The LCM of 54,90 and a third number is 1890 and their HCF is 18 . what is the third number
The Correct Answer and Explanation is:
To find the third number, let’s denote the three numbers as:
- A = 54
- B = 90
- C = the unknown number
We are told:
- LCM(54, 90, C) = 1890
- HCF(54, 90, C) = 18
We can use the relationship between the product of numbers, their LCM, and HCF:LCM(A,B,C)×HCF(A,B,C)=LCM(A,B)×LCM(C,HCF(A,B))\text{LCM}(A, B, C) \times \text{HCF}(A, B, C) = \text{LCM}(A, B) \times \text{LCM}(C, \text{HCF}(A, B))LCM(A,B,C)×HCF(A,B,C)=LCM(A,B)×LCM(C,HCF(A,B))
But an easier method here is to proceed step-by-step.
Step 1: Find LCM and HCF of 54 and 90
We start by prime factorizing both numbers:
- 54=2×3354 = 2 \times 3^354=2×33
- 90=2×32×590 = 2 \times 3^2 \times 590=2×32×5
So,
- HCF(54, 90) = 2 × 3² = 18
- LCM(54, 90) = 2 × 3³ × 5 = 270
Step 2: Use LCM of all three numbers
We are told that:LCM(54,90,C)=1890\text{LCM}(54, 90, C) = 1890LCM(54,90,C)=1890
We already know that:LCM(54,90)=270\text{LCM}(54, 90) = 270LCM(54,90)=270
So we want:LCM(270,C)=1890\text{LCM}(270, C) = 1890LCM(270,C)=1890
We now find a number C such that:LCM(270,C)=1890\text{LCM}(270, C) = 1890LCM(270,C)=1890
We use the formula:LCM(a,b)=a×bHCF(a,b)\text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)}LCM(a,b)=HCF(a,b)a×b
Let’s solve:270×CHCF(270,C)=1890\frac{270 \times C}{\text{HCF}(270, C)} = 1890HCF(270,C)270×C=1890270×C=1890×HCF(270,C)270 \times C = 1890 \times \text{HCF}(270, C)270×C=1890×HCF(270,C)
Now divide both sides by 270:C=1890×HCF(270,C)270C = \frac{1890 \times \text{HCF}(270, C)}{270}C=2701890×HCF(270,C)
Let’s try HCF(270, C) = 18 (we are told that HCF of all three numbers is 18):C=1890×18270=34020270=126C = \frac{1890 \times 18}{270} = \frac{34020}{270} = 126C=2701890×18=27034020=126
✅ Final Answer: The third number is 126
Explanation
To find the third number when the least common multiple (LCM) of three numbers is 1890 and their highest common factor (HCF) is 18, we use number theory concepts involving prime factorization and properties of LCM and HCF.
Let’s denote the three numbers as 54, 90, and an unknown number C. We start by finding the LCM of 54 and 90. Using prime factorization, 54 equals 2×332 \times 3^32×33, and 90 equals 2×32×52 \times 3^2 \times 52×32×5. The LCM is the product of the highest powers of all primes involved: 2×33×5=2702 \times 3^3 \times 5 = 2702×33×5=270. Their HCF is the product of the lowest powers of common primes: 2×32=182 \times 3^2 = 182×32=18.
Now we are told that the LCM of all three numbers is 1890. Since the LCM of 54 and 90 is already 270, we need to find a number C such that the LCM of 270 and C is 1890. We use the identity:LCM(a,b)=a×bHCF(a,b)\text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)}LCM(a,b)=HCF(a,b)a×b
Substituting a=270a = 270a=270, LCM=1890LCM = 1890LCM=1890, and assuming the HCF is 18 (as given), we solve for C:C=1890×18270=126C = \frac{1890 \times 18}{270} = 126C=2701890×18=126
Finally, we verify that the HCF of 54, 90, and 126 is 18, and the LCM is indeed 1890, confirming our result. Therefore, the third number is 126.
