Let gcd(m,n) = p.
then m = pq and n = pr for some q and r and gcd(q,r) = 1
gcd(m,n) = p as we have chosen
lcm(m, n) = pqr as q and r are co-primes
gcd(m,n) + lcm(m,n) = m + n
$=>p + pqr = pq + pr$
$=>1 + qr = q + r$
$=>qr - q -r + 1= 0$
$=>(q-1)(r-1) = 0$
q =1 mean n is divisible by m
or r =1 meand m is divisible byn
hence proved
as we see that the roots are doubleing in term to term so multiply numeraator and denominator by $(\sqrt[16]{5}-1)$ we get
$(\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)} +1)^{48}$
= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)(\sqrt[16]{5}-1)} +1)^{48}$
= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[8]{5}-1)} +1)^{48}$ using $a^2-b^2$ formula for last 2 terms
= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[4]{5}-1)} +1)^{48}$ using $a^2-b^2$ formula for last 2 terms-
= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} +1)^{48}$ using $a^2-b^2$ formula for last 2 terms-
= $(\frac{(4*\sqrt[16]{5}-1)}{4}+1)^{48}$
= $(\sqrt[16]{5})^{48} = 5^3 = 125$
Fir example if = 5 aand a = 7 then a-b = 2 and a + b = 12. Concatenating we get 212,'
Solution:
because c is prine so $a+b$ is odd so $a-b$ ( as it is $a+b -2b$) is odd. as c is a 5 digit number $a-b$ is 2 digit and$ a+ b$ is 3 digit, mallest $a-b$ is 11 and $a+b$ is odd staring from 101 putting the values we get $a+b = 113$ and $a -b = 11$ and hence c = 11113.
We need to find the mininal and maximal of $ab+bc+ca$
we have $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ bc+ ca)$
so $ a^2 + b^2 + c^2 + 2(ab+ bc+ ca) > = 0$
puttig $a^2+b^2+c^2 = 1$ we get
$ 1 + 2(ab+ bc+ ca) > = 0$
or $(ab+bc+ca) >= - \frac{1}{2}$
Further to find tthe maximum we have $(a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 +b^2+c^2 - ab - bc - ca)$
or $2(a^2 +b^2+c^2 - ab - bc - ca) >= 0$
or $ab+bc+ca <= a^2+b^ + c^2$
or $ab+bc+ca <= 1$
so we have $ab+bc + ca \in [-.5 .. 1]$
We have $16p + 1$ is a cube say $x^3$
So $16p= x^3-1 = (x-1)(x^2+ x + 1)$
As $x^2+x+1 = x(x+1) +1$ is odd we must have
$16 | x-1$
So $x = 16k+1$ for some integer k
so $p = k(x^2 + x + 1)$
the above cannot be prime unless k =1 and this gives x = 17 and p = 307
We need to see if 307 is prime and it is so
Hence the only solution is p = 307 (giving $16 * 307 +1 = 17^3$
in the 2 patterns the 1st number is same.
now 1st pattern has common diffrerence 4 and 2nd one 3
the minimum common number difference shall be LCM(4,3) or 12
so 13th common number shall be 1 * 12 * (13–1) = 145
We are given f a and b are roots of $x^2-3x+1=0$
so $a + b = 3\cdots(1)$
$ab=1\cdots(2)$2
now $ $(a^a + b^b)(a^b+ b^a)$$
= $a^{a+b} + (ab)^a + (ba)^b b^(b+a)$
= $a^3 + 1 + 1 + b^3$ putting the value of a+b and ab from (1) and (2)
=$a^3 + b^3 + 2$
= $(a+b)^3 - 3ab(ab+b) + 2$ using formula for $a^3+b^3$
= $3^3 - 3 . 1 . 3 + 2 = 20$ putting values from (1) ansd (2)
In the above expression x not dependent on y term so let us maximize $\sqrt{1-y^2}$ and this is {maxium at y= 0 and the value is 1.
Putting the same we get $x + \sqrt{1-x^2}$
As we have $\sqrt{1-x^2}$ we can put $x=\sin\,t$ ( $0 \le t \le \frac{\pi}{4}$ to get
$x + \sqrt{1-x^2}= \sin\, t + \cos \, t $
We need to maximize $\sin\, t + \cos \, t $ so let us get in form $r \cos\, w \sin \, t + r \sin\, w \cos \, t$
which is $r \sin ( t+w)$ and maximum value is |sqrt{2}$
So let $r\cos\, t = 1$ and $r \sin\, t = 1$
By squaring both and adding
$r^2 = 2$ or $r=\sqrt{2}$
Hence largest valus is $\sqrt{2}$
We have multipying by $x(y+2)$
$\frac{2}{y(y+2)}= \frac{1}{3 * 2^x}$
or $ 3 * 2^{x+1} = y(y+2)$
let z = x+1 so
$ 3 * 2^{x+1} = y(y+2)$
LHS is even so y is even because if y is odd then y(y+2) is odd
3 is a factor of y or y + 2
let us atke the 2 cases one by one
case 1 :3 is a factor of y
so $y = 3 * 2^a$ and $y+2 = 2^b$ for some inetger a and b
$3 *2^a + 2 = 2^b$
or $2(3 * 2^{a-1} + 1) = 2^{b+1}$
as RHS is even so $3*2^{a-1}$ is odd or a = 2
so y = $3 *2^a = 6$ and $y+2 = 8$ from this $3 * 2^{x +1} = 48 = 3 * 2^4$ ot x = 3
case 2 :3 is a factor of y +2
so $y+ 1 = 3 * 2^a$ and $y-2 = 2^b$ for some inetger a and b
$3 *2^a - 2 = 2^b$
or $2(3 * 2^{a-1}-1 = 2^{b+1}$
as RHS is even so $3*2^{a-1}$ is odd or a = 2
so y = $3 *2^a = 6$ and $y -2 = 4$ from this $3 * 2^{x +1} = 24 = 3 * 2^3$ ot x = 2
So 2 solutions are (3,6) and (2,4)
We ahve $a^2-b^2 = 24\cdots(1)$
$b^2-c^2 = 16\cdots(2)$
Because $a^2-b^2 = 24$ so both a and b are odd or both ae even,
So we have a = b + 2n for some n
To put an upper bound on be a = b+ 2 beause a is at least 2 more than b
So we have $a^2-b^2 = (b+2)^2 - b^2 = 24$
Or $2b + 4 = 24$ or $b= 5$
For the lower bound $b^2 =17$ so $b ge 5$
So we get b = 5 and putting the values
So we get $a= 7, b = 5, c= 3$
For eah dice there are 6 outcomes
for 7 dices there are $6^7$ outcomes
so there are $5^7$ out comes
10 can come in followin ways
6 ones and a 4 this can come in 7 ways(4 at any of 7 places)
5 ones one 2 and one 3 this can come in 7 * 6 = 42 ways( 2 at any of 7 places and and then 3 in any of the next 6 paces)
4 ones and 3 twos = $7 \choose 3$ = 35 places
so total number of places = 7 + 42 + 35 = 84.
so probability = $\frac{84}{6^7}$
let the 3rd number be t
so $t^2 + (10x+y)^2 = (10y+x)^21$
or $t^2 = (10y+x)^2 - (10y+x)^2 = 99(y^2 - x^2) = 99(y-x)(y+x)= 11 . 3^2(y+x)(y-x)$
y+x is less than 18 and y-x is less than 10
for the RHS to be a perfect square we must have y +x = 11 and y-x is to be perfect squiare and odd
because 11 cannot be a factor of y-x and if y +x is odd then y-x has to be odd because y-x = y + x - 2x
so y+ x = 11 and y-x = 1 giving y = 6 and x = 5
hence $t^2 = 99 * 11 * 1$ or t = 33
so triple is (33,56,65)
$3n^2-2019$ is divisible by 2016
Both are mutiple of 3 so $n^3-667$ is divisible by 672
Or $n^3-667+672$ that is $n^3-1$ is divisible by 672
Now $n^3-1= (n-1)(n^2+n+1)$ and 672 = 32 * 21 (product of even and odd)
As $n^2+n+1 = n(n+1) +1 $ is always odd so we have (n-1) multiple of 32.
