[Math] Clock losing time puzzle

puzzle

The question goes as:

A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?

My approach:

In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.

In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:

$$2 + 4 + 6+ 8 + \dots + n = 720 $$

The RHS is $720$ because I assumed they'll meet after 12 hours.

With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM} $.

Is this correct?

EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:

$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.

Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.

Does this seem correct?

Best Answer

It is ok. Let $f(t)$, and $g(t)$ be the functions in hours that indicate the hours of the two clocks, we consider the time starts at that 10:00. Then we have: $$ \begin{aligned} f(t) &=t\left( 1-\frac 3{120}(t+1)\right)\ ,\\ g(t) &=t\left( 1-\frac 5{120}(t+1)\right)\ . \end{aligned} $$ This is so because we expect interpolation polynomials of degree two (as in physics a uniformly accelerated movement), and the values in $0,1,2$ shoud (only) fit. In our case we have $f(0)=0$, $f(1)=1-6/120=1-3/60$, and $f(2)=2(1-9/120)=2-9/60$. Similarly for $g$. Then we solve the equation $f(t)-g(t)=12$. (The clock period is $12$ hours. We will soon see why this multiple of $12$ gets first hit.) This equation has the solution:

sage: var('t');
sage: f(t) = -3/120*t^2 + (1-3/120)*t
sage: g(t) = -5/120*t^2 + (1-5/120)*t
sage: (f(t)-g(t)).factor()
1/60*(t + 1)*t
sage: solve( f(t)-g(t) == 12, t )
[t == -1/2*sqrt(2881) - 1/2, t == 1/2*sqrt(2881) - 1/2]
sage: ( 1/2*sqrt(2881) - 1/2 ).n()
26.3374738006393

So the exact solution is $\frac 12(\sqrt{2881}-1)$. We solve the same equation as $\displaystyle\underbrace{2+4+\dots+2n}_{=n(n+1)}=720$ in the OP. (Stated in terms of the $n$, which also counts the hours.)

Here is the statistic of true hours, the times of the two clocks, and the difference.

sage: for t in [0..28]:
....:     print ( "t=%2s wall=%5.2f table=%5.2f    ::    diff = %8.5f min"
....:             % ( t, f(t), g(t), 60*(f(t)-g(t)) ) )
....:     
t= 0 wall= 0.00 table= 0.00    ::    diff =  0.00000 min
t= 1 wall= 0.95 table= 0.92    ::    diff =  2.00000 min
t= 2 wall= 1.85 table= 1.75    ::    diff =  6.00000 min
t= 3 wall= 2.70 table= 2.50    ::    diff = 12.00000 min
t= 4 wall= 3.50 table= 3.17    ::    diff = 20.00000 min
t= 5 wall= 4.25 table= 3.75    ::    diff = 30.00000 min
t= 6 wall= 4.95 table= 4.25    ::    diff = 42.00000 min
t= 7 wall= 5.60 table= 4.67    ::    diff = 56.00000 min
t= 8 wall= 6.20 table= 5.00    ::    diff = 72.00000 min
t= 9 wall= 6.75 table= 5.25    ::    diff = 90.00000 min
t=10 wall= 7.25 table= 5.42    ::    diff = 110.00000 min
t=11 wall= 7.70 table= 5.50    ::    diff = 132.00000 min
t=12 wall= 8.10 table= 5.50    ::    diff = 156.00000 min
t=13 wall= 8.45 table= 5.42    ::    diff = 182.00000 min
t=14 wall= 8.75 table= 5.25    ::    diff = 210.00000 min
t=15 wall= 9.00 table= 5.00    ::    diff = 240.00000 min
t=16 wall= 9.20 table= 4.67    ::    diff = 272.00000 min
t=17 wall= 9.35 table= 4.25    ::    diff = 306.00000 min
t=18 wall= 9.45 table= 3.75    ::    diff = 342.00000 min
t=19 wall= 9.50 table= 3.17    ::    diff = 380.00000 min
t=20 wall= 9.50 table= 2.50    ::    diff = 420.00000 min
t=21 wall= 9.45 table= 1.75    ::    diff = 462.00000 min
t=22 wall= 9.35 table= 0.92    ::    diff = 506.00000 min
t=23 wall= 9.20 table= 0.00    ::    diff = 552.00000 min
t=24 wall= 9.00 table=-1.00    ::    diff = 600.00000 min
t=25 wall= 8.75 table=-2.08    ::    diff = 650.00000 min
t=26 wall= 8.45 table=-3.25    ::    diff = 702.00000 min
t=27 wall= 8.10 table=-4.50    ::    diff = 756.00000 min
t=28 wall= 7.70 table=-5.83    ::    diff = 812.00000 min

This is showing also how "realistic" the table clock makes time a better time. I really need this clock! There is a point between $t=26$ and $t=27$ where we hit the $12$ hours, i.e. $720$ minutes. The solution is indeed located in this interval.

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