2016/030) Let $R= (5\sqrt{5} + 11)^{2n+1}$ and $f= R- \lfloor R \rfloor $ where $\lfloor R \rfloor$ is greatest integer functon. Prove that $Rf = 4^{2n+1}$

we have $(5\sqrt{5} - 11) = \frac{(5\sqrt{5} - 11)(5\sqrt{5} - 11)}{5\sqrt{5} + 11}$
$= \frac{125-121}{5\sqrt{5} + 11} = = \frac{4}{5\sqrt{5} + 11} < 1$
$R-r =   (5\sqrt{5} + 11)^{2n+1} -  (5\sqrt{5} - 11)^{2n+1}$
$= \sum \limits_{i=0}^{2n+1}{2n+1\choose i} (5\sqrt{5})^i 11^{2n+1-i} -\sum \limits_{i=0}^{2n+1}{2n+1\choose i} (5\sqrt{5})^i  (-11)^{2n+1-i} $
$= 2\sum \limits_{i=0}^n{2n+1\choose i} (5\sqrt{5})^{2i} 11^{2n+1-1}$
$= 2\sum \limits_{i=0}^n{2n+1\choose i} 125^i 11^{2n+1-1}$ an integer
so $r = R -\lfloor R \rfloor$
So $r = f = (5\sqrt{5} - 11)^{2n+1}$
Hence
$Rf = (5\sqrt{5} +  11)^{2n+1}(5\sqrt{5} - 11)^{2n+1}$
$= ((5\sqrt{5} + 11)(5\sqrt{5} - 11))^{2n+1} = (125-121)^{2n+1} = 4^{2n+1}$

2016/029) Solve $2\log_x a + \log_{ax} a + 3\log_b a =0$ where $a > 0$ and $b=a^2 x$

We have $\frac{2}{\log_a x}  + \frac{1}{\log_a ax} + \frac{3}{\log_a a^2 }x= 0$
$=> \frac{2}{\log_a x} + \frac{1}{\log_a x + 1} + \frac{3}{\log_a x + 2}=0$
$=>  \frac{2}{p } + \frac{1}{1+p } + \frac{3}{2 + p}= 0$ where $p = \log_a x$
$=>2(1+p)(2+p) + p(2+p) + 3p(1+p) = 0$
$=> 4 + 6p + 2 p^2 + 2p + p^2 + 3p + 3p^2 = 6p^2 + 11p + 4$
$ = 6p^2 + 8p + 3p+4 = (3p+4) (2p+1)=0$
$=> p = \frac{-4}{3}$ or $\frac{-1}{2}$
Hence $x= a^{-\frac{4}{3}}$ or $a^{\frac{-1}{2}}$

2016/028) Let $a,b$ be roots of equation

Let $a,b$ be roots of equation $x^2-10cx+11d=0$ and $c,d$ be roots of equation
$x^2-10ax-11b=0$ then find the value of $a+b+c+d$ when a,b,c,d are all distinct.

we have $a,b$ are roots of $x^2-10cx + 11d = 0$
Hence $a+b = 10c\cdots(1)$
$ab= 11d\cdots(2)$
similarly
$c+d= 10a\cdots(3)$
$cd = 11b\cdots(4)$
from (1) and  (3)
$a+b+c+d = 10(c+a)$ or $b+d = 9(a+c)\cdots(5)$
from (2) and (4)
$abcd = 121bd$ or $ac = 121\cdots(6)$
now because a and b satisfy $x^2-10cx + 11d = 0$
we have
$a^2-10ca + 11d = 0$
similarly
$c^2 10ca + 11b = 0$
add to get $(a+c)^2 -2ac - 20ac +11(b+d) = 0$
or $(a+c)^2 - 22 * 121 + 99(a+c) = 0$
So $a + c = - 22\, or\,  121$
$a + c = -22$ means $a = c = -11$  wich is inadmissible
so $a + c = 121$
so $a + b+ c+ d = 10(a+ c) = 1210$

2016/027) Solve for integers $a,b,c$ given $a^2+b^2+c^2 + a + b+ c = 1$

Multiplying by 4 and adding 3 on both sides we get
$(4a^2+4a+1) + (4b^2 + 4 b+ 1) + (4c^2 + 4c+1) = 7$
or $(2a+1)^2 + (2b+1) ^2 + (2c+1)^2   = 7$
as for n odd $n^2 = 1 \, mod \, 8$
we have $LHS = 3\, mod \, 8$ and $RHS = 7 \,mod \, 8$ so no solution

2016/026) find the highest 3 digit prime factor of ${2000 \choose 1000}$

we have
${2000 \choose 1000} = \frac{2000!}{1000!1000!}$
so the prime p occurs $ \lfloor \frac{2000}{p} \rfloor - 2\lfloor \frac{1000}{p} \rfloor $ times
now if p is 3 digit $> \frac{2000}{3}$ or $>666$ then
$ \lfloor \frac{2000}{p} \rfloor = 2 $
$ \lfloor \frac{1000}{p} \rfloor = 1 $
so $ \lfloor \frac{2000}{p} \rfloor - 2\lfloor \frac{1000}{p} \rfloor =0 $
if  is $< 666$
$ \lfloor \frac{2000}{p} \rfloor - 2\lfloor \frac{1000}{p} \rfloor >= 1 $
so largest p is largest prime $< 666$ and it is $661$

