#### nanaseailie

##### New member

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guys please help me out the easy way to approach this question.

- Thread starter nanaseailie
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guys please help me out the easy way to approach this question.

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Don't hesitate to try stuff (play with the numbers). You might recognize a pattern or realize a relationship that helps out.

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sure thing! it is easy to say that 50% of 800 will give the answer.

Don't hesitate to try stuff (play with the numbers). You might recognize a pattern or realize a relationship that helps out.

I see your tips ! but I'm still wondering if there is a mathematical way in determining the boundary?

or is it just observation and trial and error again?

I can think of $999 as the upper boundary for the price

As

\(\displaystyle \dfrac{x}{100} \cdot y = 400, \text { where } x,\ y \in \mathbb Z, \ 9 < x < 100, \text { and } 99 < y < 1000.\)

This is a problem of exclusion. Much of the exclusion can be done by common sense, but it can also be done by mathematical formalism. For example, we can exclude from possible y values any value that does not exceed 400. If that is not clear to your unassisted intuition, you can find it mathematically as follows.

\(\displaystyle x = 99 \implies \dfrac{99}{100} < 1 \implies 400 = \dfrac{x}{100} \cdot y < 1 \cdot y = y \implies 400 < y < 1000.\)

You have just eliminated 301 possible values of y without any trial and error.

Using similar logic you can deduce \(\displaystyle 40 < x < 99.\)

Finally, you can apply an even more powerful mathematical tool, namely prime factorization.

\(\displaystyle y \in \mathbb Z \text { and } {x}{100} \cdot y = 400 \implies y = \dfrac{40000}{x} \implies x \text { evenly divides } 40000 = 2^6 \cdot 5^4.\)

The only numbers that divide 40000 evenly must themselves be divisible only by some combination of powers of 2 and 5.

\(\displaystyle 40 < x < 100,\ 2 \ | \ x, \text { and } 5 \ | \ x \implies 50 \text { or } 80.\)

Obviously x = 50 works. How about x = 80?

\(\displaystyle \dfrac{80}{100} \cdot y = 400 \implies 8y = 4000 \implies y = 500.\)

Only 2 trial and error efforts are needed.

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When you use mere trial and error, what are you looking for that would result in an "error"? By thinking more deeply as you try things, you can see more powerful methods. In particular, you will be thinking about factors.sure thing! it is easy to say that 50% of 800 will give the answer.

I see your tips ! but I'm still wondering if there is a mathematical way in determining the boundary?

or is it just observation and trial and error again?

I can think of $999 as the upper boundary for the price