0Example:
I. 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
II. 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 = 180seconds = 180/60 = 3 minutes.
So the bells will ring at same time again at 5.03pm
III. 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 = 120days.
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.