2016/025) find the least possible value of a + b where a and b are positive and 11 divides $a+ 13b$ and 13 divides $a + 11 b$

$11$ divides $a + 2b$ and hence $11$ divides $6a + 12b$ or $11$ divides $6a + b$
$13$ divides $a - 2b$ and hence $13$ divides $6a - 12b$ or $13$ divides $6a + b$
so $6a + b$ is divisible by $11$ and $13$ and hence $143$
say $6a +b = 143 t\cdots(1)$
$6a + 6b = 143t + 5b = 144 t + 6b - ((t+b)$
so $t + b$ is divisible by $6$ and hence $t + b > 6 \cdots(2)$
$6(a+b) = 143t + 5b = 138 t + 5(t+b) >=168$
hence $a + b >= 28$
From (2) putting $t = 1$ we get $b= 5$ and from (1) we get $a = 23$ so $a=23$ and $b=5$ satisfies
the condition so $a+b$ lowest value is 28

2016/024)integrate $\int_{x = 0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} dx$

as we have $\int_{x = a}^{b} f(a) = \int_{x = a}^{b} f(b)$
$\int_{x = 0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} dx$
$=\int_{x = 0}^{\pi}\frac{e^{\cos(\pi-x) }}{e^{\cos (\pi-x) }+e^{-\cos (\pi- x) }} dx$
$=\int_{x = 0}^{\pi}\frac{e^{^-\cos x}}{e^{- \cos (x) }+e^{\cos  x}} dx$
$=\int_{x = 0}^{\pi}\frac{e^{^-\cos x}}{e^{\cos (x) }+e^{-\cos  x}} dx$
hence
$\int_{x = 0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} dx= \int_{x = 0}^{\pi}\frac{e^{^-\cos x}}{e^{\cos (x) }+e^{-\cos  x}} dx$
$=\frac{1}{2}\int_{x = 0}^{\pi}(\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} + \frac{e^{^-\cos x}}{e^{\cos (x) }+e^{-\cos  x}})dx$
$=\frac{1}{2}\int_{x = 0}^{\pi} 1 dx = \frac{\pi}{2}$

2016/023) Show that $\lim_{n\to\infty} \frac{1}{n+1} + \frac{1}{n+2}\frac{1}{n+3} +\cdots + \frac{1}{6n} = ln 6$

we have $\lim_{n\to\infty} \frac{1}{n+1} + \frac{1}{n+2}\frac{1}{n+3} +\cdots + \frac{1}{6n}$
$= \lim_{n\to\infty} \sum_{r=1}^{5n} \frac{1}{n+r}$
$= \lim_{n\to\infty}\frac{1}{n} \sum_{r=1}^{5n} \frac{1}{1+\frac{r}{n}}$
$=\int_{x = 0}^{5}\frac{1}{1+x}dx =  \bigl[ln (1+x)\bigr]_1^5 = \ln 6 - \ln 1 = \ln 6 $

2016/022) Show that $a^b+b^a > 1$ for a , b positive

if a or b are above 1  we are done,.
so let us assume both $< 1$
by Bernoulli Inequality  we have
$(1+x)^n \>= 1+nx$
so $(1+\frac{a}{b})^{\frac{1}{a}}  \ge 1+\frac{1}{b}$
so $(1+\frac{a}{b})^{\frac{1}{a}} > \frac{1}{b}$
or $1+\frac{a}{b} > (\frac{1}{b})^a$
or $\frac{a+b}{b} > (\frac{1}{b})^a$
or $\frac{b}{a+b} < b^a$
similarly
$\frac{a}{a+b} < a^b$
Adding, we get $1 < a^b + b^a$
or $a^b+b^a > 1$

2016/021) w,x,y,z are positive real numbers that satisfy :

$xyz+xy+xz+yz+x+y+z=1\cdots(1)$
$wxy+wx+wy+xy+w+x+y=9\cdots(2)$
$wxz +wx+wz+xz+w+x+z=9\cdots(3)$
$wyz +wy+wz+yz+w+y+z=5\cdots(4)$

Solution

from (1)

$xyz+xy+xz+yz+x+y+z+1=2$
or $(x+1)(y+1)(z+1)=2\cdots(5)$
similarly from (2) (3) and (4)
$(y + 1)(x + 1)(w + 1) = 10\cdots(6)$
$(z + 1)(x + 1)(w + 1) = 10\cdots(7)$
$(y+1)(y+1)(w+1) = 5\cdots(8)$
From (5) (6) (7) and (8) we get
$(x+1)(y+1)(z+1)(w+1) = 10\cdots(9)$
dividing (9) by (5), (6), (7), (8) we get
$(w+1) = 5, y+1 = z+ 1 = 1, x + 1 = 2 => x = 1, w = 4, y= z= 0$