# Solving $\frac{1}{a}+\frac{1}{b}=\frac{1}{200}\;$

How many ordered pairs of integers $(a,b)$ are there such that $$\frac{1}{a}+\frac{1}{b}=\frac{1}{200}\;?$$

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–  lab bhattacharjee Feb 20 '13 at 10:33

## 2 Answers

Taking LCM and simplifying yeilds $\left(a-200\right)\left(b-200\right)=200^2$

$200^2=2^6\cdot 5^4$ The total number of solution of the equation would be number of ways in which we can split $2^6\cdot 5^4$ into two numbers(product of the two numbers is $2^6\cdot 5^4$)

Which would be the number of non-negative integer solutions to $x+y=6$(exponent of $2$) which is $7$ multiplied by number of non-negative integrer solutions to $x+y=4$(exponent of $5$) which is $5$ Hence the final answer would be $\boxed{35}$

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Hint: $(a-200)(b-200)=200^2$

If $\frac1a+\frac1b=\frac1k\implies (b-k)(a-k)=k^2\implies (a-k)\mid k^2$

But as $a\le b,a-k\le b-k\implies (a-k)^2\le (b-k)(a-k)=k^2$

$\implies -k\le a-k\le k$

If $k=200, -200\le a-200\le 200$

The positive divisors of $200^2$ are $(1,5,5^2,5^3,5^4)\cdot(1,2,2^2,2^3,2^4,2^5,2^6)$

If we take $5^4=625>200$

If we take $5^3=125<200$ the $2^r\le \frac{200}{125}\implies 0\le r\le0\implies r=0$

If we take $5^2=25$ the $2^r\le \frac{200}{25}\implies 0\le r\le3$

If we take $5$ the $2^r\le \frac{200}5\implies 0\le r\le5$

If we take $5^0$ the $2^r\le \frac{200}1\implies 0\le r\le6$

So, we have $1+4+6+7=18$ positive divisors of $200^2$ which are $\le 200$

If we include negative divisors, there will be $18\cdot2=36$ of them

Though $0\mid200^2$ if $a-200=0,b-200=\frac{200^2}0$

Hence, there should be $36$ ordered pairs.

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How many ordered pairs can be there?? –  user62947 Feb 20 '13 at 10:10
@ManojPandey, please find the edited answer. –  lab bhattacharjee Feb 20 '13 at 10:11
@Ethereal, have gone through the last part of the answer: "If we include negative divisors"? –  lab bhattacharjee Feb 20 '13 at 10:44
Why do you insist that the factor be less than $200?$. Factoring as $625 \cdot 64$ gives $a=825, b=264$ and $\frac 1{825}+\frac 1{264}=\frac 1{200}$ just fine. We are looking for ordered pairs. –  Ross Millikan Mar 23 '13 at 17:00