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 Apr20 comment How to find $p(t)$ when $m$ varies linearly with $t$? @YvesDaoust Thank you for reminding me of the title. My original problem is a bit more complex, and this question is very simplified, which is why, after simplifying the question a lot, it turned out to be a simple differential equation, not what the title originally stated (as Tom-Tom wrote in his answer). Now it is just a matter of rewriting my original equation as a differential equation. I guess I must learn how to express my problems in a more simple manner to be able to solve them more easily. Apr20 comment How to find $p(t)$ when $m$ varies linearly with $t$? Updating the question to clarify now, thank you. Apr20 comment How to find $p(t)$ when $m$ varies linearly with $t$? @YvesDaoust The mass decreases as time passes. The mass isn't defined based on what $t$ is at $p=p(t)$. I might have been a bit unclear in the question. Apr20 comment How to find $p(t)$ when $m$ varies linearly with $t$? Now I understand why the sum doesn't work. Since the time is squared, it increases much faster than the mass decreases, so that the limit when $k \to \infty$ does not resemble the correct position. Do you agree? Apr20 comment How to find $p(t)$ when $m$ varies linearly with $t$? I have an object that is pushed by a force $F$. As time passes, the mass of the object decreases, so that over time the acceleration of the object increases. What I was thinking was that for an infinitesimal timeframe, the difference in $p$ is given by what is inside the summation in the question. So by letting $\lim_{k→\infty}$ I would expect to get the actual $p$. Apr20 comment How to find $p(t)$ when $m$ varies linearly with $t$? Thank you for such an in depth answer! The second case is the one I was trying to solve. I was just trying to solve the problem in the wrong way. Dec7 comment Is there one method of adding and subtracting without a calculator? Yes, but $80-(-70)$ would by my algorithm be interpreted as $80+70$ since, as I see it, this method cannot be applied without running a else/elseif for these four conditions: +A+B, +A-B (|A|>|B|?), -A+B, -A-B. Dec7 comment Is there one method of adding and subtracting without a calculator? It probably is correct. I'm going to test my algorithms with a recursive fibonacci generator, so if the numbers are correct, the algorithms are probably correct as well. Dec7 comment Is there one method of adding and subtracting without a calculator? Hmm... I've been calculating on a paper for a while now. As I see it, one can make a few rules. If you always do A-B. Given A-B where |A|>|B|. One can calculate with as many digits as the number with the most digits, and then neglect all the 1's in the answer to get the right answer. Dec7 comment Is there one method of adding and subtracting without a calculator? This method is just fantastically smart and yet not logically difficult, at least not how you explained it! Dec6 comment Is there one method of adding and subtracting without a calculator? Yes, that is true. I've searched the internet, but I haven't found anything that isn't compiled, obfuscated and embedded in much other code so I can't extract it. + It is copyrighted.. Dec6 comment Is there one method of adding and subtracting without a calculator? Thanks for your response! Actually, I'm asking this question because I'm making a computer method that is simply going to add, subtract, multiply and divide numbers. I need it because the language I'm writing in thinks any number greater than $9007199254740992$ is infinite. Great answer! Dec6 comment Is there one method of adding and subtracting without a calculator? I agree with you on the multiplication/division, but not on the addition/subtraction. Dec6 comment Is there one method of adding and subtracting without a calculator? @ChrisEagle I'm not an english speaker, so I'm not very experienced with english mathematical terms. Dec6 comment Is there one method of adding and subtracting without a calculator? @HenningMakholm Exactly! I've already thought about this solution, but it again brings us back to start as you Henning stated. Sep28 comment Why does this equation have different number of answers? Please add your comments as an answers, as it is more useful for other people that come from Google. :) Sep28 comment Why does this equation have different number of answers? @GerryMyerson But if there are "traps" like this, how does one discover them when the problems get more complex than just the simple problem I wrote? Sep28 comment Why does this equation have different number of answers? @PeterTamaroff and Sep28 comment Why does this equation have different number of answers? @PeterTamaroff I feel really stupid right now... I was just too caught up trying to make it become $x=2x$... Sep28 comment Why does this equation have different number of answers? But then you can do $\frac{1}{x} = \frac{2}{x} | \times x$ and get $2=1$.