Futher we must have $n^3 \equiv 1 \pmod 3$ and $n^3 \equiv 1 \pmod 7$
For $n^3 \equiv 1 \pmod 3$ we take mod 3 and get 0 for 0 1 or 1 and 2 for 2 so n must be 1 mod 3
Or f n-1 must be mutiple of 3
So $n \equiv 1 \pmod {96}$
Futher we must have $n^3 \equiv 1 \pmod 7$ and $n^3 \equiv 1 \pmod 7$
for $n^3 \equiv 1 \pmod 7$ we take mod 7 and get 0 for 0,1 for 1,2,4, and 6 for 3,5,6
So we must have n = 1/2/4 mod 7
n = 96k + 1
k =1 gives n = 97 , mod 7 = 6 so not a sulution
k = 2 gives n = 193, mod 7 = 4 so is a solution
So ans smallest n = 193
We have $\frac{1}{m} + \frac{1}{n} = \frac{1}{143}$
or $143n + 143m = mn$
or mn - 143m - 143n = 0
this if of the form $m(n-143) - 143n = 0$
To solve these type of probem add and $143^2$ to both sdes to get
$m(n-143) - 143(n- 143)= 143^2$
or $(m-143)(n-143) = 143^2= 11^2 * 13^2$
we get the following pairs as solutions (n-143,n-143) = (1,20449), (11, 1859), (13,1573), (121,169),(143,143), (169,121), (1573,13), (1859,11),(20449.1)
or (n,m) = (144, 20592), (11,2002),(13,1716), (264, 312), (286,286), (312,264),(1716,13),(2002,11),(20592.144)
there are 9 pairs out of which 8 are distinct
Let $\frac{a}{b}$ be rational and so let it be $\frac{n}{m}$ where n and m are integers
so $\frac{a}{b}= \frac{n}{m}$ or $b= \frac{ma}{n}$
now let $t_n$ the the nth term
so $t_(n+1) = a + ma = a(1+m)$
to get $a(1+m)^2$ we need to add $a(m^2 + 2m)$ or $am(m+2)$ as $nb = ma$ so $nb(m+2) = am(m+2)$ and so we get this term as well
to get $a(1+m)^k$ we need to add $a((1+m)^k -1)$ or $amf(m)$ as $nb = ma$ so $nbf(m) = amf(m)$ and so we get this term as well
so we get the terms in GP
Let n ba me number.
now $36 = 2 ^ 2 * 3^2$
$gcd(n, 36) = 2$ means 2 is a facttor bf n but is is not a factor of n. and 3 is not a factor of it
So nwe take the multiple of 2 and remove from this mutipls of 4 and multiples of 6.
because we need to remove multiple of 3 so we need to remove multiple of 6 (as they are multiple of 2)
To find the number of numbers
the number of numbers which are mutilple of 2 = $\lfloor\frac{999}{2}\rfloor - \lfloor{99}{2}\rfloor = 449 - 49 = 500$
the number of numbers which are mutilple of 4 = $\lfloor\frac{999}{4}\rfloor - \lfloor{99}{4}\rfloor = 249 - 24 = 225$
the number of numbers which are mutilple of 6 = $\lfloor\frac{999}{6}\rfloor - \lfloor{99}{6}\rfloor 166 - 16 = 150$
multiple of 4 and 6 both contain multiples of 12(LCM of 4 and 6).
mutiples of 12 will be in both the lists that is multiple of 4 and 6.
the number of numbers which are mutilple of 12= $\lfloor\frac{999}{12}\rfloor - \lfloor{99}{12}\rfloor 83 - 8 = 75$
so number of mutiples of 4 and 6 are 225 + 150 -75 = 300
so number of numbers whose GCD with 36 =2 is 450 - 300 = 150
We have $ab + a + b + 82 = z^2$
or $ab + a + b + 1 + 81 = z^2$ we do so that we can facrot LHS
or $(a+1)(b+1) = z^2 - 81 = (z-9)(z+9)$
we can choose a + 1 = z - 9 and b+1 = z + 9 (this shall not give all solutions but some
so a = z - 10 and b = z + 8
they have to be co-prime so both are odd say z = 2 m + 1
so a = 2m - 9 and b = 2m + 9
for these to be co-prime there should not a factor of 18 (that is the difference)
so 2m - 9 shall not have a factor 3 or m must not have a factor 3
so a = 2m - 9 and b = 3m + 9 m > 5 and of the forn $3k \pm 1$
Now let us take GCD(N,N-101)
GCD(N,N=101) = GCD(N,101)
There are 2 cases either N is multiple of 101 or not
If N is multiple of 101 say N= 101m
So N-101 = 101(m-1) so $N(N-101) = 101^2m(m-1)$ which is not a perfect square
So N is not a multiple of 101 so GCD(N,N-101) = 1 so N and N-101 both are perfect squares
Say $N= x^2$ and $N-101$ = y^2
$x^2-y^2 = 101$ or $(x+y)(x-y) = 101$
So x + y = 101 and x-y =1 solving these we get x = 51 and y = 50
So $N = x^2 = 51^ = 2601$
We have for $n^{th}$ number it is n 1s followed by n-1 5's and ending with 6
of it is n 1's follwed by n 5's and add 1
n '1s $\frac{1}{9}(10^n-1)$
n 5's is $\frac{5}{9}(10^n-1)$
so the number is $\frac{1}{9}(10^n-1) * 10^n + \frac{5}{9}(10^n-1) + 1$
$= \frac{1}{9}(10^{2n}-10^n + 5* (10^n-1) + 9)$
$= \frac{1}{9}(10^{2n} + 4* 10^n + 4)$
$=\frac{1}{9} (10^n + 2)^2$
$=(\frac{1}{3}(10^n+2))^2$
which is peffect square
We have $2^4+2^7 + 2^n = 2^n + 144 = m^2$ where m is positive
or $2^n = m^2 - 144 = (m+12)(m-12)$
as 2 is a prime so both m+ 12 and m-12 powers of 2
now difference (m+12)- (m-12) = 24
so powers of must have a difference 24
as $2^5 = 32$ so $2^6-2^k \ge 2^6 - 2^5 > 32$ (for any k less of equals 5)
so we need to check for candidates 2,4,8,16,32 and get 8 and 32.
