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
The HCF and LCM of two numbers are 12 and 72, respectively. If the ratio of the two numbers is 2 ∶ 3, then the larger of the two numbers is:
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Joseph visits the club on every 5 th day, Harsh visits on every 24 th day, while Sumit visits on every 9 th day. If all three of them met at the club on a Sunday, then on which day will all three of them meet again?
What is the least number which when divided by 12,20 and 24 leaves in each case a remainder of 8?
Which of the following is a pair of co-primes?
