0

I have two numbers: 1536 and 2048, I would like to reduce these numbers to as close as 600 as possible while retaining their ratio. How can I achieve this in mathematics?

S-K'
  • 115
  • What exactly do you mean with reduce? – Git Gud Apr 21 '13 at 17:18
  • Keep the current ratio of 1536:2048 but turn the numbers into say 548:600 – S-K' Apr 21 '13 at 17:19
  • 3
    Do you want the ratio to be exactly the same, or do only need it close to the original ratio? – ShreevatsaR Apr 21 '13 at 17:34
  • Anyway for this particular example, as the ratio is $3 : 4$ and both $3$ and $4$ divide $600$, you can make either the earlier or latter number exactly $600$: the ratio $1536 : 2048$ is equal to $450 : 600$ and to $600 : 800$. See my answer for a general method. – ShreevatsaR Apr 21 '13 at 17:54

3 Answers3

4

$\dfrac{1536}{2048}=k$

$\dfrac{x}{600}=k$ or $\dfrac{600}{x}=k$

$\dfrac{1536}{2048}=\dfrac{x}{600}$ or $\dfrac{600}{x}=\dfrac{1536}{2048}$.

Solve for $x$, you will still have ratios constant.

Inceptio
  • 7,881
3

You can continually divide both numbers by a common factor.

$2048:1536 = 1024:768 = 512:384 = 256:192 = 128:96 = 64:48 = 32:24 = 16:12 = 8:6 = 4:3$

Notice how I simply continued to divide both numbers by a common factor of $2$. The closest to 600 with these integers was $512:384$.

If you want exactly $600$ for the first term, you could divide both numbers in the original ratio by $\frac{2048}{600}$, and that would give $600:450$.

If it's the other number that you want to be $600$, then you would divide both sides by $\frac{1536}{600}$ instead, which would give you $800:600$.

Cisplatin
  • 4,675
  • 7
  • 33
  • 58
2

Here is a systematic recipe (algorithm) for doing what you want, assuming you want the ratio to remain exactly the same. (Otherwise, there are solutions involving continued fractions!)

  • First, reduce the given ratio to lowest terms, and call the ratio $p/q$. This involves dividing both the numerator and denominator by their gcd. In other words, if you have a fraction $\frac{a}{b}$, in lowest terms it is $\frac{p}{q} = \frac{a\,/\gcd(a,b)}{b\,/\gcd(a,b)}$.

  • If $q > 600$, stop. You're already as close as you can get.

  • Else, consider the multiples of $q$ closest to $600$. Specifically, you need to consider the two numbers (possibly equal) $l = \lfloor \frac{600}{q} \rfloor$ and $h = \lceil \frac{600}{q} \rfloor$.

  • For each of these two, consider the corresponding ratios. That is, consider the ratios $\frac{lp}{lq}$ and $\frac{hp}{hq}$. Pick whichever is "better" according to you.

The gcd can be found with the Euclidean algorithm. For example, Python code:

def gcd(a, b):
    return abs(a) if b == 0 else gcd(b, a%b)

For the given example $1536/2048$: you have $\gcd(1536, 2048) = 512$, so the fraction in lowest terms is $\frac{1536/512}{2048/512} = \frac{3}{4}$. So $q = 4$, and you want $600 / q$, which is $150$. Multiplying both numerator and denominator by $150$ gives the ratio $\frac{150 \times 3}{150 \times 4} = \frac{450}{600}$, so $450 : 600$ is the ratio with denominator closest to $600$.

(Note: if it's the numerator that you want closest to 600, then do the procedure (finding $l$ and $h$) with $p$ in place of $q$. In this case, that gives $600/3 = 200$, so the ratio with numerator closest to $600$ is $600 : 800$, as $\frac{200 \times 3}{200 \times 4} = \frac{600}{800}$. Also, of course, the whole thing works with any number in place of $600$.)

ShreevatsaR
  • 41,374