this gives m = 20 and putting m = 20 we get n= 8
Let $y^2- my + 1 = 0\cdots(1)$
so $y^2 = my -1$
or squaring both sides $y^4 = m^2y^2 - 2my + 1 = m^2y^2 - 2(y^2) + 1= (m^2-2)y^2$
or $y^4 = (m^2-2)y^2 + 1\cdots(2)$
putting y =x and m = 3 we get
$x^2 - 3x + 1$
and from (2) we have $x^4 - 7x^2 +1 = 0\cdots(3)$
putting $y = x^2$ and m = 7 in (1) we get (3) and from(2) we get
$x^8 - 47x^4 +1 = 9\cdots(4)$
putting $y = x^4$ and m = 47 in (1) we get (4) and from(2) we get
Because 4x is even so we have a is odd
So let a = 2n + 1
$4x= (2n+1)^3 + 3 = 4n^2 + 4n + 4$
or $x= n^2 + n + 1 = n(n+1) + 1$
now n can be 0/1/2 mod 3
if n is one of the form $(3k, 3k+1, 3k-1$
If n is of the form 3k then $x-1 = 3k(3k+1)$ divisible by 6 so x of the orm 6m + 1
if n is of the form 3k + 1 then $x-1 = (3k+ 1)(3k+2) = 9k^2 + 9k^2 + 2 = 9k(k+1) + 2$ remainder 2 when devided by 6
so x = 6m + 3 but is divisible by 3 so not admissible
if n is 3k -1 then
$x-1 = (3k-1)(3k)$ of the form 6m or x+ 1of the form 6m + 1
So it is of the form 6m + 1(or 6y + 1) and hence proved
Multiply both sides by x+ y to get
$\frac{x+y}{x} + \frac{x+y}{y} = 1$
or $ (1+ \frac{y}{x}) + (\frac{x}{y} + 1) = 1$
or $ \frac{y}{x} +\frac{x}{y} + 2 = 1$
or $\frac{y}{x} + \frac{x}{y} = -1$
$(a-5)^2 + (b-7)^2 = 4$ is a circle whose centre is $(5,7)$ and radius is 2
We need to find minimum of $(a+7)^2 + (b+2)^2$ that is distance of the point A with coordinates a,b fom (-7,-2).
This is minimum when the point lies on the straight line from (-7,2) , (5,7) .
this distane from (-7,2) to (5,7) = $\sqrt{(5+7)^2 + (7-2)^2} = 13$
the distance of a from (5,7) is 2
so distance of point from (-7,2)is 13 - 2 = 11
so minmum of $(a+7)^2 + (b+2)^2$ is 121
Let us express 328 as sum of power of 2 (as 2,4,8 all are power of 2)
$328 = 256 + 64 + 8 = 2^8 + 2^ 6 + 2^3$
Only $2^6$ or $2^3$ are power of 8
So let us consider these 2 cases
1) $2^6 = 8^2$ and 8 is not power of 4 so $8 = 2^3$ and $256=4^4$ giving a = 3, b = 4, c= 1
2) $2^3 = 8^1$ and this gives 2 cases
case 1 $4^4 = 256, 2^6 = 64$ giving a = 6, b = 4, c = 1
case 2 $2^8 = 256, 4^3 = 64$ giving a = 8, b = 3, c = 1
The above 3 are solutions
The 2020 numbers are in a circle. Let the numbers be $a_1$ to $a_{2020}$ now startting wth $a_k$ we have 4 number consecutive $a_k,a_{k+1} , a_{k+2}$, $a_{k+3}$ for k from 1 to 2017 and other 4 consecutive number $a_{2018},a_{2019}, a_{2020}.a_1$ so on
now let us compute the value sum of outer numbers - sum of innter numbers calling $S_n$
$S_{1} = a_{1} - a_{2} - a_{3} + a_{4}$
$S_{2} = a_{2} - a_{3} - a_{4} + a_{5}$
...
...
$S_{2020} = a_{2000} - a_{1} - a{2} + a_3$
If we add all of these then we have
$\sum_{n=1}^{2020}S_{n} = 0$
For the sum of 2020 numbers be zero either all are zero (that means some numbers are same which is not true ) or at least one number is positive say $S_{k}$ then $a_{k}$ is the starting point
we have $3^{32}-1 = (3^{16}+ 1)(3^{16}-1) = (3^{16}+ 1)(3^{8}+1) (3^{8}- 1)$
$ = (3^{16}+1)(3^8+1)(3^4+1)(3^4-1)$
out of the above $3^4+1 = 82$ and %3^4-1=80$ are between 75 and 85 and product is 80 * 82 = 6560
We have $x^2 = 1-x\cdots(1)$
multiply both by x to get
$x^3 = x -x^2 = x - (1-x) = 2x -1\cdots(2)$
$x^5 = x^3 .x x^2 = (2x-1)(1-x) = -2x^2 +3x -1 = -2(1-x) + 3x - 1 = 5x-3$ (using (1))
or $x^5 + 8 = 5x + 5$
or $\frac{x^5+8}{x+1} = 5$
We have let $t_n = \frac{1}{t^3-1} + \frac{1}{2}$
$= \frac{1}{2}\frac{t^3+1}{t^3-1}$
=$\frac{1}{2}\frac{(t+1)(t^2-t+1)}{(t-1)(t^2+t+1)}$
Now given value = $\prod_{n=2}^{100} \frac{1}{2}\frac{(t+1)(t^2-t+1)}{(t-1)(t^2+t+1)}$
$= \frac{1}{2^{99}} \prod_{n=2}^{100} \frac{(t+1)}{(t-1)}\frac{(t^2-t+1)}{(t^2+t+1)}$
now let us compute the 2 products
$ \prod_{n=2}^{100} \frac{(t+1)}{(t-1)}= \frac{3}{1}.\frac{4}{2}.\frac{5}{3}\cdots \frac{98}{96}. \frac{99}{97}. \frac{100}{98}. \frac{101}{99}$
In the above product the numerator of $n^{th}$ term is same as denoniator of $n+2^{nd}$ term giving
$ \prod_{n=2}^{100} \frac{(t+1)}{(t-1)} = \frac{100 * 101}{1 * 2}\cdots(1)$
let us compute 2nd product
$ \prod_{n=2}^{100} \frac{(t^2-t+1)}{(t^2+t+1)}= \frac{3}{7} ( \frac{ 7}{13} \cdots \frac{9901}{10101}$
In the above product the numerator of $n^{th}$ term is same as denoniator of $n-1^{st}$ term giving
$ \prod_{n=2}^{100} \frac{(t^2-t+1)}{(t^2+t+1)}= \frac{3}{10101}\cdots(2)$
using (1) and (2) we get the required result
$= \frac{1}{2^{99}}\frac{3}{10101} * 5050= \frac{2525}{2^{98} * 3367}$
We know that 2^m * 2 is a power of 2
so if we chose $x^3 = y^4 = 2^t$ for some t then we shall get power of 2
so let $x = 2^{4m}$ and $y = 2^{3m}$
so $x^3 + y^ 4= 2^{12m} + 2^{12m} = 2^{12m+1}$
Now there is only one 1ssure remanins that is 12m + 1 has to be multiple of 31
that is not a problem as gcd(31,12) is 1 and we shall use exteneded euclidean algoirithm to find the same
now 31 = 2 * 12 + 7 we take the highest multiple of 12 and leave the remainder
or $7 = 31 - 2 * 12\cdots(1)$
12 = 7 * 1 + 5
or $5 = 12 - 7 * 1 = 12 - (31 - 2 * 12)$ using (1)
or $5 = 12 * 3 - 31\cdots(2)$
now $ 7 = 5 + 2$
or $2 = 7- 5 = (31- 2 * 12) - (12 * 3 -31) = 2 * 31 - 5 * 12$ (uinsg (1) and (2)
or $2 = 2 * 31 - 5 * 12$
now $5 = 2 * 2 + 1$
or $1= 5 - 2 * 2$
or $1 = (12 * 3 - 31) - 2 * ( 2 *31 - 5 *12) = 13 * 12 - 5 * 31$
so if m is 13 or any number 31n + 13 we get 12m+ 1 is a multiple of 31
m = 31n + 13 gives 12m +1 = 12(31n + 13) = 31(12n + 5)
so $x = 2^{4(31n + 13)}$, $y = 2^{3(31n + 13)}$ and $z = 2^{12n+5}$ is parametric solution and hence infinite solutions.
clearly n > 7.
