
$ \lfloor \sqrt{n} \rfloor = \lfloor sqrt{n+1} \rfloor \cdots \lfloor \sqrt{n+ 2023} \rfloor$
as these are in increasing order we have
$ \lfloor \sqrt{n} \rfloor = \lfloor \sqrt{n+ 2023} \rfloor$
The 1st term should be as low as possible so n is a perfect square $x^2$
So we have $ x = \lfloor \sqrt{x^2+ 2023} \rfloor$
As we have the smallest $k=(x+1)^2$ such that $ x < \sqrt{k}$
so we have $x^2 + 2023 \lt x^2 + 2x + 1$ or $2024 \lt 2x $ so x = 1013 and
smallest $n = 1013^2= = 1026169$
We have factoring all 4
$60 = 2^2 * 3 *5 $
$150 = 2 * 3 * 5^2$
$30 = 2 * 3 * 5$
$900 = 2^2 * 3^2 * 5^2$
Because the GCD is 30 so the number has to be multiple of 30.
Now let us consider highest power of 2 coming from the numbers 900 has 2 and 60 has so 3rd number need not have (might have but not necessary)
Consider highest power of 3 coming from the numbers 900 has 3 and 60 and 150 have 2 so 3rd number must have so it has to be multiple of $2 * 3^2 * 5$ that is 90
Consider highest power of 5 coming from the numbers 900 has 2 and 150 has so 3rd number need not have (might have but not necessary)
So 3rd number has to be multiple of 90 and it has to de divisor of 900 say 90m
90m is a factor of 900 or is a factor of 10 that is 1 or 2 or 5 or 10
giving 3rd number one of 90,180,450,900
Above are roots of equation $9x^2 - 18x + 5 = 0$ and $9x^2 - 18x + 5 = 0$
the discriminant of of the equations are $18^2 - 4 * 9 * 5 = 324-180 = 144$
so roots are real and product of roots of $9x^2 - 18x + 5 = 0$ is $\frac{5}{9}$ and same is the product of other equation and so we get overall product $\frac{25}{81}$
Let the sides be a, a-d and a + d
we have $(a-d)^2 + a^2 = (a-d)^2$
or $a^2-2ad + d^2 + a^2 = a^2 + 2ad + d^2$
or $a^2 -4ad=0$
or $a(a-4d)=0$
so $a=4d$
area of the $\triangle$ is $\frac{a(a-d)}{2} = 24$
or $\frac{4d * 3d}{2} = 24$
or d= 2 and hence sides are 6,8, 12
we have $81^{\sin ^2x} + 81^{\cos^2x} = 30$
Or $81^{\sin ^2x} + 81^{1-\sin^2x} = 30$
or $81^{\sin ^2x} + \frac{81}{81^{\sin^2x}} = 30$
Let $81^{\sin ^2x}=y$
So we get $y + \frac{81}{y}=30$
or $y^2-30y + 81=0$
or $(y-3)(y-27)=0$
Or $y=3$ or $y=27$
$y=3$ mean $81^{\sin ^2x}=3$ or $sin^2x = \frac{1}{4}$ or $sin^2x = \pm \frac{1}{2}$ or $x= \pi \pm \frac{n\pi}{6}$
$y=27$ mean $81^{\sin ^2x}=27$ or $sin^2x = \frac{3}{4}$ or $sin^2x = \pm \frac{\sqrt{3}}{2}$ or $x= \pi \pm \frac{n\pi}{3}$
hence combining both we get $x= \frac{n\pi}{2}\pm \frac{\pi}{6}$
The number is of the form $2^x3^y5^zm^{30}$ because the it has 2 ,3 and 5 as factors and $m^{30}$ is a $30^{th}$ power because it is a square, cibe and fith power so multiplying by this shall satisfy the criteria.
