In a book store, each of the word of the glowsign board “MODERN BOOK STORES” is visible after 5/2, 17/4 and 41/8 seconds respectively. Each of them is put off for 1 second. Find the time after which one person can see a completely visible glowsign board.
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Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Mar 21 '20 at 08:41
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Please state the rules of the game more clearly. What is the state at $t=0$? What do the $5/2$ seconds mean, and is the put off time included here? Etcetera. – Christian Blatter Mar 21 '20 at 10:40
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Hint : Find the lcm of the three numbers. – Peter Mar 21 '20 at 14:15
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Hint $ $ As explained in this thread
$$\rm\ lcm\left(\frac{a}b,\frac{c}d\right) = \frac{lcm(a,c)}{gcd(b,d)}\ \ \ if\ \ \ \gcd(a,b) = 1 = \gcd(c,d)\qquad$$
Thus $\,{\rm lcm}(\frac{5}2,\frac{17}4)=\frac{35}2,\, $ and $\ {\rm lcm}(\frac{35}2,\frac{41}8)=\frac{3485}2\ $ is the least common multiple of all three,
since $\,{\rm lcm}(a,b,c) = {\rm lcm}({\rm lcm}(a,b),c),\,$ by lcm is associative.
Or put all fractions over a common denominator of $8$ then take the lcm of the numerators.
Bill Dubuque
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The most basic (and simple answer for this case) would be to first change the denominator of all three fractions to their lcm, which is 8. Taking the lcm of the changed numerators, we have : lcm(5×(8÷2),17×(8÷4),41)=lcm(20,34,41) =13940
Putting the fraction back together and adding a one(for the second when the lights are put off), we have: ($\frac{13940}{8}$+1) sec=1743.5 sec≈29.04min
Then the final answer is 1743.5 sec(or 29.04 min if you prefer)
ilikecats
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