now if t ia factorial then 3!5! must be product of sum k sucesssive numbers from 8
3!5! = 6 * 120 = 720 = 8 * 90 = 8 * 9 * 10
so n! = 10 * 9 * 8 * 7! = 10!
or n = 10
Let n be sum of 3 squares so
$n= a^2 + b ^2 + c^2$
we need to show that $n^2$ is sum of 3 squares
from given condition
$n^2 = (a^2+b^2+c^2)^2 = a^4 + b^4 + c^4 + 2 a^2b^2 +2 b^2 c^2 + 2 c^2 a^2$
$= a^4 + b^4 + c^4 + 2 a^2b^2 -2 b^2 c^2 - 2 c^2 a^2 + 4 b^2c^2 + 4 c^2 a^2$
$=(a^2+ b^2 -c^2)^2 + (2bc)^2 + (2ca)^2$
sum of 3 sqaures and hence proved
We are given
$\frac{a^2}{b+c} = 1$
or $a^2 = b + c$
or $a^2 + a = a + b + c$
or $a(a+1) = a + b + c$
or $\frac{1}{a+1} = \frac{ a}{a + b+ c}\cdots(1)$
similarly $\frac{1}{b+1} = \frac{a}{a + b+ c}\cdots(2)$
and $\frac{1}{c+1} = \frac{a}{a + b+ c}\cdots(3)$
adding all 3 above we get $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = \frac{a+b+c}{a+b+c} = 1$
We have 2001 is divisible by 3 and not 9. And if n Is above 5 then n! is divisible by 9. So $2001 + n!$ is divisible by 3 but not 9. so it cannot be a perfect square. so checking from 0 to 5 we get y = 45 and n =4
because 7 is a prime so as per Fermats Little theoren
$ 7 | 2^6-1$
now as $2^6$ leaves remainder 1 after dividing by 7 so it may be taht for some factor a of 6 $7 | 2^a-1$
we need to check for 1,2,3
so see $2^1-1 = 1$ $2^2-1 = 3$ and $2^3 - 1 = 7$ out f these 3 3 satisfies
as so k = 3m for all m satisfies.
Further $2^1+1 = 3$ $2^2+1 = 5$ and $2^3 +1 = 9$ out f these 3 none is divsible by 7 so there is no n such that $ 7 | 2^n+1$
proved
we see that $(2+\sqrt{3})(2-\sqrt{3}) = 4 - 1 = 1$
so $\frac{1}{(2+\sqrt{3}} = 2 - \sqrt{3}$
Let $(2+\sqrt{3})^x = y$ so $(2-\sqrt{3})^x = \frac{1}{y}$
so $y +\frac{1}{y} = 4$
or $y^2 - 4y + 1 = 9$
ot $y = \frac{4 \pm \sqrt{16-4}}{2}$
or $y= 2 \pm \sqrt{3}$
If we take $y = 2 + \sqrt{3}$ then $(2+\sqrt{3})^x = 2 + \sqrt{3}$ or $x= 1$
Taking $y = 2 - \sqrt{3}$ then $(2+\sqrt{3})^x = 2 - \sqrt{3}$ or $x= -1$
so $x \in \{ 1, -1\}$
Without loss of generality set n = 2x ( x need noy be integer so we can choose
p = m - 3x, q = m - x, r = m + x, s = m + 3x
now $pqrs+n^2 = (m-3x)(m-x)(m+x)(m+3x) + 16x^4$
$= (m- 3x)(m+ 3x)(m-x)(m+x) + 16x^4$
$= (m^2-9x^2)(m^2- x^2) + 16x^4$
$= m^4 - 10x^2m^2 + 9x^4 + 16x^4$
$=m^2 - 10x^2m + 25x^4 = (m^2 -5x^2)^2$ so it is a perfect square
as it is integer square root is also integer
The sum of 23 numbers is 1000 so the numbers can be 12 numbers 77 and 76, so the LCM>=77.
the LCM is lower if 13th number is a factor of 12 numbers (which are same)
let us look at factors of 1000. and one factor is minumum 13.
we have 50 * 20. 7 numbers can be 100 and 6 can be 50 giving LCM = 100
40 * 25, 12 numbers 80 and one number 40 giving LCM = 80
25 * 40 12 numbers can be 75 but 13th number is 100 LCM = 300
20 * 50 12 numbers 80 and one number 40 giving LCM = 80
So lowest LCM = 80 (12 numbers 80 and one number 40)
we have
$(1+ \frac{1}{x})^{x+1} = (\frac{x+1}{x})^{x+1}$
$=(\frac{x}{x+1})^{-(x+1)}$
$=(1- \frac{1}{x+1})^{-(x+1)}$
$= (1+ \frac{1}{-(x+1)})^{-(x+1)}$
comparing with the RHS we get -(x+1) = n or x = -(n+1)
Because 1995 is divisible by 3 so all the powers $ ≥2 $ of 1995 shall be divisible by 9.
So sum of digits of the result must be divisible by 9.
We get sum of digits of given number = 160 + X.
Smallest multiple of 9 above 160 is 162 which gives X= 2 and next multiple of 9 gives X = 11.
so X = 2
From the given condition we have
$(x-3) = \sqrt[3]3 + \sqrt[3]{3^2}$
squaring both sides
$(x-3)^3 = (\sqrt[3]3)^3 + ( \sqrt[3]{3^2})^2 + 3 \sqrt[3]{3} \sqrt[3]{3^2}( \sqrt[3]{3} + \sqrt[3]{3^2})$ using $(a+b)^3 = a ^3 + b^3 + 3ab(a+)$
or $(x-3)^3 = 3 + 9 + 9 ( x-3)$ using $( \sqrt[3]{3^2} + \sqrt[3]{3^2} = x- 3$
or $(x^3 - 9x^2 + 27 x - 27 = 12 + 9x - 27 = 9x + 12$
or $x^3 - 9x^2+ 18x -12 = 0$
We have numerator = $8^x - 2^x = (2^3)^x - 2^x = (2^(3x) - 2^x = 2^x(2^{2x} - 1)= 2^x(2^x + 1)(2^x-1)$
Denominator = $6^x - 2^x = 3^x(2^x-1)$
So $\frac{8^x - 2^x }{6^x - 3^x} = \frac{2^x(2^x+1)}{3^x} = 2$
Or $2^x(2^x+1) = 2 * 3^x$
x cannot be zero as in the original question x = 0 is not possible
So we must ahve $2^x = 2$ and $2^x+ 1 = 3^x$ by equating even on both sides and odd on both sides
1st equuation gives x = 1 and it satisfies 2nd equation as well.
so Solution x = 1
Let the sides be a -b , a , a+b
as the tringle is right angled so $(a-b)^2 + a^2 = (a+b)^2$
or $a^2-2ab + b^2 + a ^2 = a^2 + 2ab + b^2$
or $a^2= 4ab$
or $a= 4b$
as the triangls is right angled so aea = $\frac{1}{2} a(a-b) = \frac{1}{2} 4b(4b-b) = 6b^2 = 24$ or b= 2
the smallest sde = 3b = 6.