$2^{x+1}$ has to be perfect square and $2^{x}$ is a perfect cube and a fifth power.
so $x+1 \equiv 0 \pmod 2\cdots(1)$
$x \equiv 0 \pmod 3\cdots(2)$
$x \equiv 0 \pmod 5\cdots(3)$
we can solve the same by chinese remainder theorem but we use a short method
From (2) and (3)
$x \equiv 0 \pmod {15} \cdots(4)$
from (1) and (4) we have checking multiple of 15 that is 0 and 15
$x \equiv 15 \pmod {30} \cdots(5)$
similarly we evaluate y
$3^{y+1}$ has to be perfect cube and $3^{x}$ is a perfect square and a fifth power.
so $y+1 \equiv 0 \pmod 3\cdots(6)$
$y \equiv 0 \pmod 2\cdots(7)$
$y \equiv 0 \pmod 5\cdots(8)$
From (7) and (8)
$y \equiv 0 \pmod {10} \cdots(9)$
from (6) and (9) we have checking mutiple of 10 that is 0,10,20
$y \equiv 20 \pmod {30} \cdots(10)$
similarly we evaluate z
$5^{z+1}$ has to be perfect fifth cube and $5^{x}$ is a perfect square and a perfect cube.
so $z+1 \equiv 0 \pmod 5\cdots(11)$
$z \equiv 0 \pmod 2\cdots(12)$
$z \equiv 0 \pmod 3\cdots(13)$
From (12) and (13)
$z \equiv 0 \pmod {6} \cdots(14)$
from (11) and (14) we have checking mutiple of 6 that is 0,6,12,18,24
$z \equiv 24 \pmod {30} \cdots(10)$
we can take principal vaues and get
$a=2^{15}3^{20}5^{24}m^{30}$
we have
$2n^3+3n^2 + n = n(2n^2+3n +1) = n(2n+1)(n+1)$
$= n((n-1) + (n+2))(n+1) = (n-1)n(n+1) + n(n+1)(n+2)$
$(n-1)n(n+1)$ is product of 3 consecutive integers and hence divisible by 6 and so is $n(n+1)(n+2)$ and hence the sum is divisible by 6
As $x^2 + y^ 2 = 1$ we can chose $x = \sin\, t , y = \cos\, t$
$3x + 4 y= 3 ]sin\, t + 4 \cos\, t$
To convert $3x + 4 y= 3 ]sin\, t + 4 \cos\, t$ to the form $A \sin (x+ t)$
$A \sin (x+t) = A \sin\,t \cos\, x + A \cos\, t \sin\, x$
We can choose $3 = 5 \cos\, x$ and $4 = 5\ sin\, x$ (as $3^2 + 4^2 = 25 = 5^2$)
$= 5 \cos\, x \sin\, t + 5 \cos\, t \sin\, x = 5 \sin (x-t)$
It is maximum when $\sin (x-t) = 1$ and maximum value = $5$
minimum when $\sin (x-t) = -1$ and minimum value = $-5$
We have $n^2+ 3n + 2= (n+1)(n+2)$
This is divisible by 2 for any n
For this to be divisible by 6 either n+1 or n+ 2 is divisible by 3 that is n should not be divisible by 3
The number of numbers that is divisible by 3 is $\lfloor \frac{2024}{3} \rfloor = 674$
So there are 664 numbers for which it is not divisible by 6 or there are 2024-674=-1350 numbers that are divisible by 6
We have
$\frac{2+3z-4z^2}{2 - 3z+ 4z^2} \in R $
So $\frac{2+3z-4z^2}{2 - 3z+ 4z^2} -1 \in R $
Or $\frac{6z}{2 - 3z+ 4z^2} \in R $
Or $\frac{2 - 3z+ 4z^2}{6z} \in R $
Or $\frac{2 - 3z+ 4z^2}{z} \in R $
Or $\frac{2}{z} -3 + 4z \in R$
Or $\frac{2}{z} + 4z \in R$
Let $z = x + iy$ and $y\ne 0$
So we have $\frac{2}{x+iy} + 4(x+iy) \in R$
Or $\frac{2(x-iy)}{x^2+y^2} + 4(x+iy) \in R$
Or $\frac{-2y}{x^2+y^2} + 4y = 0$ as imaginary part has to be zero
So as we have y is not zero dividing by y we get $x^2+y^2 = \frac{1}{2}$ or $|z|^2=\frac{1}{2}$
