No One post any problem !!!!!!
Ok ,so I think, I can post one.
Problem - $\boxed {14}$
Let $n$ is a natural number.How many solution are there in ordered positive integers pairs $(x,y)$ to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
Solution - $13$ (may be I'm correct :? ) At first proof a nice lemma, the product of all divisors of a number $N$ is "$N^{k/2}$" where $k$ is the number of divisors. if we consider only the proper divisors, then according to the condition we get, $\frac{N^{k/2}}{N}=N$ so $k=4$ Now $k=4=(3+1)=(1+1)(1...
you can check one by one. but the another way is that to check only squire number. Now notethat, $x=4$ implies $\lfloor \sqrt {x +15} \rfloor - \lfloor \sqrt x \rfloor ═1$ because $16<19<25$ so $ 4 <\sqrt { 4+15} <5$ now checking one by one squire number you will see that for $x=49$ you can calculat...
Problem $\boxed {12}$
How many different $4 \times 4$ arrays whose entries are all $1$'s and $-1$'s have the property that the sum of entries in each row is $0$ and sum of entries in each column is $0$
we know $1+2+3+...+n = \frac{n(n+1)}{2}$ Let, $S=1^k+2^k+3^k+...+n^k$ We know that for any odd $n$ $a+b|a^n+b^n$ so $1+(n-1)|1^k+(n-1)^{k}$ $2+(n-2)|2^k+(n-2)^{k}$ . . . so all terms of this sequence is divisible by $n$ Now, $1+(n+1)-1|1^k+n^k$ $2+(n+1)-2|2^k+(n-1)^k$ . . . so $S$ is also divisible ...