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Math Help - how to get a number without a remainder

  1. #1
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    how to get a number without a remainder

    Let A and B be some number, integer or decimal.

    I want to figure out the lowest possible integer X that will yield another integer Y in the following formula.

    (A * X) / B = Y

    Using some concrete numbers, say A = 55 and B = 2.25.

    Then, applying the above formula, you get

    (55 * X) / 2.25 = Y

    By trial and error, I was able to figure out that X has to be 9 in order to yield a value for Y that is another integer.

    This is feels like a common denominator type of a problem, but I can't figure out how to get X without doing a trial and error.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by doolindalton View Post
    Let A and B be some number, integer or decimal.

    I want to figure out the lowest possible integer X that will yield another integer Y in the following formula.

    (A * X) / B = Y

    Using some concrete numbers, say A = 55 and B = 2.25.

    Then, applying the above formula, you get

    (55 * X) / 2.25 = Y

    By trial and error, I was able to figure out that X has to be 9 in order to yield a value for Y that is another integer.

    This is feels like a common denominator type of a problem, but I can't figure out how to get X without doing a trial and error.
    you want y to integer so you want to delete the denominator so

    \frac{55(x)}{2.25} = \frac{5500(x)}{225}

    \frac{5(1100)x}{5(45)} = \frac{1100(x)}{45} = \frac{5(220)x}{5(9)}

    \frac{220(x)}{9} you want to relax from 9 how ?? let x be 9 that's it and when you relax from denominator y will be a integer
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  3. #3
    MHF Contributor
    Grandad's Avatar
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    Dec 2008
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    South Coast of England
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    Hello doolindalton
    Quote Originally Posted by doolindalton View Post
    Let A and B be some number, integer or decimal.

    I want to figure out the lowest possible integer X that will yield another integer Y in the following formula.

    (A * X) / B = Y

    Using some concrete numbers, say A = 55 and B = 2.25.

    Then, applying the above formula, you get

    (55 * X) / 2.25 = Y

    By trial and error, I was able to figure out that X has to be 9 in order to yield a value for Y that is another integer.

    This is feels like a common denominator type of a problem, but I can't figure out how to get X without doing a trial and error.
    First of all, note that this is not always possible. For example, if a = \sqrt2 and b = 1, then there are no integers x, y for which \frac{ax}{b}=y.

    Why? Substituting a = \sqrt2, b = 1 and re-writing the equation, we get

    \frac{\sqrt2}{1}=\frac{y}{x},

    and it is well-known that \sqrt2 is irrational; in other words, it can't be written in the form \frac{y}{x}, where x, y are integers.

    So, when can we solve the problem? If we re-write the equation yet again, we get

    x = \frac{b}{a}\cdot y

    So the answer is that \frac{b}{a} has to be rational. It doesn't mean that individually a and b must be rational - for example, a = \sqrt2 and b = \sqrt8 - but when we divide b by a, the answer must be rational. So that means that we must be able to write

    \frac{b}{a}=\frac{p}{q}, for some integers p, q

    So, assuming that we can do that, we then express \frac{p}{q} in its lowest terms, by 'cancelling' if necessary. Then, with y = q being the smallest possible integer for which \frac{py}{q} is an integer, the resulting value of p gives the smallest possible value of x.

    Let's see how this works with the example you gave: a = 55, b = 2.25:

    \frac{b}{a}=\frac{2.25}{55}=\frac{225}{5500}=\frac  {9}{220}

    So there are our values of p and q: p = 9, q = 220. And there's your answer: x = 9.

    Grandad
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