a,b,c are in HP so $\frac{1}{a}$, \frac{1}{b}$, $\frac{1}{c}$ are in AP
or $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$
multiply both LHS and RHS by a + b + c we get
$\frac{a+b+c}{b} - \frac{a+b+c}{a} = \frac{a+b+c}{c} - \frac{a+b+c}{b}$
Or
$(\frac{a+b+c}{b} - 2) - (\frac{a+b+c}{a}-2) = (\frac{a+b+c}{c} -2) - (\frac{a+b+c}{b} -2) $
or $(\frac{a+c-b }{b}) - (\frac{b+c-a }{a}) = (\frac{a+b-c }{c}) - (\frac{c + a - b}{b} $)
Hence $\frac{b+c-a}{a}$, $\frac{c+a-b}{b}$, $\frac{a+b-c}{c}$ are in A.P
Let us put $p = \sin ^2 \theta \cos^2 \theta$
We have $t_2 = \sin ^2 \theta + \cos^2 \theta = 1 \cdots(1)$
and $t_4 = (\sin^4x + \cos^4 x) = (\sin ^2 x + \cos^2 x) - 2 \sin ^2 x +\cos^2 x = 1 - 2p$
Now we an sert
$t_n t_2 = ( \sin ^n \theta + \cos^n \theta)(\sin ^2 \theta + \cos^2 \theta) = \sin ^{(n+2)}\theta + \cos^{(n+2)} + \sin ^n \theta \cos^2 \theta + \sin ^2 \theta \cos^n \theta$
or $t_n = t_{(n+2)} + t_{(n-2)}p$
or $t_(n+2) = t_n - t(n-2)p$
Using this we have
$t_6 = t_4 - t_2p = t_4 - p = 1- 3p$
$t_8 = t_6 - t_4p = 1-3p - (1-2p ) p^2 = 1- 4p + 2p^2$
$t_{10} = t_8 - t_6p = (1-4p +2p^2) - (1-3p)p = 1 - 5p + 5p^2$
so $6t_{10} − 15t_8 + 10t_6 = 6(1-5p + 5p^2 ) - 15(1- 4p + 2p^2 ) + 10(1-3p) = 1 $
let the prime be p
we have $2019^8 \equiv -1 \pmod p$
or $2019^{16} \equiv 1 \pmod p$
so $2019^{16k} \equiv 1 \pmod p$
as per Formats Little theorem p shall be of the form 16k + 1.
any number less than 16 is not possible because $2019^8 \equiv -1 \pmod p$ and $16=2^4$ and $8=2^3$ so any factor of 16 shall not give a power =1.
so p = 16 *2 + 1 = 33 (not a prime) or 16 * 3 + 1= 49( not a prime) or 16 * 4 + 1 = 65( not a prime) or 16 * 5 + 1 = 81 (not a prime) or 16 *6 + 1 = 97.
so we check for 97
now $2019 \equiv - 18 \pmod {97}$
so $2019^2 \equiv 324 \pmod {97}$
or $2019^2 \equiv 33 \pmod {97}$
so $2019^4 \equiv 33^2 \pmod {97}$
or $2019^4 \equiv 22 \pmod {97}$
or $2019^8 \equiv 484 \pmod {97}$
or $2019^8 \equiv -1 \pmod {97}$
or or $2019^8 + 1\equiv 0 \pmod {97}$
so smallest prime factor = 97
let $f(x) = x^4+ 12x--5$
As there change of sign once in f(x) so there is one positve real toot as per descartes' rule of signs
$f(- x) = x^4- 12x--5$
As there change of sign once in f(-x) so there is one real toot as per descartes' rule of signs
so there are 2 real roots
Number of ways of choosing r objects from p objects is $\frac{(p!}{r!(p-r)!}$. so $\frac{(p!}{r!(p-r)!}$ is an integer
or $\frac{p(p-1)!}{r!(p-r)!}$ is integer . Now because p does not divide the denominator because p is prime dividing by p we get the result
we have $\sin\, 54^\circ = \cos \, 36^\circ$
let $x = 18^\circ$
so we have $\sin 3x = \cos 2x$
Or $3 \sin \, x - 4\sin ^3 x = 1 - 2 \sin ^2 x$
putting $\sin \,x = y$ we get
$3y - 4y^3 = 1 - 2y^2$
or $4y^3 - 2y^2 - 3 y + 1 = 0$
or we see that (y-1) is a factor as putting above as f(y) and find f(1) = 0
by division we get $(y-1)(4y^2 + 2y -1) = 0$
as $\sin\,18^\circ$ is not zero so $4y^2 + 2y - 1=0\cdots(1)$
we need to show $y^3 + y^2 = \frac{1}{8}$
from (1) $4y^2 = - 2y + 1\cdots(2)$
or $8y^3 = 2y(-2y + 1) = - 4y^2 + 2y = - -4y^2 + 1 - 4y^2$ (from (2))
or $8y^3 + 8y^2 = 1$
or $y^3 + y^2 = \frac{1}{8}$
proved
there are 2 cases $x^2 >= 1$
this gives $3x^2 - 4(x^2-1) + x -1 = 0$
or $-x^2 + x + 3 = 0$
or $x^2 -x - 3 = 0$
this has two slutions one in $(1,\infty$ and another is $(-\infty, -1)$
for for the other case we have $3x^2 + 4(x^2-1) + x -1 = 0$ or $7x^2 + x - 5 = 0$ f(0) = -5, f(1) = 3 f(-1) = = 1 so there are 2 roots one between 0 and 1 and another between 0 and -1.
So there are 4 solutions
we have
$x^4+3 x^2 y^2+2 y^4+4 x^2+5 y^2+3$
$= x^4+3 x^2 y^2+4 x^2 + 2 y^4 +5 y^2+3$ putting them in descending power of x
$=x^4+ x^2(3 y^2 +4) + (2y^2 + 1)(y^2 + 3)$ factoring the expression involving y only
$=x^4+ x^2((2y^2 + 1) + (y^2 +3) + (2y^2 + 1)(y^2 + 3)$ split the middle term as $3y^2 + 4 = 2y^2 + 1 + y^3 + 3$
$= x^2(x^2 + 2y^2 + 1) + (y^2 + 3)(x^2 + 2y^2 +1 ) = (x^2 +2y^2 + 1)(x^2 + y^2 + 3)$
We know as $-1 \le \sin\, x \le 1$ so $sin^5 x \ le sin ^2 x$. they are same when sin x = 0 or 1 otherwise $\sin^5 x \lt sin ^2 x$
similarly $\cos^5 x \le \cos ^2 x$. they are same when cos x = 0 or 1 otherwise $cos^5 x \ lt cos ^2 x$
so $\sin ^5 x + \cos^5 x \le \sin ^2 x + \cos^2 x$ or 1 they are same when $\sin \, x =$ 0 or and $\cos\,x$ = 0 or 1 that is one of them is zero and another 1
this is possible when $\sin\, x = 0$ and $\cos\,x = 1$ that is $x = 2n\pi$
or $\sin\, x = 1$ and $\cos\,x = 0$ that is $x = (2n + \frac{1}{2}) \pi$
hence $x \in \{ 2n\pi, (2n + \frac{1}{2}) \pi \} $
We have
$\frac{2k+1}{(k^2+k)^2} = \frac{(k+1)^2 - k^2}{(k+1)^2 k^2} = \frac{1}{k^2} - \frac{1}{(k+1)^2}$
The above is a telescopic sum and we have
$\sum_{k=1}^n \frac{2k+1}{(k^2+k)^2} = \frac{1}{1} - \frac{1}{(n+1)^2} = .9999$
so $\frac{1}{(n+1)^2} = 1- .9999 = .0001 = \frac{1}{100^2}$
n+ 1 = 100 or n = 99
Using the property of polynomial that $a-b | p(a)- p(b)$ we have between N and 21
$N-21 | P(N) - p(21)$
As $P(N) = N+ 51$
We get $N-21 | N+ 51 - 17$
or $N-21 | N + 34| or | N - 21 |(N+ 34) - (N-21) $
or $N-21 | 55$
This gives N-21 = 1 or 5 or 11 or 55 or -1 or -5 or - 11 or -55
So N = 26 or 32 or 36 or 76 or 20 or 16 or 10 or - 34 (1)
Now with 32 we get $N- 32 | N + 51 - (-247)$ or $N-32 | 330$ (2)
Finally with 37 we get $N -37 | N + 51 - 33$ | or $N- 37 | 55$
from set of (1) we get 26 or 32 or 36
from above (3) values we get N = 26
so N = 26 is the answer
We are given $a-\frac{1}{a} = b\cdots(1)$
$b-\frac{1}{b} = c \cdots(2)$
$c-\frac{1}{c} = a\cdots(3)$
From (1)$\frac{1}{a} = a-b $
dividing both sides by b
$\frac{1}{ab} = \frac{a}{b} - 1 = a(b-c) -1 $ using |(2)
similarly
$\frac{1}{bc} = \frac{b}{c} - 1 = b(c-a) -1 $
and $\frac{1}{ca} = \frac{c}{a} - 1 = -c(a-b) = 1 $
adding above 3 we have $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}= - 3 $
