




The least common multiple, or LCM, is another number that's useful in solving many math problems. Let's find the LCM of 30 and 45. One way to find the least common multiple of two numbers is to first list the prime factors of each number.
30 = 2 × 3 × 5 45 = 3 × 3 × 5
Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs.
2: one occurrence 3: two occurrences 5: one occurrence 2 × 3 × 3 × 5 = 90 <— LCM
After you've calculated a least common multiple, always check to be sure your answer can be divided evenly by both numbers.
Illustration 1: Find the LCM of 3, 9 and 21. Solution: List the prime factors of each. 3: 3 9: 3 × 3 21: 3 × 7 Multiply each factor the greatest number of times it occurs in any of the numbers. 9 has two 3s, and 21 has one 7, so we multiply 3 two times, and 7 once. This gives us 63, the smallest number that can be divided evenly by 3, 9, and 21. We check our work by verifying that 63 can be divided evenly by 3, 9, and 21.
Illustration 2: Find the LCM of 12 and 80 . Solution: List the prime factors of each. 12: 2 × 2 × 3 80: 2 × 2 × 2 × 2 × 5 = 80 Multiply each factor the greatest number of times it occurs in either number. 12 has one 3, and 80 has four 2's and one 5, so we multiply 2 four times, 3 once, and five once. This gives us 240, the smallest number that can be divided by both 12 and 80. We check our work by verifying that 240 can be divided by both 12 and 80.
L.M.C. of 12, 24 and 36 is _______________ | |||
| Right Option : C | |||
| View Explanation | |||
Micheal and Martin collect hockey cards. Micheal has 45 cards in his collection and Martin has 30 cards in his collection. If the cards in collection of both boys come in packages of the same number of cards, then how many cards of each of them can be packed of equal numbers ? | |||
| Right Option : A | |||
| View Explanation | |||
20 is written as the product of primes as: | |||
| Right Option : C | |||
| View Explanation | |||
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