At a certain factory, each of Machines A, B, and C is periodically on for exactly 1 minute and periodically off for a fixed number of consecutive minutes. The table above shows that Machine $A$ is on and uses 15 units of power every $18{ }^{\text {th }}$ minute, Machine B is on and uses 18 units of power every $15{ }^{\text {th }}$ minute, and Machine C is on and uses 12 units of power every $12{ }^{\text {th }}$ minute. The factory has a backup generator that operates only when the total power usage of the 3 machines exceeds 30 units of power. What is the time interval, in minutes, between consecutive times the backup generator begins to operate?

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At a certain factory, each of Machines A, B, and C is periodically on for exactly 1 minute and periodically off for a fixed number of consecutive minutes. The table above shows that Machine A is on and uses 15 units of power every 18^{th} minute, Machine B is on and uses 18 units of power every 15^{th} minute, and Machine C is on and uses 12 units of power every 12^{th} minute. The factory has a backup generator that operates only when the total power usage of the 3 machines exceeds 30 units of power. What is the time interval, in minutes, between consecutive times the backup generator begins to operate?

Explanation	Calculations	Help
$We observe that Machine A is on at minute t when t is a multiple of 18, Machine B is on at multiples of 15, and Machine C is on at multiples of 12.$	$t = 18 k for some integer k Machine A on if t = 18 k t = 15 m for some integer m Machine B on if t = 15 m t = 12 n for some integer n Machine C on if t = 12 n$
$We define U (t) as the total power usage at minute t by adding 15 units if Machine A is on, 18 units if Machine B is on, and 12 units if Machine C is on.$	$U (t) = 15 \times (A on at t) + 18 \times (B on at t) + 12 \times (C on at t)$
$We find when Machines A and B turn on together by computing the least common multiple of 18 and 15.$	$18 = 2 \times 3^{2} 15 = 3 \times 5 LCM (18, 15) = 2 \times 3^{2} \times 5 = 90$	Theory & Tactics Method Card DIV6-A Click to view full details
$We check U (t) at key candidate times to find the first time it exceeds 30.$	$U (36) = 15 + 12 = 27 and 27 < 30 U (60) = 18 + 12 = 30 and 30 is not greater than 30 U (90) = 15 + 18 = 33 and 33 > 30 LCM (18, 15, 12) = 180 U (180) = 15 + 18 + 12 = 45 and 45 > 30$	Theory & Tactics Method Card DIV6-A Click to view full details
$We see the generator begins operating at t = 90, 180, 270, \dots, so the interval between consecutive start times is 90 minutes.$	$Interval = 90 minutes$

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90

At a certain factory, each of Machines A, B, and C is periodically on for exactly 1 minute and periodically off for a fixed number of consecutive minutes. The table above shows that Machine A is on and uses 15 units of power every 18^{th} minute, Machine B is on and uses 18 units of power every 15^{th} minute, and Machine C is on and uses 12 units of power every 12^{th} minute. The factory has a backup generator that operates only when the total power usage of the 3 machines exceeds 30 units of power. What is the time interval, in minutes, between consecutive times the backup generator begins to operate?

Explanation	Calculations	Help
$We observe that Machine A is on at minute t when t is a multiple of 18, Machine B is on at multiples of 15, and Machine C is on at multiples of 12.$	$t = 18 k for some integer k Machine A on if t = 18 k t = 15 m for some integer m Machine B on if t = 15 m t = 12 n for some integer n Machine C on if t = 12 n$
$We define U (t) as the total power usage at minute t by adding 15 units if Machine A is on, 18 units if Machine B is on, and 12 units if Machine C is on.$	$U (t) = 15 \times (A on at t) + 18 \times (B on at t) + 12 \times (C on at t)$
$We find when Machines A and B turn on together by computing the least common multiple of 18 and 15.$	$18 = 2 \times 3^{2} 15 = 3 \times 5 LCM (18, 15) = 2 \times 3^{2} \times 5 = 90$	Theory & Tactics Method Card DIV6-A Click to view full details
$We check U (t) at key candidate times to find the first time it exceeds 30.$	$U (36) = 15 + 12 = 27 and 27 < 30 U (60) = 18 + 12 = 30 and 30 is not greater than 30 U (90) = 15 + 18 = 33 and 33 > 30 LCM (18, 15, 12) = 180 U (180) = 15 + 18 + 12 = 45 and 45 > 30$	Theory & Tactics Method Card DIV6-A Click to view full details
$We see the generator begins operating at t = 90, 180, 270, \dots, so the interval between consecutive start times is 90 minutes.$	$Interval = 90 minutes$

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90