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There are two analog clocks. Clock A works as normal, but clock B goes backward and seven times as fast. How many times in 24 hours will clock B show the right time (same as A). Meaning how many times do the short hands align?

I found similar questions at How to visualize this modular arithmetic problem? and One clock gains, another clock loses time. When will both clocks next show the same time?.

But I'm having a hard time using the same logic since one clock works backward. The right answer is 16 if you don't count the last alignment since it's not the same day anymore.

I solved the problem by just drawing clocks and tables. It was not too time-consuming for this question, but I would like to know a more efficient way of solving the problem using modular arithmetic. For example, transferring the problem into a congruence.

Stephen21212
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Let $x$ be the number the short hand of clock $A$ represents, then $$x \equiv12-7x \text{ (mod $12$)} $$ or $$ 8x \equiv 0 \text{ (mod $12$)}$$ since we only consider $24$ hours, $x \in [0, 24)$, and there are $$8x = 0, 12, 24, ..., 12*15$$ or $$x = 0, \frac{12}{8}, \frac{24}{8}, ..., \frac{12*15}{8}$$ a total of $16$ solutions.

Lab
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