Word Problem Involving the Least Common Multiple of 2 Numbers

Using Prime Factors to Find the Least Common Multiple of Two Numbers

The two numbers are written as products of their prime factors.
The product of the maximum occurrences of each prime factor in the numbers gives the least common multiple of the two numbers.

Example

Find least common multiple (lcm) of 21 and 48

Solution

Step 1:

The prime factors of 21 and 48 are 21 = 3 × 7

48 = 2 × 2 × 2 × 2 × 3

Step 2:

Maximum occurrences of the prime factors are 2(4 times); 3(1 time); 7(1 time)

Step 3:

Least common multiple of 21 and 48 = 2 × 2 × 2 × 2 × 3 × 7 = 336

Problem 1:

A bell rings every 18 seconds, another every 60 seconds. At 5.00 pm the two ring simultaneously. At what time will the bells ring again at the same time?

Solution

Step 1:

A bell rings every 18 seconds, another every 60 seconds

Prime factorizations of 18 and 60 are

18 = 2 × 3 × 3

60 = 2 × 2 × 3 × 5

Step 2:

LCM is the product of maximum occurrences of each prime factor in the given numbers.

Step 3:

So L C M (12, 18) = 2 × 2 × 3 × 3 × 5 = 180 seconds = 180/60 = 3 minutes.

So the bells will ring at same time again at 5.03pm

Problem 2:

A salesman goes to New York every 15 days for one day and another every 24 days, also for one day. Today, both are in New York. After how many days both salesman will be again in New York on same day?

Solution

Step 1:

A salesman goes to New York every 15 days and another every 24 days

Prime factorizations of 15 and 24 are

15 = 3 × 5

24 = 2 × 2 × 2 × 3

Step 2:

LCM is the product of maximum occurrences of each prime factor in the given numbers.

Step 3:

So L C M (12, 18) = 2 × 2 × 2 × 3 × 5 = 120 days.

So both salesmen will be in New York after 120 days.

Problem 3:

What is the smallest number that when divided separately by 20 and 48, gives the remainder of 7 every time?

Solution

Step 1:

Prime factorizations of 20 and 48 are

20 = 2 × 2 × 5

48 = 2 × 2 × 2 × 2 × 3

Step 2:

LCM is the product of maximum occurrences of each prime factor in the given numbers.

Step 3:

So L C M (20, 48) = 2 × 2 × 2 × 2 × 3 × 5 = 240

The required number is 240 + 7 = 247