We have a prime number is 2 or 3 of of the form $6n\pm 1$ for $n \gt 0$ let us compute $n^2+2$
$2^2 + 2 = 6 = 2 * 3$ not a prime
$3^2+ 2 = 11$ is a prime
$(6n\pm 1)^2+ 2 = (36n^2 \pm 12 n + 1) + 2 = (36n^2 \pm 12 n + 3) = 3(12n^2 \pm 4 n + 1)$ which is not a prime
hence 11 is the only prime
We have $n^{th}$ triangular number $t_n= \sum_{k=1}^n k = \frac{n(n+1)}{2}$
So we must have $\frac{n(n+1)}{2} \equiv -1 \pmod {11}$
Or $n(n +1) \equiv -2 \pmod {11}$
Or $n^2 + n + 2 \equiv 0 \pmod {11}$
Or $4 n^2 + 4n + 8 \equiv 0 \pmod {11}$ (the purpose of doing this is to covert to perfect square as evident from next line)
Or $(2n+1)^2 + 7 \equiv 0 \pmod {11}$
Or $(2n+1)^2 = \equiv -7 \pmod {11}$
Or $(2n+1)^2 = \equiv 4 \pmod {11}$
Let us find the square mod 11 for n = 0 to 5 we get (0,0),(1,1),(2,4),(3,9),(4,5),(5,3)$
So the numbers are 2 and 9 ( that is 11 -2)
$2n + 1 \equiv 2 \pmod {11}$ gives n = 6 and $2n + 1 \equiv 9 \pmod {11}$ gives n = 4
So we have the triangular numbers are $t_{11k+4}$ and $t_{11k+6}$ for any non negative k
We have by definition triangular number $t_k = \frac{k(k+1)}{2}$
So $t_{3k}+t_{4k+1}$
$= \frac{(3k(3k+1)}{2} + \frac{(4k+1)(4k+2)}{2}$
$= \frac{(9k^2+3k}{2} + \frac{16k^2 + 12 k + 2}{2}$
$= \frac{25k^2+15k + 2}{2}$
$= \frac{(5k+1)(5k + 2)}{2}= t_{5k+1} $
Then
a) exactly one real root is -ve
b) all real roots are positive
c) exactly one real root is positive
d) no real root is positive.
Solution
we have
$x^{2021} + x^{2020} + \cdots+x - 1=0$
or $x^{2021} + x^{2020} + \cdots+x + 1=2$
Note that 1 is not a root of this equation.
Multiplying both sides by $x-1$ we get
$x^{2022} - 1 = 2(x-1)$
or $f(x) = x^{2022} - 2x + 1 = 0$
Note that as we have multiplied by x-1 so x = 1 so there is at least one positive root that is 1
As there is change of sign two times as per Descarte rule there are two or zero positive roots root but as it has at least one positive root so there are two positive roots so original equation has one positive root.
Now $f(-x) = x^{2022} + 2x + 1 = 0$
As there is no change of sign so there is no -ve root
So the equation has has one root and it is positive.
So answer is (c)-
We have the terms 8,15,24,35
Let us compute 1st order difference 7,9,11 which is an AP
Next difference is 13 and so next term is 35 + 13 = 48
So the nth term of the given sequence is quadratic say $an^2+bn + c$
Putting n =1 ,n= 2 and n =3 we get following
$a+b + c = 8\cdots(1)$
$4a+2b + c = 15\cdots(2)$
$9a+3b + c = 24\cdots(3)$
Subtracting (1) from (2) we get
$3a+b = 7\cdots(4)$
Subtracting (2) from (3) we get
$5a+b = 9\cdots(5)$
Subtracting (4) from (3) we get $2a=2$ or $a=1$
Putting $a=1$ in (4) we get $b=4$ and putting in (1) we get $c=3$
So $n^{th}$ term = $n^2+4n + 3$
we get the same results by putting n=1,2,3,4 so on
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
(just an observation and not required for the result) 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.
