Show that the number of $5$-tuples $(a, b, c, d, e)$ such that $abcde$ = $5(bcde + acde + abde + abce + abcd)$ is odd Show that the number of 5-tuples $(a,b,c,d,e)$ of positive integers such that $$abcde=5(bcde+acde+abde+abce+abcd)$$ is odd.
 A: The given relation is $$\frac15 = \frac1a+\frac1b+\frac1c+\frac1d+\frac1e$$
When $a=b=c=d=e=25$, this is obviously satisfied, and this gives the only n-tuple when all the numbers are the same.  
Suppose there is a solution where exactly one number is distinct.  WLOG then we have
$$\frac15 = \frac1a + \frac4b  \implies a > 5, b > 20$$
So let $a = 5+x, b =20+y$ where $x, y$ are natural numbers.  Simplifying the equation, we get $xy = 100 = 2^2 5^2$, which must have $3^2=9$ solutions.  Discarding the solution which would equate $a=b$, we are left with $8$ solutions for $(a, b)$, and hence $8 \times 5 = 40$ n-tuples.  
Suppose there is another solution, in which $k>1$ of the numbers are distinct.  Then that contributes through permutations $\dfrac{5!}{(5-k)!}$ an even number of n-tuples.  Hence the total number of solutions remains odd.  
Similarly for other patterns - viz. any solution of form $\frac1a+\frac2b+\frac2c$ would generate $\dfrac{5!}{2!2!} = 30$ n-tuples which is even, and $\frac2a+\frac3b$ would generate $\dfrac{5!}{2!3!} = 10$ n-tuples which is also even.  This covers all cases.
A: Let's look at the number of occurrences of each number. The number of times the numbers occur must be one of the following: $\{(1,1,1,1,1), (1,1,1,2), (1,2,2), (1,1,3),(2,3),(1,4),(5)\}$. Now, for each distribution, we can calculate the number of ways we can permute the values of $a$, $b$, $c$, $d$ and $e$. For all possibilities, except for $(1,4)$ and $(5)$, that number is even, so it doesn't contribute to the 'oddness' of the total number of solution. Now, we only have to look at solutions where all numbers are the same or where only one number is different. I suppose the number of solutions will be odd in one of these cases and even in the other case, but I think you can solve that for yourself.
(I hope the first line is clear to you, if not, please ask for clarification.)
A: that is obviously !
if one of them is an even number, for example $a$ is even , then :
$abcde$ is even 
but $5(bcde+acde+abde+abce+abcd)=odd\cdot(odd+even+even+even+even)=odd$
then $odd=even$ , a contradiction !
if there exists two numbers are both even, then :
we can assume $a,b$ are even numbers , then :
$ab(cde-5de-5ce-5cd)=5(bcde+acde)$$\Longrightarrow$$\frac{cde}{5}-de-ce-cd=\frac{cde}{a}+\frac{cde}{b}$$\Longrightarrow$
$de+ce+cd=cde(\frac{1}{5}-\frac{1}{a}-\frac{1}{b})$$\Longrightarrow$$even=odd$
a contradiction holds !
if three numbers are even, then :
$de(\frac{1}{5}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c})=e+d$$\Longrightarrow$$even=odd$
...... the same methods to four even numbers and five even numbers !
