Two Broken Clocks

it should be noticed that they will show the same time after the first clock has gained 12 hours with respect to the first clock 1 day gain for first clock is 50 minutes 12 hour gain will come in 72/5 days ...but as the given condition it should be a multiple of 24 as he checks the clock after an interval of 24 hours so we have to multiply it with 5 in order to get a whole number.... so minimum number of days is 72 days ..!!!!

Since one clock will be 20 minutes faster and the other clock will be 30 minutes slower on the next day, we can conclude that the time difference between the time of the clocks will increase by 50 minutes after 1 day. If you want to let the two clocks showing the same time again, the time difference should be multiple of 12 hours, which is multiple of 720 minutes. Since 720 can't be divided by 50, we take the L.C.M. of 720 and 50 which is 3600. $3600÷50=72$ , therefore Liz has to wait 72 days until she can see the two clocks showing the same time again.

$The~ first~ clock~ lags~ by~ {\large \color{#D61F06}{-}} 1/3~ hr/day, ~seond ~lags ~by ~1/2 ~ hr/day \\ trying ~to ~cover ~ up ~12hr.\$ $\text{ If this can be done in n days, (1/2 - 1/3) * n=12. } \ \therefore n= 72.\\ \text{We check. First lead 1/3 * 72=24hr . Second lag 1/2 * 72=36hr }\\ So \ 24\equiv 0 (mod 12) ,\ \ -36 \equiv 0 (mod 12).\\ \text{In 72 days, the first clock showed 73 days, the second 70.5 days.}\\ \ \ \ \\ \text{I had first worked out a long way. }\\ \text{First showing more by 2 hours in 6 days and second showing less by 3hr in 6 days.}\\ \text{Thus difference of 1 hr per 6 days. Difference of 12 hours in 72 days. } \\ \text{Note Liz checks after every 24 hr . And the above method is based on this logic.} \\ \text{Seeming contradiction is because the counting is cyclic. Some one with Number theory } \\ \text{may explain the reason 1/3 is not added to 1/2 but subtracted. }\\ \text{Our normal formula is Final - Initial.}$
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