Let the number of numbers be n and it starts at a
So we have the sum $= an + frac{n(n-1)}{2} = 2024$
Or $2an + n(n-1) = 4048$ and $a\ge 1$
Or $n(2a+n-1)= 4048 = 16 * 253 = 2^4 * 23 *11$
Now out of n and 2a+n-1 one is even and one is odd so n = 16 ( 2a +n -1 = 253) or 1 ( 2a +n -1 = 4048) or 11 ( 2a +n -1 = 368) or 23 ( 2a +n -1 = 176)
Clearly n = 23 is the largest giving 2a + 22 = 176 and a = 77
So largest number of consecutive positive integers is 23 and it starts with 77
LCM of 48 and 60 (that is 12 * 4 and 12 *5) = 240
Now LCM 1680 = 240 * 7
So
3rd number should have a factor 7 and it should be multiple of 12(else
GCD cannot be be 12) and can have other that is 3rd factor that is a
factor of 240/12 or 20 it could be 1,2,4,5,10,20. as GCD of 48 and 60 is 12 so taking a higher multiple shall not change GCD
So 3rd number could be 84/168/324/420/840/1680 that is any of these 6 numbers
We shall use 2 facts
$\cos 3x = 4 \cos^3 x - 3 \cos\, x\cdots(1)$
And
if $a+b+c=0$ then $a^3+b^3+c^3 = 3abc\cdots(2)$
We are given
$\cos\, A +\ cos\, B + \cos\, C = 0\cdots(3)$
Hence $\cos^3 A +\ cos^3 B + \cos^3 C = 3 \cos\, A \cos\, B \cos\, C\cdots(4)$
Now $\cos 3A + \cos 3B +\cos 3C$
$ = 4 \cos^3 A - 3 \cos\, A$ + $ 4 \cos^3 B - 3 \cos\, B $ + $ 4 \cos^3 C - 3 \cos\, C$ using(1)
$=4(\cos^3 A +\ cos^3 B + \cos^3 C - 3 (\cos\, A + \cos\, B + \cos\, C)$
$ =4 * 3 \cos\, A \cos\, B \cos\, C- 3 *0 $ using (3) and (4)
$ =12 \cos\, A \cos\, B \cos\, C$
Proved
We have a,b,c are in AP so
$2b = a+ c$
Hence $a+2b -c = a + a+c -c = 2a\cdots(1)$
Let the $26^{th}$ digit be k
so $N = 1111\cdots( 50\, times) + (k-1) * 10^{24}$
or $N = \frac{10^{50}-1}{9} + (k-1) * 10^{24}$
or $9N = 10^{50}-1 + 9 * ((k-1) * 10^{24})$
because $GCD(9,13)=1$ so if n is divisible by 13 9N is divisible by 13.
as 13 ia prime we have
$10^{12} \equiv 1 \pmod {13}$
so
$10{24} \equiv 1 \pmod {13}\cdots(1)$
and
$10{48} \equiv 1 \pmod {13}$
hence
$10{50} \equiv 100 \pmod {13}$
or
$10{50} \equiv 9 \pmod {13}$
$9N = 10^{50}-1 + 9 * ((k-1) * 10^{24}) \pmod {13}$
$= 9-1 + 9 * ((k-1) * 1) \pmod {13}$
$= 9k -1 \pmod {13}$
this is zero and because k is a digit trying from 0 to 9 we get k=3 satisfies the condition
so the $26^{th}$ digit is 3
LHS
If n = 2 we have $10^2+3 = 103$ this is divisible by 103 .
Hence $10^2 + 3 = 0 \pmod {103}\cdots(1)$
We know that 103 is a prime but we shall use this as a starting point to show that there are infinite numbers of the form $10^n+3$ is divisible by 103.