We first note that $3-x \ge 0$ and $1+x <=0$ as both are under square root and so we have $-1 \le x \le 3$
Secondly we have the LHS is monotonically decreasing as x increases
so let uf have $f(x) = \sqrt{3-x} - \sqrt{1+x}$
We have the LHS is monotonically decreasing as x increases and if is defined in $[-1,3]$
now $f(-1) = \sqrt{3-(-1)} - \sqrt{1+{-1}} = 2$ and $f(3) = \sqrt{3-3} - \sqrt{1+3} = -2$
so we need to solve for $\sqrt{3-y} - \sqrt{1+y} = \frac{1}{2}$
and then $y \le x \le -3$
we need to solve
$\sqrt{3-y} - \sqrt{1+y} = \frac{1}{2}$
square both sides to get $3 - y + 1 + y - 2\sqrt{(3-y)(1+y)} = \frac{1}{4}$
or $4 - \frac{1}{4} = 2\sqrt{(3-y)(1+y)}$
square both sides to get $\frac{225}{64} = (3-y)(1+y)$
or $\frac{225}{64} = (3-y)(1+y)$
putting y = z + 1 we get $\frac{225}{64} = (2-z)(2+z) = 4 - z^2$
or $z^2 = \frac{31}{64}$
so $z = \pm \frac{\sqrt{31}}{8}$
so $y = 1 \pm \frac{\sqrt{31}}{8}$
now whether we show take the plus or minus we check that $f(1) = \sqrt{2} - \sqrt{2} = 0$ wich does not sattisfy the condition so we must have -ve numebr and hence
$y = 1 - \frac{\sqrt{31}}{8}$
or $-1 \le x \le 1 - \frac{\sqrt{31}}{8}$
or
We have by AM GM inequaity among $\frac{x}{y}$ ,$\frac{y}{z}$, $\frac{z}{x}$
we have
$\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} >= \sqrt[2]{\frac{x}{y} * \frac{y}{z} * \frac{z}{x}}$
or $\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} >= \sqrt[2]{1}$
or $\frac{x}{y} + \frac{y}{z} + \frac{z}{x}>= 3$
this is greater than 3 if all are not equal and is 3 if all are equal
or $\frac{x}{y} = \frac{y}{z} = \frac{z}{x}$
or x = y = z (any integer)
we have
$x^2 + y^2 + z^2 + x + y + z = 1$
or $x^2 + x + y^2 + y + z^2 + z = 1$
we shall complete the square so let us multiply 4 to avoid fraction
$(4x^2 + 4x ) + (4y^2 + 4y) + (4z^2 + 4z) = 4$
adding 1 to each term on the left
$(4x^2 + 4x + 1) + (4y^2 + 4y + 1) + (4z^2 + 4z+ 1) = 7$
or $(2x+1)^2 + (2y +1)^2 + (2z+1)^2 = 7$
multipying by LCM of denomiantor of 2x+1,2y+1,2z+1 we get
$p^2 + q^2 + r^2 = 7m^2$ for integers p,q,r,m.
this has a least m which has a solution
let m be even say 2n
then all a,b,c have to be even because RHS is divisible by 4 and if any one to 3 numbers is odd then working in mod 4 and summing we get a remainder of 1to 3 when divided by 4.
so all are even
so m cannot be even
so m is odd.
because m is say m = 2k+1
so $7m^2 = 7(2k+1)^2 = 7 * 4k^2 + 7 * 4k + 1$
ot $7m^2 = 7 * 4k(k+1) + 7$
which leaves a remainder 7 when divided by 8
a number if odd on squaring leaves a remainder 1 when divided by 8 and if even then remainder is 0
sum of 3 such numbers (0 or 1) cannot be 7.
so the equation has has no solution and hence original equation has no solution.
we are given
$m^2 = n +2 \cdots(1)$
$n^2 = m + 2\cdots(2)$
$4 mn - m^3 - n^3$
$= 4mn - m(m^2) - n(n^2)$
$= 4mn - m(n+2) - m(n+2)$ using (1) and (2)
$= 2mn - 2(m+n)$
or $4 mn - m^3 - n^3= 2mn - 2(m+n)\cdots(3)$
we need to commte mn and m + n
sutarcting (2) from (1)
$m^2 - n^2 = n - m$
as n-m is not zero divding by n-m we have
$m+n = -1\cdots(4)$
adding (1) and (2)
$m^2 + n^2 = (m+n) + 4 = -1 + 4 = 3\cdots(5)$ putting value of m + n from (4)
or
Hence $2mn = (m+n)^2 - (m^2+n^2) = 1- 3 = -2$
putting the values from (4) and (6) in (3) we get
$4 mn - m^3 - n^3= 2mn - 2(m+n) = -2 -2 (-1) = 0$
We have both $16^{x^2+ y}$ and $16^{y^2+x}$ positive.
So we can apply AM GM inquality gettting
$\frac{16^{x^2+y} + 16^{y^2+x}}{2}\ge (16^{x^2+y} 16^{y^2+x})^{\frac{1}{2}}$
or $16^{x^2+y} + 16^{y^2+x}\ge 2 (16^{x^2+y+ y^2+x})^{\frac{1}{2}}$
we are given LHS = 1 so
$ 1 \ge 2 (16^{x^2+y+ y^2+x})^{\frac{1}{2}}$
or $ 1 \ge 16^{\frac{1}{4}} (16^{x^2+y+ y^2+x})^{\frac{1}{2}}$
or $ 1 \ge (16^{x^2+y+ y^2+x+ \frac{1}{2}})^{\frac{1}{2}}$
or $x^2 + y + y^2 + x + \frac{1}{2} <=0$
Now we combine line terms and complete square to get
$(x^2 + x + \frac{1}{4}) + (y ^2 + y + \frac{1}{4}) <= 0$
or $(x+\frac{1}{2})^2 + (y + \frac{1}{2})^2 <=0$
it is sum of 2 squares so cannot be be -ve so
both terms and sum has to be zero givng $x = y = \frac{-1}{2}$
Note I picked the question at https://www.youtube.com/watch?v=iyX-q3D91Ck&t=474s and here provide a diiferent solution
Let $P(x)= x^3-24x^2+183x -440\cdots(1)$
as a,b,c are roots of the equation P(x) = 0 so we have following 2 observations
$P(x) = (x-a)(x-b)(x-c) \cdots(2)$
and comparing oefficient of $x^2$ we have
$a+b+c = 24\cdots(3)$
Now area of the triangle say A = $\sqrt{s(s-a)(s-c)(s-c}$
where s is semiperimeter of triangle
so $s=\frac{a+b+c}{2}= \frac{24}{2}$ putting value of a+b+c from (3)
or $s= 12\cdots(4)$
from (1) putting s for x we get $P(s) = (s-a)(s- b)(s-c)$
So A = $\sqrt{s(s-a)(s-c)(s-c)} = \sqrt{sP(s)}$
so using (1) and above
$A^2 = sP(s) = s(s^3 - 24s^2 + 183 s - 440) $
putting the value of s = 12 from (4) we get
$A^2= 12 *(12^3- 24 * 12^2 + 183 * 12 - 440) = 12(12^3 - 2 * 12^3 + 183 * 12 - 440)$
Or $A^2 = 12(- 12^3 + 183 * 12 - 440) = 12(- 1728 + 2196 - 440) = 12 * 28 = 336$
or $A = \sqrt{336} = \sqrt{4^2 * 21} = 4 \sqrt{21}$
Let us fix the lower bound of n
now $n^2 -10n - 22 = (n^2 - 10n + 25) - (22+25)$ adding 25 for completing the square
$= (n-5)^2 - 47 > = 0$ as it has to be positive
so $(n-5)^2 >= 57$
or $(n-5) >= \sqrt{47}$
or $ n > 5 + \sqrt{47}$
as n is integer so n > 5 + 6$ as $6 < \sqrt{47} < 7$ so 5 + 6 is the largest value which does nt satisfy the condition
so $n >11$
now n = 12 this gives product of digits = 2 and $n^2 -10n -22 = 12^2 - 10 * 12 - 22 = 2$ so they are same
so 12 is a sloution
for $n > 12$ we have $22 < 2n$ and hence $n^2 - 10n - 22 > n^2 - 10n - 2n$
or $n^2 - 10n - 22 > n(n-12)$ and as $n-12$ is positive we have
$n^2 - 10n -22 >n$ for $ n > 13\cdots((1)$
We shall prove a little lemma:
For for any n the product of the digits of n is less then n
proof:
let n be a k digit number starting with a.
