2016/059) Show that in 10 consecutive numbers there is at least one number which is co-prime to other 9 numbers

Out of 10 consecutive numbers 5 numbers are divisible by 2 and not more that 4 numbers are divisible by 3 out of which maximum 2 numbers are odd and divisible by by 3 that makes 7, 2 numbers are divisible by 5 out of which is even so there are maximum one number is divisible by 5 and neither 2 nor 3 and that nakes 8 and maximum 2 numbers are divisible by 7 out of which is only one is odd maximum one number is divisible by 7 and neither 2 nor 3 nor 5 and that makes maximum 9 numbers that are divisible by one of 2,3,5,7. so there is at least one number which is not divisible by 2,3,5 or 7 so the lowest prime factor of the same is 11 and it cannot devide any other number of the set. so this number is co-prime to rest 9.

2016/058) Show that $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots\frac{1}{2n-1}$

=$1-\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\cdots -\frac{1}{2n-1}$
 We have RHS = $\sum_{k=1}^{n}\frac{1}{2k-1} - \sum_{k=1}^{n}\frac{1}{2n}$ $=\sum_{k=1}^{n}\frac{1}{2n-1} + \sum_{k=1}^{n}\frac{1}{2k} - 2 \sum_{k=1}^{n-1}\frac{1}{2k}$ $=\sum_{k=1}^{2n-1}\frac{1}{k}- \sum_{k=1}^{n-1}\frac{1}{k}$ $=\sum_{k=n}^{2n-1}\frac{1}{k}=LHS$

2016/057)If a,b,c and d are in G.P then show that $(b-c)^2+(c-a)^2+(d-b)^2=(a-d)^2$

a,b,c,d are in GP let common ratio be x $b = ax,c = ax^2,, d = ax^3$ LHS = $(ax-ax^2)^2 + (ax^2-a)^2 + (ax^3- ax)^2$ $= a^2(x^2 (1-x)^2) + (x^2-1)^2 + x^2(x^2-1)^2)$ $= a^2(x^2(1-2x+x^4) + ( 1- 2x^2 + x^ 4) + x^2(x^4-2x^2+ 1)$ $= a^2(1- 2x^3+x^6)$ $= a^2(1-x^3)^2$ $= ( a- (ax^3))^2$ $= (a-d)^2$

2016/056) If $n^4+an^3+bn^2-8n+1 $ is a perfect square for all integer values of n. find a,b

$n^4+an^3+bn^2-8n+1 $ is a perfect square for all integer values of n so this is square of a polynomial.
hence
$n^4+an^3+bn^2-8n+1 = (n ^2 + mn + c)^2 $
comparing the constant term we have $1=c^2$ or $c = \pm 1$
so we need to consider 2 cases
c = 1 and c= -1
case 1
c = -1
this give $(n^2 + mn + 1)^2 = n^4 + mn^3 + n^2(m^2+2) + mn +1 = n^4+an^3+bn^2-8n+1$
comparing coeffcients on both sides $m = a, b = m^2 + 2, - m = -8$ giving $a= 8, b = 66$
case 2
c = -1
this give $(n^2 + mn - 1)^2 = n^4 + mn^3 + n^2(m^2-2) - mn +1 = n^4+an^3+bn^2-8n+1$
comparing coefficients on both sides $m = a, b = m^2 - 2, - m = -8$ giving $a= -8, b = 62$
so we have 2 sets $a=8, b= 62$ and $a= -8,b=66$

2016/055) what is the smallest number with 101 factors

a number of the form $p_1^{q_1} * p_2^{q_2} * p_3 ^{q_3}\cdots$ has $(1+q_1)(1+q_2)(1 + q_3) \cdots$ factors
now 101 is prime so smallest number = $2^{100}$

2016/054) Let $a,b,c$ be rational and one of the roots of $ax^3+bx+c=0$ is equal to product of other two roots. Prove that this root is rational.

we can devide by a to get $x^3+dx+e=0$ where $d=\frac{b}{a},d=\frac{c}{a}$ $d,e$ are rational.
Let $\alpha,\beta,\alpha\beta$ be the three roots
so we get using vieta's relations
$\alpha+\beta+\alpha\beta= 0\cdots(1)$
$\alpha\beta+\alpha \alpha\beta + \beta\alpha\beta  = d\cdots(2)$
$\alpha\beta\alpha\beta= \alpha^2\beta^2= -e\cdots(3)$
from (2)
$\alpha\beta+ \alpha\beta(\alpha+\beta)  = d$
or $\alpha\beta+ \alpha\beta(\alpha+\beta)  = d$
or $\alpha\beta+ \alpha\beta(-\alpha\beta)  = d$ (Using (1)
or $\alpha\beta - (\alpha\beta)^2  = d$
or $\alpha\beta + e  = d$
or $\alpha\beta = d - e$
hence the root  $\alpha\beta$ is rational

2016/053) Solve the system of equations in real

$4x^2+25y^2 +9z^2 - 10xy -15yz - 6xz = 0\cdots(1)$
$x+y+z=5\cdots(2)$

Solution
from (1) we have
$8x^2+50y^2 +18z^2 - 20xy -30yz - 12xz = 0$
or  $(4x^2 - 20xy + 25y^2) + (25y^2 - 30yz + 9z^2) + (9z^2 - 12xz + 4x^2)= 0$
or $(2x-5y)^2 + (5y-3z)^2 + (3z-2x)^2=0$
above is sum of 3 squares and hence each of them is zero ior $2x = 5y = 3z=k$ (say)
so $x= \frac{k}{2}$, $y= \frac{k}{5}$,$z= \frac{k}{3}$
putting in (2) we get
$\frac{k}{2} + \frac{k}{5}+  \frac{k}{3} = 5$
or $\frac{31k}{30} = 5$
or  $k = \frac{150}{31}$
so $x= \frac{75}{31}$, $y= \frac{30}{31}$,$z= \frac{50}{31}$