As 103 is a prime so using Fermats little theorem
$10^{102} = - 1 \pmod {103}$
Squaring both sides we get
$10^{204} = 1 \pmod {103}$
So from (1) and above we have
$10^{204n + 2} + 3= 0 \pmod {103}$
except for n = 0 when the value is 103 all other values are above 103 and have a factor 103 so not a prime
So there are infinitely many composite numbers as they are divisible by 103. We have not taken all cases and have found infinitely many composite numbers. Hence proved
We shall use Euler's theorem that is a generalization of Fermat's little theorem: For any n and any integer a coprime to n, one has
$a^{\phi(n)} \equiv -1 \pmod n$
Where $\phi(n)$ is number of numbers that is co-prime to n,
Let us compute $\phi(49)$
If $n = p_1^{k_1}p_2^{k_2}p_3^{k-3} ... $ when each $p_i$ is prime then
$\phi(n)=n(1-\frac{1}{p_1})(1-\frac{1}{p_2})(1-\frac{1}{p_3})$
As $49= 7^2$ so $\phi(49) = 49 * (1-\frac{1}{7}) = 49 * \frac{6}{7} = 42$
Hence
$6^{\phi(49)} \equiv -1 \pmod {49}n$
or $6^{(42)} \equiv -1 \pmod {49}$
Squaring both sides
$6^{(84)} \equiv 1 \pmod {49}$
Dividing by 6 we get
$6^{(83)} \equiv \frac{1}{6} \pmod {49}\cdots(1)$
Where $\frac{1}{6}} is not fraction but inverse of 6 mod 49
Similarly
$8^{(83)} \equiv \frac{1}{8} \pmod {49}\cdots(2)$
Adding (1) and(2) we get
$6^{(83)}+ 8^{(83)} \equiv \frac{1}{6} + \frac{1}{8} \pmod {49}\cdots(2)$
Now $\frac{1}{6} + \frac{1}{8} \pmod {49}$
$= \frac{8 + 6}{48} \pmod{49}$
$= \frac{14}{48} \pmod{49}$
$= \frac{14}{-1} \pmod{49}$ as $48 \equiv -1 \pmod {49}$
$= -14 \pmod{49}$
$=35$ taking positive number
Hence remainder = 35
Probability that you choose one unfair coin = $\frac{1}{1000}$
iI you choose one unfair coin the probability that all 10 heads come = 1
So Probability that you choose one unfair coin and all heads come = $\frac{1}{1000}$
Probability that you choose one fair coin = $\frac{999}{1000}$
If you choose one fair coin the probability that all 10 heads come = $\frac{1}{2^{20}} = \frac{1}{1024}$
So Probability that you choose one fair coin and all heads come = $\frac{999}{1024*1000}$
So probability that all heads come = $\frac{999}{1024*1000} + \frac{1}{1000} = \frac{2023}{1024000}$
So probability that coin is unfiar = $\frac{\frac{1}{1000}}{\frac{2023}{1024000}}=\frac{1024}{2023}$
We have $\sin \, ax + \cos \, bx = \frac{1}{2} (\sin (a+b) x + \sin (a-b) x$
Knowing that
$\frac{d}{dx} \ cos \, nx = - n \sin \, nx$
Hence $\int (\sin \, ax \cos \, bx) dx = \int (\frac{1}{2} (\sin (a+b) x + \sin (a-b) x)dx $
= $\frac{\cos(a+b)x}{a+b} - \frac{\cos(a-b)x}{a-b} + C $
Proof;
We shall use the fact that if x and y are co-prime then xy is a perfect cube if both x and y are perfect cubes.
Let the 1st number be x so the product is $x(x+1)(x+2)(x+3)$
Either x is odd or even
If x is odd then we have $GCD(x,x+2) = GCD(x+1, x+2) = GCD(x+2,x+3) = 1$
For x + 2 to be a perfect cube minimum x is 6
So $GCD(x+2,x(+1)(x+2)) = 1$
So we need to prove that $x(x+1)(x+3)$ is not a perfect cube.
