2021/027) Show that for all real numbers $x,\,y,\,z$ such that $x+y+z=0$ and $xy+yz+zx=-3$, the expression $x^3y+y^3z+z^3x$ is a constant.

We have (given)

$x+y+z=0 \cdots(1)$

$xy+yz+zx=-3\cdots(2)$

From (1)

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy +yz + zx) = 0$

or $x^2 + y^2 + z^2 + 2(-3) = 0$

or $x^2 + y^2 + z^2 = 6\cdots(3)$

Further from (1)

$x+y = -z\cdots(4)$

$y+z = -x\cdots(5)$

$z+x = -y\cdots(6)$

Now let us prove that

$x^3y+ y^3 z + z^3 x = xy^3 + yz^3 + zx^3\cdots(7)$

to prove the same

$x^3y+ y^3 z + z^3 x - (xy^3 + yz^3 + zx^3)$

$= (x^3y - xy^3) + (y^3z - yz^3) + (z^3 x - zx^3)$

$= xy(x^2 - y^2) + yz(y^2 - z^2) + zx(z^2 - x^2)$

$=xy(x+y)(x-y) + yz(y+z)( y-z) + zx(z+x)(z-x)$

$=xy(-z)(x-y) + yz(-x) (y-z) + xz(-y) (z-x)$ using (4), (5), (6)

$= - xyz(x-y) - xyz(y-z) - xyz(z-x)$

= 0

so (7) is true

Now $(x^2 + y^2 + z^2)(xy + yz + zx) = 6 * (-3) $ putting values  from above

Or $x^3y + x^2yz + zx^3 + xy^3 + y^3 z + y^2zx + z^2yx + yz^3 + z^3 x = 18$

or $(x^3y + y^3 z + z^3x) + (xy^3 + yz^3 + zx^3 ) + (x^2yz + xy^2z + xyz^2) = - 18$

or  $(x^3y + y^3 z + z^3x) + (x^3y + y^3z + z^3x ) + (x^2yz + xy^2z + xyz^2) = - 18$ (from (7)

or  $2(x^3y + y^3 z + z^3x) + xyz(x+y+z) = - 18$

or $2(x^3y + y^3 z + z^3x) + xyz. 0 = - 18$ from (1)

or $2(x^3y + y^3 z + z^3x)=  - 18$ from (1)

or $(x^3y + y^3 z + z^3x)  = - 9$

Which is a constant

Hence proved

 

2021/026) Find positive integer n for which $2^n + n | 8^n + n$

We have $8^n + n ^3 = (2^n)^3 + n^3$

So

$2^n + n | 8^n + n^3$

using above and the given condition

$2^n + n |  n^3-n$

Now there are 2 cases

1) $n^3-n = 0$ which gives n = 1 ( as other roots n = -1 and 0 are not positive

2) or $n^3-n > 0$ as for positive integer n >=2 ( n= 1 as already considered) 

as $2^n + n | n^3-n$ so we have $n^3- n \ge 2^n + n$

or $n^3 \ge 2^n + 2n$

we can show by induction that for $n > 10 $ $2^n > n^3$ 

so $2 \le n \ le 9$

by checking out the values from 2 to 9 we get

$ n \in \{2,4,6\}$

so we ahve the solution set

$n \in \{ 1,2,4,6\}$ 

2021/025) Solve the following $\lfloor x\lfloor x\rfloor \rfloor = 10$ in positive x

 Let $\lfloor x \rfloor = n$

So we have $ n \le x < n+1$

So 

$ nx \le \lfloor x \lfloor x \rfloor \rfloor <  x(n+1) $

Ss $n^2 \ le nx $ and $(n+1)^2 > x(n+1)$ we have

$ n^2 \le \lfloor x \lfloor x \rfloor \rfloor <  (n+1)^2 $

Or

$ n^2 \le 10 < (n+1)^2 $

And $ 10 \le nx < 11$

Or n= 3 and putting n = 3 we have $\frac{10}{3} \le x < \frac{11}{3}$

2021/024) if $a^3+b^3+c^3=3abc$ and a,b, c are different then prove that $\frac{(b+c)^2}{3bc} + \frac{(c+a)^2}{3ac}+\frac{(a+b)2}{3ab}=1$

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

$\frac{(b+c)^2}{3bc} + \frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}= \frac{a^3+b^3+c^3}{3abc} = \frac{3abc}{3abc}=1$

 

2021/023) Find n when $2^n + 105$ is a perfect square

 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  

2021/022)if $S(n)$ is sum of digits of n find n for which $n + S(n) = 2021$

 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 

2021/021)There are two values of a for which the equation $4x^2 +ax+8x+9 = 0$ has only one solution for x.

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)

2021/020) If a,b,c are three successive terms of an A.P., then prove that$ a^2+8bc=(2b+c)^2$

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 = $a^2 + 8bc = (2b-c)^2 + 8bc = 4b^2 - 4bc + c^2 + 8bc = 4b^2 + 4bc + c^2 = (2b+c)^2$

2021/019) If $A = \log_{12} 18$ and $B = \log_{(24)}54$ then evaluate, AB + 5(A - B)?

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 $

.

2012/018) For integer n if 4 is not a factor of n the prove that $5 | 1^n + 2^n + 3^n + 4^n$

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