the value of $n >= a* 10^{k-1}\cdots(2)$
the number that starts with a shall give the largest product when all rest (k-1) digits will be ragest that is 9
so product of digits $ < a * 9^{k-1}\cdots(3)$
using (2) and (3) we get product of digts < the value n < value of exptession.
end of lemma
using the lemma and (1) we have for $n >12$ it is not possible for the expression . same as product of digits.
we have clearly $m > n$
taking $2^n$ common we have $2^n(2^{m-n} - 1) = 2016$
clearly $2^{n}$ is even and $2^{m-n}-1$ is odd
so let us factor 2016 to be a power of 2 multiplied by odd
$2016 = 32 * 63$
so n= 5 and $63 + 1 = 64= 2^6$ which is a power of 2
so m -n = 6 or m = 11
so solution m= 11, n = 6
if is a square let it be $x^2$
$37p + 4= x^2$
or $37p = (x^2-4) = (x-2)(x+2)$
as p is prime so one number is 37 and another is prime or one number is 37p and another number is 1
if x -2 = 1 then x +2 is 5 so it is not 37p
difference between x -2 and x+ 2 is 4 so numbers are 37 and 41 or 37 and 33.
but 33 is not prime so p = 41
We have $n^{th}$ term
$= \frac{2 + 4 + \cdots 2n }{n!} = \frac{n(n+1)}{n!}$
So then given exprsssion
$\frac{2}{1!} + \frac{2+4}{2!} + \frac{2+4+6}{3!}+ \cdots$
$=\sum_{n=1}^{\infty} \frac{n(n+1)}{n!}$ putting the valut from above
$=\sum_{n=1}^{\infty} \frac{n(n-1) + 2n }{n!}$ writing n(n+1) s n(n-1) + 2n
$=\sum_{n=1}^{\infty} (\frac{n(n-1)}{n!} + \frac{2n }{n!})$
$=\sum_{n=1}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{2n }{n!}$
$=\sum_{n=2}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{2n }{n!}$ we can convert the lowerr limit from n = 1 to n =2 because for n- 1 we have n(n-1) = 0 which can be dropped
$=\sum_{n=2}^{\infty} \frac{1}{(n-2)!} + 2 \sum_{n=1}^{\infty} \frac{1 }{(n-1)!}$
$=\sum_{n=0}^{\infty} \frac{1}{n!} + 2\sum_{n=0}^{\infty} \frac{1 }{n!}$ by changing the limit
$=3 \sum_{n=0}^{\infty} \frac{1}{n!}$
= 3e by using expansion form for e
Let us have $\tan^{-1}\frac{1}{5} = \alpha$
$\tan^{-1}\frac{1}{7} = \beta$
$\tan^{-1}\frac{1}{3} = \gamma$
$\tan^{-1}\frac{1}{8} = \delta$
we need to find $\alpha + \beta + \gamma + \delta$
Let us add 2 at a time $\alpha + \beta$ , $\gamma +\delta$ and then $\alpha + \beta + \gamma + \delta$
We have
$\tan \alpha = \frac{1}{5}\cdots(1)$
and $\tan \beta = \frac{1}{7}\cdots(2)$
So $\tan (\alpha + \beta) = \frac{\tan (\alpha) + \tan (\beta)}{1 - \tan (\alpha) \tan (\beta)}$
$= \frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \frac{1}{7}}$
$ = \frac{\frac{12}{35}}{ \frac{34}{35}} = \frac{12}{34} = \frac{6}{17}$
We have
$\tan \gamma = \frac{1}{3}\cdots(3)$
and $\tan \delta = \frac{1}{8}\cdots(4)$
So $\tan (\gamma + \delta) = \frac{\tan (\gamma) + \tan (\delta)}{1 - \tan (\gamma) \tan (\delta)}$
$= \frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \frac{1}{8}}$
$ = \frac{\frac{11}{24}}{ \frac{23}{24}} = \frac{11}{23}$
now $\tan (\alpha + \beta + \gamma + \delta)$
$=\frac{\tan (\alpha + \beta) + \tan (\gamma + \delta)}{1 - \tan (\alpha + beta) \tan (\gamma + \delta)}$
$= \frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \frac{17}{23}}$
$ = \frac{\frac{6 * 23 + 11 * 17}{17 * 23}}{1- \frac{66}{17 * 23}}$
$ = \frac{6 * 23 + 11 * 17}{17 * 23 - 66}$
$=\frac{ 325}{325} = 1$
so $\tan (\alpha + \beta + \gamma + \delta) = \tan \frac{\pi}{4}$
but we need to show that $\alpha + \beta + \gamma + \delta = \frac{\pi}{4}$
as it could be $\frac{5\pi}{4}$
but as $\alpha,\beta,\gamma, \delta$ each is less than $\frac{\pi}{4}$ so sum is less than $\pi$ so it is $\frac{\pi}{4}$
We have 34x + 34 y = 43 y + 34 y = 77 y
so 34(x+y) = 77 y
or $x+ y = \frac{77 * y}{34}$
34 is coprime to 77 , x+ y is integer so $\frac{y}{34}$ is integer
$x+y = 77 * \frac{y}{34}$ has a factor 77 and 77 is composite so x + y is composite
To keep the calculation simple let n = 10^{100}$
So $10^{100} + 3 = n + 3$
So $10^{20000} = n^{200}$
We need to find remainder when $n^{200}$ is divided by n+ 3
Now $n^{200} = n^{200} - 3^{200} + 3^{200}$
$n^{200} - 3^{200}$ is divisible by n + 3 and $3^{200} = 9^{100}$ whcih is less than $n+3$
So we need to find the unit digit of quotient of $n^{200} - 3^{200}$ divided by n + 3
we have $n^{200} - 3^{200} = (n+3)\sum_{k=0}^{199}(-1)^k n^k 3^{199-k}$
Hence
$ \frac{n^{200}-3^{200}}{n+3}= \sum{k=0}^{199}(-1)^k n^k 3^{199-k}$
putting back $n= 10^{100}$
we get $ \frac{10^{20000}-3^{200}}{10^{100}+3}= \sum{k=0}^{199}(-1)^k (10^{200})^k 3^{199-k}$
All the trerms except when k = 0 have a factor 100 so we get unit digit is unit digit of $-3{199}$
now $3^4$ ends with 1 so $3^{199} = (3^4)^{49} * 3^3 $ ends with unit digit of $3^3$ or 7 so if we negate it it ends with 10 - 7 = 3
so ans 7
Note: I picked the question from https://www.youtube.com/watch?v=uqaRfADGOtg and I provide my solution for the same in a diffrerent way.