$x(x+1)(x+3) = x^3 + 4x^2 + 3$
We have $x(x+3) - (x-1)^2 = x^3+3x - x^2 -2x -1= x-1 > 0$ as x is minimum 6
So $x(x+1)(x+3) >= (x+1)^3$
$x^3 + 4x^2 + 3$ is below $(x+2)^3 = x^3 + 6x^2 + 12x + 8$
As it is between 2 cubes it cannot be a cube
Similarly we can prove taking $x +1$ when $x$ is even
1,2,4,6
We know that $ x+ y | x^3+y^3 $
Hence $2^n + n | (2^3)^n + n^3$
as $2^n + n | 8^n + n $
so $2^n + n | n^3-n $
so we must have $ n^3-n =0 $ or $2^n + n < n^3-n $
$n^3-n= 0$ gives n = -1,0, 1 and out of theses only 1 is solution
we need to solve $2^n + n < n^3-n $ or $2^n < n^3-2n $
let us find an upper bound for n putting a condition
$2^n < n^3$ for $n \lt 10$
putting n from 1 to 9 we see that $n \in \{1, 2,4,6\} $ satisfy the case and there is no other solution
We are given
$x^2-y=111\cdots(1)$
$y^2-x=111\cdots(2)$
From (1) and (2)
$x^2-y = y^2 -x$
Or $x^2-y^2 + x -y = 0$
Or $(x-y)(x+y) + (x-y) = 0$
Or $(x-y)(x+y+1) = 0$
As $ x\ne y$ dividing by x-y we get $x+y+1=0\cdots(3)$
Or $y = -(x+1)$
Putting the value in (1) we get $x^2 + (x+ 1) = 111$ or $x^2 + x - 110 = 0$
Or $(x-10)(x+11) = 0 $
Or $ x= 10$ or $x = -11$
Putting in (3) if $x= 10$ then $y = -11$
If $x= -11 $ then $y = 10$
So Solution set $x=10,y= -11$ or $x=-11,y=10$
To show that 123123 is a factor of $2^{60}-1$ we need to show that each prime factor of 123123 to the highest power is a factor
Now let us factor 123123
$123123= 123 * 1001 = 3 * 41 * 7 * 11 * 13$
As each prime occurs once we need to prove each of 3,7,11,13,41 divides $2^{60}-1$
We know $x^y-1 | x^{my}-1$ for any m
We have $2^2-1 | 2^{60}-1$ and $3| 2^{2} -1$ using FLT(Fermat's Little Theorem) so $3| 2{60}^-1$
$2^6-1 | 2^{60}-1$ and $7| 2^{6} -1$ using FLT(Fermat's Little Theorem) so $7| 2{60}^-1$
$2^{10}-1 | 2^{60}-1$ and $11| 2^{10} -1$ using FLT(Fermat's Little Theorem) so $11| 2{60}^-1$
$2^{12}-1 | 2^{60}-1$ and $13| 2^{12} -1$ using FLT(Fermat's Little Theorem) so $13| 2{60}^-1$
Now we need to prove for 41
using FLT we know $41| 2^{40} -1$
So let us find $GCD(2^{(60}-1,2^{40}-1)$
$GCD(2^{(60}-1,2^{40}-1)= GCD((2^{(60}-1) - (2^{40-1},2^{40}-1)$
$=GCD((2^{60} - 2^{40}), 2^{40}-1)$
$=GCD((2^{40}( 2^{20}-1), 2^{40}-1)$
$=GCD((2^{20}-1), 2^{40}-1)$ as second number is odd so removing even factors of 1st number
$= 2^{20}-1$ as this is a factor of $ 2^{40}-1$
now $= 2^{20}-1)= (2^{10}-1)(2^{10}+1)$
$= 1023 * 1025$ as $41 | 1025$ so $41 | 2^{60}-1 $
as each of 3,7,11,13,41 divides $2^{60}-1$ so 123123 is a factor
We have $a^3-a= a(a^2-1) = a(a-1)(a+1) = (a-1)a(a+1)$
As $(a-1)a(a+1)$ is product of 3 consecutive if is s divisible by 6 say 6m for some m
So $a^3 = a + 6m\cdots(1)$
Similarly
$b^3 = b + 6p\cdots(2)$
and
$c^3 = c + 6q\cdots(3)$
Adding (1),(2) and (3) we get $a^3+b^3+c^3 = (a+b+c) +6(m+p+q)$
So if $(a+b+c)$ is divisible by 6 then $a^3+b^3+c^3$ is divisible by 6
We have $6^2=36$ that is it ends with 6.
