Understanding the LCM Concept
The Least Common Multiple (LCM) of a set of numbers is the smallest positive integer that is divisible by each of the numbers in the set.
To find the LCM, we often use the prime factorization method. We find the prime factors of each number and then take the highest power of each prime factor that appears in any of the factorizations. The product of these highest powers is the LCM.
Prime Factorization Step
First, let's find the prime factorization of the known numbers and the given LCM:
- Prime factorization of 40:
$40 = 4 \times 10 = (2 \times 2) \times (2 \times 5) = 2^3 \times 5^1$
- Prime factorization of 78:
$78 = 2 \times 39 = 2 \times 3 \times 13 = 2^1 \times 3^1 \times 13^1$
- Prime factorization of the LCM (7800):
$7800 = 78 \times 100 = (2 \times 3 \times 13) \times (10^2) = (2 \times 3 \times 13) \times (2 \times 5)^2$
$= (2 \times 3 \times 13) \times (2^2 \times 5^2) = 2^{1+2} \times 3^1 \times 5^2 \times 13^1 = 2^3 \times 3^1 \times 5^2 \times 13^1$
LCM Requirement Analysis
The LCM is given as $7800 = 2^3 \times 3^1 \times 5^2 \times 13^1$.
Let's combine the prime factors of 40 and 78:
Factors from 40: $2^3, 5^1$
Factors from 78: $2^1, 3^1, 13^1$
Combined highest powers from 40 and 78: $2^{\max(3,1)} \times 3^{\max(0,1)} \times 5^{\max(1,0)} \times 13^{\max(0,1)} = 2^3 \times 3^1 \times 5^1 \times 13^1$.
Comparing this with the LCM ($2^3 \times 3^1 \times 5^2 \times 13^1$), we see that the number 'x' must provide the missing factor $5^2$. The powers of other prime factors in 'x' cannot exceed the powers in the LCM ($2^3, 3^1, 13^1$).
Checking the Options
Now, let's check the prime factorization of each option to see which one satisfies the condition:
| Option (Number x) | Prime Factorization of x | Combined LCM Factors (40, 78, x) | Matches LCM (7800)? |
|---|---|---|---|
| 180 | $180 = 2^2 \times 3^2 \times 5^1$ | $2^3 \times 3^2 \times 5^1 \times 13^1$ | No (Incorrect power of 3, missing $5^2$) |
| 200 | $200 = 2^3 \times 5^2$ | $2^{\max(3,1,3)} \times 3^{\max(0,1,0)} \times 5^{\max(1,0,2)} \times 13^{\max(0,1,0)} = 2^3 \times 3^1 \times 5^2 \times 13^1$ | Yes |
| 127 | $127^1$ (Prime) | $2^3 \times 3^1 \times 5^1 \times 13^1 \times 127^1$ | No (Includes extra factor 127) |
| 152 | $152 = 2^3 \times 19^1$ | $2^3 \times 3^1 \times 5^1 \times 13^1 \times 19^1$ | No (Includes extra factor 19, missing $5^2$) |
Conclusion
By analyzing the prime factorizations, we found that option 2, which is 200 ($2^3 \times 5^2$), provides the necessary prime factors to achieve the LCM of 7800 when combined with the factors of 40 and 78.
The LCM calculation for 40, 78, and 200 is:
LCM($2^3 \times 5^1$, $2^1 \times 3^1 \times 13^1$, $2^3 \times 5^2$) = $2^3 \times 3^1 \times 5^2 \times 13^1 = 7800$.
Therefore, 200 can be the value of x.
Important Questions from LCM and HCF
Six bells begin to toll together and toll, respectively, at intervals of 3, 4, 6, 7, 8 and 12 seconds. After how many seconds, will they toll together again?
A and B are two prime numbers such that A > B and their LCM is 209. The value of A 2 - B is:
Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
Calculate the HCF of \(\frac{12}{5}\) , \(\frac{14}{15}\) and \(\frac{16}{17}\) .
Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?
