Three lighthouse beacons flash at regular intervals. Beacon A flashes every 24 seconds, Be… — ACT

Three lighthouse beacons flash at regular intervals. Beacon A flashes every $24$ seconds, Beacon B flashes every $36$ seconds, and Beacon C flashes every $s$ seconds, where $s$ is a positive integer. If all three beacons flash at the same instant, and the next time they all flash at the same instant is exactly $6$ minutes later, what is the number of possible values for $s$ ?

Answer: 12

Answer

To find the number of possible values for

s

, we convert the joint flashing time to seconds (

6 \text{ minutes} = 360 \text{ seconds}

). The least common multiple (LCM) of the three intervals must equal this joint interval:

\text{LCM}(24, 36, s) = 360

. Writing the prime factorizations gives

24 = 2^3 \cdot 3^1

36 = 2^2 \cdot 3^2

, and

360 = 2^3 \cdot 3^2 \cdot 5^1

. For any positive integer

s = 2^a \cdot 3^b \cdot 5^c

, the power of each prime in the LCM is the maximum of the powers in the individual prime factorizations. This gives the constraints:

a \in \{0, 1, 2, 3\}

(4 options),

b \in \{0, 1, 2\}

(3 options), and

c = 1

(1 option). Multiplying these possibilities gives

4 \times 3 \times 1 = 12

possible values for

s

Step-by-Step Solution

Convert the joint interval to seconds.

360

seconds

The individual intervals are given in seconds, so the joint interval must be converted to the same unit to perform calculations.

Set up the LCM equation.

\text{LCM}(24, 36, s) = 360

The beacons will flash together at intervals that are multiples of all three individual intervals. The first time they flash together again represents the least common multiple.

Find the prime factorizations of the known numbers.

24 = 2^3 \cdot 3^1

36 = 2^2 \cdot 3^2

, and

360 = 2^3 \cdot 3^2 \cdot 5^1

Prime factorization allows us to analyze the relationship between the individual numbers and their least common multiple.

Analyze the exponents of the prime factors of

s

For

s = 2^a \cdot 3^b \cdot 5^c

, we must have

a \in \{0, 1, 2, 3\}

b \in \{0, 1, 2\}

, and

c = 1

The exponent of each prime factor in the LCM is the maximum of the exponents of that prime factor in the numbers being combined. Since the LCM has

2^3

, the maximum exponent of 2 must be 3, which is already satisfied by

24 = 2^3 \cdot 3^1

, so

a

can be any integer from 0 to 3. Since the LCM has

3^2

, the maximum exponent of 3 must be 2, which is already satisfied by

36 = 2^2 \cdot 3^2

, so

b

can be any integer from 0 to 2. Since the LCM has

5^1

and neither 24 nor 36 has a factor of 5,

s

must provide exactly one factor of 5 (

c=1

Calculate the total number of combinations for

s

4 \times 3 \times 1 = 12

Since the choice of each exponent is independent, we multiply the number of choices for each prime factor's exponent.

Key Concept

Using prime factorizations to determine the relationship between numbers and their least common multiple (LCM).

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