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Anne is given the equation $1000 / x = y$

Anne know that $x$ is a positive whole number and $y$ is a $2$-digit positive whole number.

How many combinations of numbers of x and y are possible to solve this equation.

A. 3 B. 4 C. 5 D. 6

Is there a quicker method to working this out other than the way I did (testing likely numbers in the range of x being 11 to 100).

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  • $\begingroup$ The positive divisors of $100$ are $1,2,4,5,10,20,25,50,100$. Some clearly don't give a two-digit quotient. $\endgroup$ – André Nicolas Aug 3 '14 at 5:59
  • $\begingroup$ You've included a factoring tag. Do you know the factors of $100$? How many of these factors happen to be $2$-digit numbers? $\endgroup$ – Adriano Aug 3 '14 at 6:00
  • $\begingroup$ This question (at least now) uses 1000 rather than 100 $\endgroup$ – Thoth19 Aug 3 '14 at 6:36
  • $\begingroup$ Yes, sorry, I made a typo. It should have read 1000 rather than 100. $\endgroup$ – user1495024 Aug 4 '14 at 4:58
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Starting from the prime decomposition of 1000,

$1000 = 10 * 10 * 10 = (2*5)^3=2^3*5^3$

X is some combination of the prime factors (2 and 5) which can each occur 0 to 3 times. If $ x=2^a*5^b $ then $y=1000/x =2^(3-a)*5 (3-b)$ (where a and b are integers on the range [0, 3])

For example $ x=2^3*5^1=40 $ and $ y=2^0*5^2=25 $ is a solution while $x=2^2*5^0=4 $ and $ y=2^1*5^3=250 $ is not.

Using this method, there's five solutions: x={ 20, 25, 40, 50, 100} and y={50, 40, 25, 20, 10}.

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Factors of $100$ will be made up of $2$ once, twice or not at all, and $5$ once, twice or not at all. There are therefore $3\times3=9$ factors altogether. One of these is $10$; of the other eight, four are less than $10$ and therefore too small; and $100$ is too big. There remain four possible values for $y$.

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