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
$(w+1) = 5, y+1 = z+ 1 = 1, x + 1 = 2 => x = 1, w = 4, y= z= 0$