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LCM(a, b) is 2520, where a and b are natural numbers. If a=120, what is the smallest possible value of b?

User Ujell
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Final answer:

The smallest possible value of b is found by factoring the LCM (2520) and the given number a (120), and then identifying the prime factors that are in the LCM but not in a. The smallest b that includes these missing factors is 21.

Step-by-step explanation:

To determine the smallest possible value of b when the least common multiple (LCM) of a and b is 2520, and we know a is 120, we need to find a number b that when multiplied by 120 gives 2520 without introducing new prime factors that are not in 120.

First, let's factorize 120 and 2520 to their prime factors:

  • 120 = 23 × 3 × 5
  • 2520 = 23 × 32 × 5 × 7

Looking at the prime factors of 2520, we can see that 2520 includes all the prime factors of 120. Additionally, 2520 has a prime factor of 7 and an extra 3 compared to 120. Therefore, to obtain the LCM of 2520 with 120, b must at least have the prime factors that are in 2520 but not in 120.

The smallest number that has these is b = 3 × 7 = 21. When we multiply this with 120, we get the LCM of 2520 without introducing any new prime factors. Therefore, the smallest possible value of b is 21.

User Ivan Cachicatari
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