3
$\begingroup$

I have this problem:

Find three odd consecutive numbers with the property that the product of the first one and the third one minus the product of the first one and the second is greater by eleven than the third one.

I have solved the problem with the following equation:

$(2x+1)(2x+5)-(2x+1)(2x+3)=11 + (2x+5).$

Solution is $x=7$ which gives $15,17,19$ as the requested numbers.

Now If use simpy $x, x+2, x+4$ to denote these numbers I also obtain the same solution with the equation:

$x(x+4)-x(x+2)=11+(x+4)$.

Solution is $x=15$ so the numbers are $15, 17, 19$.

So my doubt is why I didnt have the need to express the numbers as a proper odd number $(2x+1)$ and it works simply with $x$.

Could be the case that there exist some other three (even) numbers that satisfy the conditions and so it's ok to use the proper expression for an odd number. Or can I always use $x, x+2, x+4, x+6$ to denote consecutive odd numbers without getting into any problems.

My question is why I was able to find the same solution using $x, x+2, x+4$ to denote the numbers, why I didn't have to use the proper $2x+1$. This tells me I can always use $x, x+2, x+4..$ to denote consecutive odd numbers to solve this kind of problems. Probably it's just a coincidence.

$\endgroup$
3
  • $\begingroup$ Wolfram alpha actully gives 2 different solutions, wolframalpha.com/input/… and wolframalpha.com/input/?i=x(x%2B4)-x(x%2B2)%3D11%2B(x%2B4) $\endgroup$
    – Ovi
    Jul 22, 2016 at 2:50
  • $\begingroup$ But it is a little weird, if anything I'd actually expect them to give the same solution, not different ones. $\endgroup$
    – Ovi
    Jul 22, 2016 at 2:51
  • $\begingroup$ Oh that was dumb of me because it doesn't matter because they do give the same sequence of numbers, duh. $\endgroup$
    – Ovi
    Jul 22, 2016 at 2:53

1 Answer 1

Reset to default
2
$\begingroup$

Your first equation requires that the numbers be odd if $x$ is a whole number. Your second does not. If you find a whole solution to the first, you are done. In theory, you could find a whole solution to the second in which $x$ was even. You would have to check the solution(s) you find to make sure the numbers were odd. Alternatively, you could look at your second equation and note that if $x$ is even all the terms are even except for the $11$, so the equation will fail. This guarantees that all solutions will have $x$ odd, satisfying the constraint. There is nothing wrong with the second approach if you check the solution(s) you find to be odd, or prove that all solutions are odd. It will find all the solutions, and maybe more.

$\endgroup$
2
  • $\begingroup$ Just a side note: even if you assume $x$ is even and use $2x$, $2x+2$, $2x+4$, it gives $x = \dfrac {15}{2}$, so the same $15, 17, 19$ [WolframAlpha][1] [1]: wolframalpha.com/input/?i=2x*(2x%2B4)-2x*(2x%2B2)%3D11%2B(2x%2B4) $\endgroup$
    – Ovi
    Jul 22, 2016 at 3:03
  • $\begingroup$ @Ovi: we can also solve either system over the reals, then check that the solution results in odd integers. Working over the integers allows new tools, such as divisibility. This problem doesn't need it as it reduces to a linear equation. $\endgroup$
    – Ross Millikan
    Jul 22, 2016 at 3:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .