1
$\begingroup$

Let the number $A$ have an unknown, positive value. The numbers $B$ and $C$ have defined values which are known. If we keep subtracting $B$ from $A$ as long as the difference is positive we get some smallest positive value. Let $C$ be the sum of all such positive differences.

As an example, let $$A = 7.2,\quad B = 0.5.$$ The largest multiple of $B$ we can subtract from $A$ and remain positive is $14B$. Hence, $$C = \sum_{k = 1}^{14} (A - kB) = 48.3.$$

Given $B$ and $C$, how can I express $A$ with a formula? Can't figure it out.

$\endgroup$
  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos May 5 '18 at 10:38
  • $\begingroup$ Thanks for your edit $\endgroup$ – Adelka Smrčková May 5 '18 at 10:41
  • $\begingroup$ Your wording is confusing. I can subtract B from A for as long as I want and A will still be positive. And $(x - 0.5) + (x - 1) + (x - 1.5) + (x - 2)...$ implies that the series never stops. $\endgroup$ – steven gregory May 5 '18 at 10:59
  • 1
    $\begingroup$ Presumably from your example, what you are actually saying is that you subtract all positive multiples of $B$ from $A$ and then add up all the positive results of these subtractions to give $C$. $\endgroup$ – Henry May 5 '18 at 22:41
0
$\begingroup$

I have a partial answer.

As mentioned in the comments, your wording is a little confusing. First you say that $C$ is the smallest positive number you can get by subtracting multiples of $B$ from $A$, but then you say that $C$ is the sum of all such positive differences. I will answer the latter, since it agrees with your example.

I will also assume that $B$ is positive.

We can subtract integer multiples of $B$ from $A$ until $A - kB$ is negative. The "last" one before that happens is $A - mB$, where $$A - (m + 1)B \leq 0 < A - mB.$$ These inequalities are equivalent to $$A - B \leq mB < A,$$ or $$\frac{A}{B} - 1 \leq m < \frac{A}{B}.$$ These inequalities are equivalent to $$m = \left\lceil \frac{A}{B} - 1 \right\rceil = \left \lceil \frac{A}{B} \right \rceil - 1,$$ where $\lceil x \rceil$ is the ceiling of $x$, the smallest integer greater than or equal to $x$.

Therefore, from your definition, $$C = \sum_{k = 1}^m (A - kB),$$ where $m = \lceil A / B \rceil - 1$. This sum evaluates nicely: $$C = mA - B \frac{m (m + 1)}{2}.$$ Assuming that $C > 0$, we will have $$A = \frac{C}{m} + B \frac{m + 1}{2} = \frac{C}{\lceil A / B \rceil - 1} + B \frac{\lceil A / B \rceil}{2}.$$

I am not immediately sure how to solve this equation for $A$, or even if it can be explicitly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.