We see that $n!$ grows at a rate much faster than $n^3$
For solving this we can fix the bound for n and check on the same.
as we see $6^3 = 216$ and $6! = 720$ which is greater
by chceking the values of n from 0 to 5 we get n = 5 satsfies the equation
so Ans $n= 5$
We have $GCD(19! + 19, 20!+19)$
$= GCD( 19!+ 19 , 20(19! + 19) - (20!+ 19))$ using gcd(a,b) = gcd(a, na -b))
$= GCD(19! + 19, 20! + 20 * 19 - 20! -19)$
$= GCD(19! + 19, 20 * 19 - 19)$
$= GCD(!9! + 19, 361)$
Now $361 = 19^2$
$19! + 19 $is 1$9(18! + 1)$
As 19 is a prime as per wilson's theorem 18! + 1 is divisible by 19
So $19! + 19$ is divisible by $19^2 = 361$
so GCD = 361
we have $\sin^3 x + \cos ^3 x= \frac{1}{\sqrt{2}}$
letting $\sqrt{2} \sin\,x = s$ and $\sqrt{2} \cos\, x = c$
we have $s^3 + c^3 = 2 \sqrt{2} \frac{1}{\sqrt{2}}= 2$
or $s^3 + c^3 = 2\cdots(1)$
and $s^2 + c^2 = 2\cdots(2)$
we are interested to find $\sin\,2x$ or sc
squaring (1) we get $s^6 + c^6 + 2s^3c^3 = 4$
and cubing (2) we get $s^6 + c^6 + 3s^2c^2(s^2 + c^2) = 8$
or $s^6 + c^6 + 6s^2c^2 = 8$
from (3) and (4) we have letting sc = x
$2x^3 - 6x^2 + 4 = 0$
or $x^3 - 3x^2 + 2$ this gives x = 1
and deviding by x-1 for factoring we get $(x-1)(x^2 - 2x -2) = 0$
so x = 1 or $1\pm \sqrt{3}$ but $1 + \sqrt{3}$ being above 1 is not admissible
so x = 1 or $1-\sqrt{3}$
$n^2−33,n^2−31$ and $n^2−29$ are 3 consecutive even numbers or 3 consecutive odd numbers
they cannot be 3 consecutive even numbers as all cannot be prime
so they are 3 cosecutive odd numbers.
so one of them is divisible by 3.
so the number has to be 3 otherwise it is not prime.
so 3 numbers are 3 , 5, 7 and $n^2-33 = 3$ giving $n = \pm 6$ or number of values of n = 2
If we choose n as a square say $m^2$ then
$n(n+1) = m^2(m^2+1) = m^4 + m^2$
now n(n+1) is reperesented as sum of 2 squares that is $(m^2)^2 + m^2$
if $m^2$ can be represented as sum of 2 squares that is $p^2 + q^2$ this is possible as as (m,p,q) form a pythagorean triple then we have
$n(n+1) = (p^2+q^2)(m^2+1) = (pm+ q)^2 + (p-qm)^2$ this is another way
if we chose $p=x^2-y^2, q = 2xy, m= x^2+y^2, n = (x^2+y^2)^2 $ then it satisfies the condition (p,q,m) form a pythagrean triple)
When $x^{1000}$ is divided by $x^2-4x + 3$ the raminder shall be a polynomial of degree 1 that is Ax + B
So $x^{1000} = (x^2 - 4x +3) P(x) + Ax + B$ where P(x) is quotient
We need to find A and B
Now $x^2-4x + 3 = (x-1)(x-3)$
So $x^{1000} = (x-1)(x-3) P(x) + Ax + B$
Putting x = 1 we get $1= A + B\cdots(1)$
Putting x = 3 we get $3^{1000} = 3A + B\cdots(2)$
Subtracting (1) from (2) we get 2A = $3^{1000} -1$
Or $A = \frac{3^{1000}-1}{2}$
Putting in (1) we get $B= 1 - A = \frac{1-3^{1000}}{2}$
So remainder = $\frac{3^{1000}-1}{2}x + \frac{1-3^{1000}}{2}$
First let us find the lengths of sides of the trinagle. Let a,b,c be sides of triangle
By triangle inequality we have as $a+b > c$ so $a+b+c > 2c$ or $8 > 2c$ or $ c< 4$
$c = 3 => a + b = 5$ giving a = 3 and b =2 or a =2 and b = 3
$c=2=> a+b=6$ giving a = b= 3
$c=1$ is not possible as it gives $a+b= 7$ give a a =3 , b= 4 invalid triangle
sides 3,3,2 say a = b= 3 and c = 2
and s (semiperimeter) = 8/2 = 4
so if A is area $A^2 = s(s-a)(s-b)(s-c) = 4 * 1 * 1 * 2 = 8$ or Area = $2\sqrt{2}$
We have
$(x-y)^2 >=0$ or $x^2 + y^2 >=2xy$
$(x-1)^2 >=0$ or $x^2 + 1 >= 2x$
$(y-3)^2 >= 0$ or $y^2 + 9 >= 6y$
Adding we get $2(x^2+y^2 + 5) >= 2(xy+x + 3y)$
Or $x^2+y^2 + 5 >= xy + x + 3y$
This is equal when x = y, y = 3, x= 1 which canot be true so $x^2+y^2 + 5 > xy + x + 3y$
Because the product of digits is 10000 so this can be represented as product os single numbers.
Let us findone such number and then see if we canmake it smaller
$10000 = 2 ^ 4 * 5 ^ 4$
So if the number has 4 2's and 4 5's then the product of the digits is 10000. This is an 8 digit number. this can be mad smaller if we can multiply 2 or more one digit number out of these and getting one digit number, multiplying 3 2's we get 8 and digits are 2, 8, 5,5,5 ,5 giviing the product 1000. the smallest number that can be formed is 255558 and this is the answer
We are given
$x+y+ z = 3\cdots(1)$
$x^3 + y^3 + z^ 3 = 3\cdots(2)$
cubing (1) we get $(x+y+z)^3 = 3^3$
or $x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x) = 27$
or $3 + 3(x+y)(y+z) (z+x) = 27$ putting thw vaue of $x^3 + y^3 + z^3 = 3$ from (2)
or $(x+y)(y+z)(z+ x) = 8$
now $((x+y) + (y+z) + (z+ x)) = 2 ( x + y + z) = 6$ usng (1)
we need to find 3 numbers whose sum is 6 and product is 8. They are (2,2,2) or (4,1,1) and other combinations fail
taking x + y = 2 , y + z = 2, z+ x = 2 and using (2) that is x + y + z = 3 we get x=1,y=1,z = 1
taking x + y = 4 , y + z = 1, z+ x = 1 and using (2) that is x + y + z = 3 we get x=5,y=z = - 4
takinn n rotaton other combinations we get x=y = -4, z = 5 as one solution and z = z = -4 and y =5 as another solution
so $\{x,y,z\} = \{2,2,2\}$ or $\{5,-4,-4\}$ (that is any permuation of the same)
