Isn't 100000 such a huge number (Part 2)?

Generally speaking, when $a$ and $b$ are positive integers and $\frac{1}{a} + \frac{1}{b} = \frac{1}{M}$ for some positive integer $M$ , then $a > M$ and $b > M$

Let $a = M + p$ and $b = M + q$ where $p$ and $q$ are positive integers

From $M(a + b) = ab$ we have $M(2M + p + q) = (M + p)(M + q)$ or $pq = M^2$

Therefore the number of solutions equals to the number of divisors of $M^2$

Since $M = 100000 = 2^5 \times 5^5 \Rightarrow M^2 = 2^{10} \times 5^{10}$ and the number of its divisors is $(10 + 1) \times (10 + 1) = \boxed{121}$

$\begin{aligned} & & \frac{1}{a} + \frac{1}{b} &= \frac{1}{100000}\\ &\Longrightarrow & ab - 100000a - 100000b &= 0\\ &\Longrightarrow & (a-100000)(b-100000) &= 10^{10} = 2^{10} \cdot 5^{10} \end{aligned}$

It is readily seen that there are $(10+1)(10+1) = 121$ such pairs.

(a+b)/ab=1/100000 =>100000(a+b)=ab =>ab-100000a-100000b=0 multiplying coefficients of a and b we get, 10^10. now add 10^10 on both sides, we get (a-100000) (b-100000)=10^10 now, (a-100000) * (b-100000)= 100000 * 100000=10^10=(2 5)^10=2^10 5^10. so, no.of factors=11 * 11=121 factors=121/2=60 2 +1=121 natural no. solutions

Substitute c for 100000 to make the following simpler.

First, multiply both sides by $abc$ .

You get: $ac+bc=ab$

Now, there is only one solution if $a=b$ , namely $a=b=2c=200000$ . So, putting aside that solution, without loss of generality we can assume $b>a$ .

In that case, b must be greater than 2c. Let's call this surplus x, which must be positive: $b=2c+x$ . Substitute that into our equation: $ac+2c^2+cx=2ac+ax$ $2c^2+cx=ac+ax$ Divide by $x+c$ $\frac{2c^2+cx}{x+c}=a$

As a is integral, $2c^2+cx$ must be divisible by $x+c$ .

As $c^2+cx$ is divisible by $x+c$ , we can subtract that from $2c^2+cx$ to get $c^2$ , which must be divisible by $x+c$ . Putting 100000 back for c, the question becomes "how many factors of 10000000000 are greater than 100000?" I then used wolframalpha , and counted manually to get 60. Each of these had a complement where $a>b$ , plus the one where $a=b$ , for a total of $\boxed{121}$ .

$\frac { 1 }{ a } +\frac { 1 }{ b } =\frac { 1 }{ { 10 }^{ 5 } } \\ \frac { a+b }{ ab } =\frac { 1 }{ { 10 }^{ 5 } } \\ ab-{ 10 }^{ 5 }a-{ 10 }^{ 5 }b=0\\ (a-{ 10 }^{ 5 })(b-{ 10 }^{ 5 })={ 10 }^{ 10 }$ Many positive factors from $10^{10} = 2^{10} \times 5^{10}$ is $(10 + 1)(10 + 1) = 121$ .

that form is same with ab-10^5a-10^5b We can make that form into this one :

(a-10^5)(b-10^5)=10^10

10^10=2^10 x 5^10

For M = p^a . p^b.......... p^n where p is prime number we can find the positive factor by (a+10)(b+1) = 11 x 11 = 121