I was recently introduced to modular arithmetic and was playing around with clock related questions, when I thought, if we ignore the minute-hand and second-hand, the hour hand will be on a whole number every 60 minutes.
But if the hour hand jumps directly a certain distance, like say 3 minutes every 1 minute (let's call it $n$), we can see, it will be on a whole number every 20 minutes as $60 \div 3 = 20$. And if 60 was not divisible by $n$, then we can multiply it by 60, for example, if $n=7$, then the answer will be $7 \times 60 = 420$ . So, the first time the hour hand will be on a whole number will be in 60 minutes (or 60 steps) and it will be on the number 7 on the clock.
Now what if the hour-hand gains a constant speed every step? For example, 3 + 5 + 7 + 9 + ...
As in, on the first step it covers 3 minutes, on second step it covers another 5 minutes, so a total of 8 minutes in two steps, and it continues on, going around the 12-hour clock until it reaches exactly a whole number.
I tried to solve it as follows:
$3 + 5 + 7 + 9 + 11 + 13 + ...$ upto m steps
taking 3 from all numbers gives
$3m + (0 + 2 + 4 + 6 + 8 + ...$ upto m steps)
$= 3m + 2 (0 + 1 + 2 + 3 + 4 + ...$ upto m steps) (taking 2 common)
$= 3m + 2 (1 + 2 + 3 + 4 + ...$ upto (m-1) steps)
$ = 3m + \frac{2(m-1)(m)}{2} $
$ = 3m + (m-1)m$
$ = 3m + m^2 - m$
$ = m^2 + 2m$
Which should be equal to $60n$ I think, (I'm a little confused at this point), but then that gives me a single equation with 2 variables. How do I solve this?
Do I need to put every possible value one by one or there can be a generalized formula here, which can help me to come up with a solution in least number of steps with other values of n, and 12 or 60 as well?
(Note: As you can probably see, I'm not very good at maths, so please try to keep it simple, while giving as much information as possible. I'm willing to study further if needed. Thank you).
To answer your mathematical question; you want to find the smallest positive integer $m$ such that $$m^2+2m=60n,$$ for some integer $n$. This will be the number of steps $m$ until you reach the first whole number $n$. Phrased in terms of modular arithmetic, you are looking to solve the congruence $$m^2+2m\equiv0\pmod{60}.$$ A very general and powerful observation is that if $m$ is a solution to the congruence, then so is $m\pm60$. So if any solution exist, then can keep adding or subtracting $60$ to get a solution in the range $0\leq m<60$. This already greatly limits the search for a solution.
In this case, because $60$ is not a very large number, and the left hand side is a fairly simple function, by some educated trial and error you can find that $$10^2+2\cdot10\equiv0\pmod{60},$$ which shows that after $10$ steps you reach the number $2$. Then it is not hard to check that no smaller positive integer $m$ satisfies the congruence.
Here is an approach that is applicable more generally: If $n$ is a positive integer with prime factorization $n=\prod_pp^{n_p}$, then by the Chinese remainder theorem the solutions to a congruence $$f(m)\equiv0\pmod{n},$$ correspond bijectively to the simultaneous solutions of the system of congruences $$f(m)\equiv0\pmod{p^{n_p}}\ \text{ for each prime $p$ dividing $n$}.$$ In this case we have $60=2^2\times3\times5$ and so it suffices to solve \begin{eqnarray*} m^2+2m&\equiv0\pmod{2^2},\\ m^2+2m&\equiv0\pmod{3},\\ m^2+2m&\equiv0\pmod{5}. \end{eqnarray*} The moduli are now so small that it is easy to simply check all possibilities; we find that \begin{eqnarray*} m&\equiv0,2\pmod{2^2},\\ m&\equiv0,1\pmod{3},\\ m&\equiv0,3\pmod{5}. \end{eqnarray*} These congruences give $2\times2\times2=8$ solutions to the original congruence, and they are $$m\equiv0,10,18,28,30,40,48,58\pmod{60}.$$ This shows that indeed $10$ is the smallest positive integer that satisfies the congruence.
If you would like to remove 'guessing' or 'trial and error' completely, then there are general methods to solve the system of simple congruences above. These use the (rather amazing) fact that modulo prime numbers you can use a lot of algebraic tricks you know for real numbers. For example, the quadratic formula still works modulo every prime number $p>2$. So the congruence $$m^2+2m\equiv0\pmod{3},$$ leads to the solutions $$m\equiv\frac{-2\pm\sqrt{2^2-4\cdot1\cdot0}}{2\cdot1}\equiv\frac{-2\pm\sqrt{4}}{2}\equiv-1\pm1\pmod{3},$$ where of course $-1+1\equiv0\pmod{3}$ and $-1-1\equiv-2\equiv1\pmod{3}$. You can solve the equation modulo $5$ in the exact same way.