2016/062) If $x= 2 + \sqrt[3]{2^2} + \sqrt[3]{2}$ then the value of $x^3 - 6x^2+6x$ is

we have  $x-  2 =  \sqrt[3]{2^2} + \sqrt[3]{2}$
hence $(x-2)^3 = 4 + 2 + 3 * 2 *(\sqrt[3]{2^2} + \sqrt[3]{2}) = 6 + 3 * 2 (x-2) = 6x - 6$
Hence $x^3 - 6x^2 + 12x -8 = 6x-6 $ or $x^3-6x^2+6x = 2$

2016/061)Find the smallest positive integer m such that 5m is an exact 5th power, 6m is an exact 6th power, and 7m is an exact 7th power. (26th Irish)

The number has to be of the form $5^a6^b7^c$
now $5^{a+1}6^b7^c$ is a 5th power so  $a+1 \equiv 0 \pmod 5$ ,$b \equiv 0 \pmod 5$,$c \equiv 0 \pmod 5$
$5^a6^{b+1}7^c$ is a 6th power so  $a \equiv 0 \pmod 6$ ,$b+1  \equiv 0 \pmod 6$,$c \equiv 0 \pmod 6$
$5^a6^b7^{c+1}$ is a 7th power so  $a \equiv 0 \pmod 7$ ,$b \equiv 7 \pmod 5$,$c+1 \equiv 0 \pmod 7$
so we need to solve for
$a+1 \equiv 0 \pmod 5$ ,$a \equiv 0 \pmod 42$ giving a = 84 (taking multiples of 42 adding 1 to be divsible by 5)
$b+1 \equiv 0 \pmod 6$ ,$b \equiv 0 \pmod 35$ giving b = 35 (taking multiples of 35 adding 1 to be divsible by 6)
$c+1 \equiv 0 \pmod 7$ ,$c \equiv 0 \pmod 30$ giving c = 90 (taking multiples of 30 adding 1 to be divsible by 7)
so the number is $5^{84}* 6^{35}*c^{90}$

2016/060) Find all integers x such that $x(x+1)(x+7)(x+8)$ is square of an integer (21st Irish math olympiad)

we see that x = 0, x = -1, x = -7 and x = -8 gives the answer zero so a perfect square
let us look for other values
we have
$x(x+1)(x+7)(x+8)$
= $x(x+8)(x+1)(x+7)$
= $(x^2+8x)(x^2+8x+7)$
$= y(y+7)$ where y is $x^2+8x$
for it to be a perfect square we see that $(GCD(y,y+7) = GCD(y,7)$
y cannot be a multiple of if 7 beacuse then y and y + 7 are consecutive multiples of 7 and as y is not zero product cannot
be a perfect square.
so y and y + 7 are coprimes and hence perfect square or -ve of perfect square
let $y = n^2$ and $y+7 = m^2$
giving $n^2+7=m^2$
or $m^2-n^2 = 7$
or $(m+n)(n-n) = 7 * 1$ hence $m+n = 7, m-n= 1$ or $m= 4,n= 3$
so $y = 16$
hence $x^2+8x-9=0$ giving $x = 1,=9$
taking -ve values we have $y= - 16 , y + 7 = - 9$
or $x^2+8x+ 16= 0 => x = - 4$
so we have x is one of -9,-8,-7, -1,0,1$

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 $x= \frac{75}{31}$, $y= \frac{30}{31}$,$z= \frac{50}{31}$