We have $a^3+b^3+c^3=3abc$ so a=b=c or a+b+c = 0
As a,b,c are different so a+b + c = 0
Hence
$\frac{(b+c)^2}{3bc}= \frac{(-a)^2}{3bc}=\frac{a^3}{3abc}$
Similarly we have
$\frac{(c+a)^2}{3ac}= \frac{b^3}{3abc}$
$\frac{(a+b)^2}{3ab}= \frac{c^3}{3abc}$
Adding these 3 we get
$\frac{(b+c)^2}{3bc} + \frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}= \frac{a^3+b^3+c^3}{3abc} = \frac{3abc}{3abc}=1$
Let $2^n + 105 = p ^2 $
Let us work mod 3. for a number to be perfect square it as to 1 or zero mod 3.
So $2^n + 105 \equiv 0 \pmod 3 $ or $2^n + 105 \equiv 0 \pmod 3 $
or $2^n \equiv 0 \pmod 3 $ or $2^n \equiv 1 \pmod 3 $
but $2^n \equiv 0 \pmod 3 $ is not possible
$2^n \equiv 1 \pmod 3 $
Now $2^x \equiv 1 \pmod 3 $ if x is even and -1 if x is odd
so n is even say 2m
so $2^{2m} + 105 = p^2$
or $p^2- 2^{2m} = 105$
or $(p+2^m)(p - 2^m ) = 105 = 105 * 1 = 35 * 3 = 15 * 7 = 21 * 5$
Now $(p + 2^m) - (p- 2^m) = 2^{m+1}$ a power of 2
from the above the combination we get
$p+2^m = 35$ $p-2^m= 3$ giving p =19 and m = 4 or n = 8
Or
$p+2^m = 15$ $p-2^m= 7$ giving p =11 and m = 2 or n = 4
Or $p+2^m = 21$ $p-2^m= 5$ giving p = 13 and m = 3 or n = 6
so n = 4 or 6 or 8
As $n+ S(n) = 2021$ so $n < 2021$.
Now highest $S(n)$ for number $n < 2021$ is 28 that is when
n = 1999.
so $ n >= 2021-28$ or $n>=1993$
Now working mod 9 we have $ n \equiv S(n) \pmod 9$
So $n + S(n) \equiv 2n \pmod 9$
so $2n \equiv 2021 \pmod 9$
or $2n \equiv 5 \pmod 9$
as 5 is odd add 9 to get even
so $2n \equiv 14 \pmod 9$
or $n \equiv 7 \pmod 9$
so we need to check for candidates between 1993 and 2021 which are 7 mod 9 and they are 1996, 2005,2014 out of which 1996 and 2014 satisfy the condition
What is the sum of these values of a? (A) -16 (B) -8 (C) 0 (D) 8
say 8+ a =m
$4x^2+ mx + 9$ has only one solution for x only if $4x^2+ mx + 9$ is a perfect square
or $4x^2+ mx + 9= (2x\pm 3)^3$
Taking $(2x+3)^2$ we get m = 12 or 8+a = 12 or a = 4
Taking $(2x-3)^2$ we get m = -12 or 8+a = -12 or a = -20
so sum of a = -16 so ans (D)
Because a, b, c are 3 successive terms of an A.P. so we have
b- a = c- b or a = 2b - c
now LHS = $a^2 + 8bc = (2b-c)^2 + 8bc = 4b^2 - 4bc + c^2 + 8bc = 4b^2 + 4bc + c^2 = (2b+c)^2$
We have
$A = \frac{\log18}{\log12} = \frac{log\,2 + 2\log\,3}{2\log\, 2 + \log\, 3}$
$A = \frac{\log 54}{\log 24} = \frac{log\,2 + 3\log\,3}{3\log\, 2 + \log\, 3}$
putting $x=\log\, 2$ and $y=\log\,3$ we get
$A= \frac{x+2y}{2x+y}$
and $B= \frac{x+3y}{3x+y}$
So
AB + 5(A - B)
= $\frac{(x+2y)(x+3y)}{(2x+y)(3x+y)} + 5(\frac{(x+2y)(3x+y) - (2x+y)(x+3y)}{ (2x+y)(3x+y)})$
= $\frac{[(x^2+5xy+6y^2) + 5(x^2 - y^2)]}{ (2x+y)(3x+y)}$
= $\frac{6x^2 + 5xy + y^2}{x^2 + 5xy + y^2} =1 $
.
4 is not a factor of n so there are 2 cases
1) n is odd 2) n is even of the form 2(2m + 1) where m in integer.
4 is not a factor of n so there are 2 cases
1) n is odd 2) n is even of the form 2(2m + 1) where m in integer.
Let us deal both the cases one by one
1) Case 1: n is odd
We have rearranging the terms $1^n + 2^n + 3^n + 4^n = (1^n + 4^n) + (2^n + 3^n) $
if n is odd then we have (a+b) is a factor of $a^n+ b^n$ so we get
(1+4) is a factor of $1^n + 4^n$ or 5 is a factor of $1^n + 4^n$
(2+3) is a factor of $2^n + 3^n$ or 5 is a factor of $2^n + 3^n$
adding the above 2 we get $5 | 1^n + 2^n + 3^n + 4^n$
2) case 2: n is even of the form 2(2m + 1)
$1^{2 * (2m+1)} + 2^{2 * ( 2* (2m + 1)} + 3^{2 * (2m + 1)} + 4 ^ {2* (3m+1)} $
$= (1^2)^{2m+1} + (2^2)^{2m + 1} + (3^2)^{2m + 1} + (4 ^ 2)^{3m+1} $
$= 1^{2m+1} + 4^{2m + 1} + 9^{2m + 1} + 16^{3m+1} $
using the arguments is case 1 we know
1 + 4 or 5 is a factor of $1^{2m+1} + 4^{2m + 1}$
And 9 + 16 or 25 is a factor of $9^{2m + 1} + 16^{2m+1} $ and hence 5 is a factor of $1^{2m+1} + 4^{2m + 1} + 9^{2m + 1} + 16^{3m+1} $ or the original expression.
hence proved
as we have proved both parts so we have proved the same