, 2015

Since $5$ is an odd integer, then $n$ must be an odd integer for the sum $n+5$ to be an even integer.

(If $n$ was an even integer, then $n+5$ would be the sum of an even integer and an odd integer, which is an odd integer.)
We first note that the product of an even integer and any other integers, even or odd, is always an even integer.

Let $N=cd(c+d)$.

If $c$ or $d$ is an even integer (or both $c$ and $d$ are even integers), then $N$ is the product of an even integer and some other integers and thus is even.

The only remaining possibility is that both $c$ and $d$ are odd integers.

If $c$ and $d$ are odd integers, then the sum $c+d$ is an even integer.

In this case, $N$ is again the product of an even integer and some other integers and so it is an even integer.

Therefore, for any integers $c$ and $d$, $cd(c+d)$ is always an even integer.
Since $e$ and $f$ are positive integers so that $ef=300$, then we may begin by determining the factor pairs of positive integers whose product is 300.

Written as ordered pairs $(x,y)$ with $x<y$, these are: $$(1,300), (2,150), (3,100), (4,75), (5,60), (6,50), (10,30), (12,25), (15,20).$$ It is also required that the sum $e+f$ be odd and so exactly one of $e$ or $f$ must be odd.

Therefore, the factor pairs whose sum is odd are: $$(1,300), (3,100), (4,75), (5,60), (12,25), (15,20).$$ There are 6 ordered pairs $(e,f)$ satisfying the given conditions.
Since both $m$ and $n$ are positive integers, then $2n>1$ and so $2n+m>m+1$.

Let $a=m+1$ and $b=2n+m$ or $a=2n+m$ and $b=m+1$ so that $ab=9000$.

We must first determine all factor pairs $(a,b)$ of positive integers whose product is 9000.

We begin by considering the parity (whether each is even or odd) of the factors $a$ and $b$.

Since 2 is even, then $2n$ is even for all positive integers $n$.

If $m$ is even then $2n+m$ is even since the sum of two even integers is even.

However if $m$ is even, then $m+1$ is odd since the sum of an even integer and an odd integer is odd.

That is, if $m$ is even, then $a$ is odd and $b$ is even or $a$ is even and $b$ is odd.

We say that the factors $a$ and $b$ have different parity since one is even and one is odd.

If $m$ is odd then $2n+m$ is odd. If $m$ is odd then $m+1$ is even.

That is, if $m$ is odd, then $a$ is even and $b$ is odd or $a$ is odd and $b$ is even and so the factors $a$ and $b$ have different parity for all possible values of $m$.

Now we are searching for all factor pairs $(a,b)$ of positive integers whose product is 9000 with $a$ and $b$ having different parity.

Written as a product of its prime factors, $9000=2^3\times3^2\times5^3$ and so $ab=2^3\times3^2\times5^3$.

Since exactly one of $a$ or $b$ is odd, then one of them does not have a factor of 2 and so the other must have all factors of 2.

That is, either $a=2^3r=8r$ and $b=s$, or $a=r$ and $b=8s$ for positive integers $r$ and $s$.

In both cases, $ab=8rs=9000$ and so $rs=\frac{9000}{8}=1125=3^25^3$.

We now determine all factor pairs $(r,s)$ of positive integers whose product is 1125.

These are $(r,s)=(1,1125),(3,375),(5,225),(9,125),(15,75),(25,45)$.

Therefore $(a,b)=(8r,s)=(8,1125),(24,375),(40,225),(72,125),(120,75),(200,45)$, or $(a,b)=(r,8s)=(1,9000),(3,3000),(5,1800),(9,1000),(15,600),(25,360)$.

Since $2n+m>m+1>1$, then the pair $(1,9000)$ is not possible.

This leaves 11 factor pairs $(a,b)$ such that $ab=9000$ with $a$ and $b$ having different parity.

Each of these 11 factor pairs $(a,b)$ gives an ordered pair $(m,n)$.

To see this, let $m+1$ equal the smaller of $a$ and $b$, and let $2n+m$ equal the larger (since $2n+m>m+1$).

For example when $(a,b)=(8,1125)$, then $m+1=8$ or $m=7$ and so $2n+m=2n+7=1125$ or $2n=1118$ or $n=559$.

That is, the factor pair $(a,b)=(8,1125)$ corresponds to the ordered pair $(m,n)=(7,559)$ so that $(m+1)(2n+m)=9000$.

Each of the 11 pairs $(a,b)$ gives an ordered pair $(m,n)$ such that $(m+1)(2n+m)=9000$.

We determine the corresponding ordered pair $(m,n)$ for each $(a,b)$ in the table below (although this work is not necessary since we were only asked for the number of ordered pairs).

$\boldsymbol{(a,b)}$
$\boldsymbol{m+1}$
$\boldsymbol{2n+m}$
$\boldsymbol{(m,n)}$

$(8,1125)$
8
1125
$(7,559)$

$(24,375)$
24
375
$(23,176)$

$(40,225)$
40
225
$(39,93)$

$(72,125)$
72
125
$(71,27)$

$(120,75)$
75
120
$(74,23)$

$(200,45)$
45
200
$(44,78)$

$(3,3000)$
3
3000
$(2,1499)$

$(5,1800)$
5
1800
$(4,898)$

$(9,1000)$
9
1000
$(8,496)$

$(15,600)$
15
600
$(14,293)$

$(25,360)$
25
360
$(24,168)$

There are 11 ordered pairs $(m,n)$ of positive integers satisfying $(m+1)(2n+m)=9000$.