combination algebra
- Thread starter nanase
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Otis
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To start, can you find a single solution? We can then experiment, to see what happens when we alter those numbers. (We need to find the bounds of each.)
Don't hesitate to try stuff (play with the numbers). You might recognize a pattern or realize a relationship that helps out.
?
Don't hesitate to try stuff (play with the numbers). You might recognize a pattern or realize a relationship that helps out.
?
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
As is frequently the case, the problem is not well worded. It should say "solve this problem without any rounding.
As always, translate the problem into mathematical language.
[MATH]\dfrac{x}{100} \cdot y = 400, \text { where } x,\ y \in \mathbb Z, \ 9 < x < 100, \text { and } 99 < y < 1000.[/MATH]
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.
[MATH]x = 99 \implies \dfrac{99}{100} < 1 \implies 400 = \dfrac{x}{100} \cdot y < 1 \cdot y = y \implies 400 < y < 1000.[/MATH]
You have just eliminated 301 possible values of y without any trial and error.
Using similar logic you can deduce [MATH]40 < x < 99.[/MATH]
Finally, you can apply an even more powerful mathematical tool, namely prime factorization.
[MATH]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.[/MATH]
The only numbers that divide 40000 evenly must themselves be divisible only by some combination of powers of 2 and 5.
[MATH]40 < x < 100,\ 2 \ | \ x, \text { and } 5 \ | \ x \implies 50 \text { or } 80.[/MATH]
Obviously x = 50 works. How about x = 80?
[MATH]\dfrac{80}{100} \cdot y = 400 \implies 8y = 4000 \implies y = 500.[/MATH]
Only 2 trial and error efforts are needed.
As always, translate the problem into mathematical language.
[MATH]\dfrac{x}{100} \cdot y = 400, \text { where } x,\ y \in \mathbb Z, \ 9 < x < 100, \text { and } 99 < y < 1000.[/MATH]
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.
[MATH]x = 99 \implies \dfrac{99}{100} < 1 \implies 400 = \dfrac{x}{100} \cdot y < 1 \cdot y = y \implies 400 < y < 1000.[/MATH]
You have just eliminated 301 possible values of y without any trial and error.
Using similar logic you can deduce [MATH]40 < x < 99.[/MATH]
Finally, you can apply an even more powerful mathematical tool, namely prime factorization.
[MATH]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.[/MATH]
The only numbers that divide 40000 evenly must themselves be divisible only by some combination of powers of 2 and 5.
[MATH]40 < x < 100,\ 2 \ | \ x, \text { and } 5 \ | \ x \implies 50 \text { or } 80.[/MATH]
Obviously x = 50 works. How about x = 80?
[MATH]\dfrac{80}{100} \cdot y = 400 \implies 8y = 4000 \implies y = 500.[/MATH]
Only 2 trial and error efforts are needed.
Dr.Peterson
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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.