If a = 2^3 × 3, b = 2 × 3 × 5, c = 3^n × 5 and LCM (a, b, c) = 2^3 × 3^2 × 5, then n =?
By BYJU'S Exam Prep
Updated on: September 13th, 2023
The value of ‘n’ is 2 if a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5 and LCM (a, b, c) = 23 × 32 × 5. To determine this, candidates have to find the LCM of a, b, and c. After finding the LCM, of a, b, and c it will be easy to get the value of n.
Given: LCM (a, b, c) = 23 × 32 × 5 (equation 1)
To find: value of ‘n’
a = 23 × 3, b = 2 × 3 × 5, and c = 3n × 5
We are aware that while assessing LCM, we use prime numbers with higher exponents to factor the number.
As a result, by using this formula and assuming that n is more than or equal to 1, we may calculate the LCM as
LCM (a, b, c) = 23 × 3n × 5 (equation 2)
Comparing equation 1 and equation 2, we will get
23 × 32 × 5 = 23 × 3n × 5
n = 2
Therefore, the value of n is 2.
Table of content
Value of ‘n’ in a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5
As discussed above, the value of n = 2 for a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5 and LCM (a, b, c) = 23 × 32 × 5. The abbreviation “LCM” stands for “Least Common Multiple.” The smallest multiple shared by two or more different numbers is referred to as the least common multiple.
When, HCF ( p, q, r) × LCM (p, q, r) is not equal to p × q × r, where p, q, r are positive integers then
LCM (p, q, r) = [p × q × r × HCF ( p, q, r)] ÷ HCF (p,q) × HCF (q,r) × HCF (p,r) and
HCF ( p, q, r) = [p × q × r × LCM ( p, q, r)] ÷ LCM (p,q) × LCM (q,r) × LCM (p,r)
Summary:
If a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5 and LCM (a, b, c) = 23 × 32 × 5, then n =?
The value of n = 2 if a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5 and LCM (a, b, c) = 23 × 32 × 5. To find the value of ‘n’ find the LCM of a, b, and c and then compare its values with LCM (a, b, c) = 23 × 32 × 5.
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