So we have all the powers of 6 end with 6
So $6^n+1$ shall end with 7
So we need to find n such that all the digits of $6^n+1$ must have all digits 7 and let it be k 7's/
So $6^n+1 = \frac{7}{9} (10^k -1 )$
Or $9(6^n+1) = 7 (10^k -1 )$for
Or $9 * 6^n + 16 = 7 * 10^k\cdots(1)$
We have $2 | 6$ so $2^2 | 6^2 $ so if $n \ge 5$ then we have
$9 * 6^n + 16 \equiv 16 \pmod {32} $ for $n \ge 5$
So $9 * 6^n + 16 \equiv m \pmod {32} $ where $m \le 16$ and not zero
For $k \ge 5$ $ 7 * 10^5 \equiv 0 \pmod {32} $
So $k \le 4$
So we need to check for n such that $9 * 6^n + 16 \le = 70000$
Or $6^n \le 7776$
Or $6^n+1 \le 7777$
We calculate for n = 1 $6^1 + 1 = 7$ meets criteria
n = 2 $6^2 + 1 = 37$ does not meet criteria
n = 3 $6^3 + 1 = 217$ does not meet criteria
n =4 $6^4 + 1 = 1297$ does not meet criteria
n = 5 $6^5 + 1 = 7777 $ meets criteria
Other value of n takes the value out of range
So $ n \in \{1,5\}$
We first factorize 437
$437 = 19 * 23$
Now let compute mod relative to19 and 23
As 19 is prime number as as per Wilson'sTheorem we have
$18 ! \equiv -1 \pmod {19}\cdots(1) $
As 23 is prime number as as per Wilson'sTheorem we have
$22 ! \equiv -1 \pmod {23} $
Now
$22 \equiv -1 \pmod {23}\cdots(2) $
$21 \equiv -2 \pmod {23} \cdots(3)$
$20 \equiv -3 \pmod {23}\cdots(4) $
$19 \equiv -4 \pmod {23}\cdots(5) $
As $22! = 22 * 21 * 20 * 19 * 18! $
So $22 ! \equiv -1 \pmod {23} $
$\implies 22 * 21 * 20 * 19 *18 ! \equiv -1 \pmod {23} $
$\implies (-1) *(-2) * (-3) * (-4) *18 ! \equiv -1 \pmod {23} $
$\implies 24 *18 ! \equiv -1 \pmod {23} $
$\implies 24 *18 ! \equiv -1 \pmod {23} $
$\implies 1 *18 ! \equiv -1 \pmod {23} $
$\implies 18 ! \equiv -1 \pmod {23}\cdots(6) $
Using (1) and (3) we get
$18 ! \equiv -1 \pmod {437} $
Proved
We have $8 = 2 * 4 = 2 * 2^2$
so $\sqrt{8} = 2 \sqrt{2}$
now $\sqrt{9} = 3$
so $\sqrt{\sqrt{9} - \sqrt{8}}$
= $\sqrt{3 - 2\sqrt{2}}$
this is of the form $\sqrt{n - 2\sqrt{n-1}}$ when n = 3
so $\sqrt{3 - 2\sqrt{2}} = \sqrt{2 + 1 - 2 * \sqrt{2} * 1}$
$= \sqrt{(\sqrt{2})^2 + 1^2 - 2 * \sqrt{2} * 1} $
$= \sqrt{2} - 1$
We have the constraint $(a-5)^2+(b-7)^2=4$ this is set of points lying in a circle with centre (5,7) and radius 2.
We need to find the minimum of square of the distance of (a,b) from (-7,-2) and this is minimum when (a,b) lies in the line from (5,7) to (-7,-2)
Distance from (5,7) to (-7,2) = $\sqrt{(5+7)^2 + (7+2)^2}= \sqrt{12^2 + 9^2} = 15$
So distance from (a,b) to (-7, -2) is $15-2 = 13$
Minimum Value of $(a+7)^2+(b+2)^2= 13^2 =169$
Note that we cannot have $x=k$ or $x = -k$
So multiplying both sides by $x^2-k^2$
We get
$((x-k) + k(x+k) + 2k) )(| x-k | -k) = 0$
Or $(x+k)(k+1))(| x-k | -k)=0$
As x cannot be -k $(| x-k | -k)=0$
So we get 2 values of x that is 2k or 0
Hence no solution
The power of digit of 3 shall have 1 or 3 or 7 or 9 as the unit digit.
Let us see some power of 3 that is 1,3,27,81(after this the sequence in the 1st digit repeats.
So we have (20n + x) where x is 1 or 3 or 7 or 9.
When we multiply by 3 we get 60n + 3 or 60n + 9 or 60n + 21 or 60n + 27 that is 60n + 3 or 60n + 9 or 20(3n+1) + 7 or 20(3n+1) + 9
So tens digit is even
Because $m > 1$ $2m + 1\ge 4$ So $2^{2m+1}$ is divisible by 16.
