2019/004) If $\frac{a}{b+ c} + \frac{b}{c+a} + \frac{c}{a+b} = 1$ then prove that $\frac{a^2}{b+ c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 0$

We are given
$\frac{a}{b+ c} + \frac{b}{c+a} + \frac{c}{a+b} = 1$
Multiplying both sides by (a+b+c) we get
$\frac{a(a+b+c)}{b+ c} + \frac{b(a+b+c)}{c+a} + \frac{c(a+b+c)}{a+b} = a+b+c$
Or
$\frac{a^2}{b+ c} + a + \frac{b^2}{c+a} + b + \frac{c^2}{a+b} + c= a+b+c$
Or
$\frac{a^2}{b+ c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 0$

2019/003) if $a^2=b+c$, $b^2=c+a$, $c^2 = a+b$ find $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}$

We have
$a^2=b+c$
Hence $a^2+a= a+b+c$
Hence $a(a+1) = a + b+ c$
or $\frac{1}{a+1} = \frac{a}{a+b+c}\cdots(1)$
Similarly $\frac{1}{b+1} = \frac{b}{a+b+c}\cdots(2)$
and $\frac{1}{c+1} = \frac{c}{a+b+c}\cdots(3)$

adding these 3 we get

  $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = \frac{a}{a+b+c} +  \frac{b}{a+b+c} + \frac{c}{a+b+c}= \frac{a+b+c}{a+b+c}=1$

2019/002) Given $x^5+\frac{1}{x^5}$ Show that $x+\frac{1}{x}=3$ where x is real

let $x + \frac{1}{x} = y$ 
so cube both sides 
$x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = y^3$ 
or $x^3 + \frac{1}{x^3} = y^3-3y$ 
and $x^2+ \frac{1}{x^2} = y^2 -2$ 
multiply both to get 
$x^5 + \frac{1}{x} + x + \frac{1}{x^5} = 123 + y = (y^3-3y)(y^2-2) = y^5 - 5y^3 + 6y$ 
or $y^5 - 5y^3 + 5y - 123 =0$ 
or $(y-3)(u^4 + 3y^3 + 4y^2 + 12y + 41) = 0$ 
gives y = 3 or complex solutions 
so $x+ \frac{1}{x} = 3$ if x is real

2019/001) Find smallest n such that n! ends with 2019 zeroes.

n! ends with k zeroes if is is divisible by $2^k$ and $5^k$ but not with $2^{k+1}$ or $5^{k+1}$.
in the product the power 2

but in n! there are lot many 2 than 5 so it should be divisible by $5^k$  but not $5^{k+1}$.
$\lfloor\frac{n}{5}\rfloor$ numbers shall be divisible by 5
$\lfloor\frac{n}{5^2}\rfloor$numbers shall be divisible by $5^2$
$\lfloor\frac{n}{5^3}\rfloor$ numbers shall be divisible by $5^3$

so number of times 5 appears in product n ! = $p(n) =  \sum_{k=1}^{\inf}\lfloor\frac{n}{5^k}\rfloor$

this shall stop when $5^k > n$ so finite number is elements are to be taken.

We need to find an estimate for n and then correct the same.

Based on the we need to estimate n and then correct the same.

out of 25 consecutive numbers 5 numbers are divisible by 5 and one of those 5 is divisible by one more 5 so out of 25 so product of 25 consecutive numbers so 25 consective number product is divisble by $5^6$

so estimate for n = $\frac{2019*25}{6} = 8412.5$ so take 8410 the highest multiple of  5 less than 8412.

so n = 8410

now p(8410) = 2100

so we have overshot by 2100- 2019 = 81

so n should be reduced by $\frac{81*25}{6} = 337$

giving n -= 8410 - 337 = 8073 or highest multiple of 5 below it 8070.

n = 8070 and p(8070) = 2014

we fall short by 5 and next we take P(8075) = 2016 and P(8080) = 2017, P(8085) = 2018, P(8090) = 2019

so correct n = 8090

TITBIT001) Some interesting property of 2019.

Happy New Year 2019.  I have not solved a  problem here and I shall be starting shortly.Here I provide some interesting property of 2019.

2019 is the smallest number that can be expressed as sum of 3 squares of prime number in 6 different ways

2019 = $7^2+11^2+43^2$
        = $7^2 + 17^2 + 41^2$
        = $13^2+13^2 + 41^2$
        =  $11^2+23^2+37^2$
        = $17^2 + 19^2 + 37^2$
        = $23^2+23^2+31^2$