$2^{2m+1} $ is double of $(2^m)^2$ and is not a square.
Now we consider 2 cases
Let us consider when n is odd
Then $n= (2k+1)$
So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1)+1$
Ss k or k+1 is even we $n^2 \equiv 1 \pmod 8$
So $2^{2m+1}$ is divisible by 8 and minimum number above $n^2$ which is divisible by 8 is $n^2 + 7$
So we have proved for the case n is odd.
Let us consider when n is even
If $n^2$ is divisible by 8 then $n^2$ is divisible by 16 and as $2^{2m+1}$ is divisible by 16 and n^2 is divisible by 16
So $2^{2m+1}-n^2 = 16k$ for some positive k
So $2^{2m+1}-n^2 \ge16$
Hence $2^{2m+1}-n^2 \gt 7$ or $2^{2m+1} \gt n^2+ 7$
If $n^2$ is even and is not divisible by 8 then n is of the form 4k+2
$n^2= 16k^2 + 16k + 4$
Or $n^2 \equiv 4 \pmod {16}$
As $2^{2m+1} = 0 \pmod {16}$
So $2^{2m+1} \ge n^2 + 12 $
Hence $2^{2m+1} \gt n^2 + 7 $
we have proved for all 3 cases hence done
We have $640= 128 * 5 = 2^7 * 5$
To make it a power of 10 we need to multiply by $5^6$ giving $10^7$
so $a=5^6$ and b = 7 are the smallest values
Here let $frac{1}{10} = x$
We have for $|x| \lt 1$
$\sum_{n=0}^{\infty} x^n= \frac{1}{1-x}$
Differentiating both sides wrt x we get
$\sum_{n=0}^{\infty} nx^{n-1}= \frac{1}{(1-x)^2}$
Multiplying both sides by x we get
$\sum_{n=0}^{\infty} nx^n= \frac{x}{(1-x)^2}$
Differenting both sides wrt x we get
$\sum_{n=0}^{\infty} n^2x^{n-1}= \frac{d}{dx} \frac{x}{(1-x)^2}$
Usimg $\frac{d} ({dx} \frac{u}{v})= \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
We get
$\sum_{n=0}^{\infty} n^2x^{n-10} =\frac{(1-x)^2 . 1 - x . (x-2)}{(1-x)^4} = \frac{x +1}{(1-x)^3}$
Multiply by x on both sides we get $\sum_{n=0}^{\infty} n^2x^n = \frac{x^2 +x}{(1-x)^3}$
Putting value of $x = \frac{1}{10}$ we get the result
Let $a = 2^{\frac{1}{3}}\cdots(1)$
So $a^3 = 2\cdots(2)$
And $\frac{1}{a} = 2^{-\frac{1}{3}}\cdots(3)$
And
$x = a+ \frac{1}{a}\cdots(4)$
Cubing both sides $x^3 = a^3 + \frac{1}{a^3} + 3 * a * \frac{1}{a} ( a + \frac{1}{a})$
Or $x^3 = a^3 + \frac{1}{a^3} + 3 * ( a + \frac{1}{a})$
Or $x^3 = 2 + \frac{1}{2} + 3 *x$ from (2), 3) and (4)
Or $x^3 = \frac{5}{2} + 3 *x$
Or $2x^3 = 5 + 6x$
Or $2x^3-6x = 5$
Proved
Let $\frac{x-a}{x-b} = m$ and $\frac{a}{b} =n $
So we get $ m + \frac{1}{m} = n + \frac{1}{n}$
Or $m^2n + n = n^2m + m$
Or $m^2n - n^2m = m - n$
Or $mn(m -n ) = m - n$
Or m-n = 0 Or $mn = 1$
case 1
m = n gives $\frac{x-a}{x-b} = \frac{b}{a}$
Or $b(x-a) = x(a-b)$ or $bx-ab = ax-ab$ or $bx = ax$ or $x(a-b)=0$ or $x = 0$
Case 2)
mn =1 gives
$\frac{x-a}{x-b}*\frac{a}{b}=1$
or $a(x-a)=b(x-b)$
Or $ax-a^2=bx-b^2$
Or $ax-bx = a^2-b^2$
Or $x(a-b) = a2-b^2$
Or $x = a + b$
