Question

If HCF of the numbers 299, 253 and a third number 'A' is 23 and their LCM is 16445, then what is the number 'A'?

• A. 69
• B. 95
• C. 115
• D. 138

Hint:

For two numbers \( a \) and \( b \), \( a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b) \). For three or more numbers, a direct formula doesn't exist, but if the HCF is \( k \), then each number can be written as \( k \times p, k \times q, k \times r, \dots \), and the LCM will be \( k \times \text{LCM}(p,q,r, \dots) \).

Answer 1

The Correct Option is C

Let the three numbers be \( N_1 = 299 \), \( N_2 = 253 \), and \( N_3 = A \).
Given:
HCF(\( N_1, N_2, A \)) = 23
LCM(\( N_1, N_2, A \)) = 16445
Step 1: Express the given numbers as multiples of their HCF.
Since the HCF of the three numbers is 23, each number must be a multiple of 23.
Let \( N_1 = 23 \times p \), \( N_2 = 23 \times q \), and \( A = 23 \times r \), where \( p, q, r \) are integers.
Calculate \( p \) and \( q \):
For \( N_1 = 299 \):
\[ 299 = 23 \times p \quad \Rightarrow \quad p = \frac{299}{23} = 13 \] For \( N_2 = 253 \): \[ 253 = 23 \times q \quad \Rightarrow \quad q = \frac{253}{23} = 11 \] So, the numbers are \( 23 \times 13 \), \( 23 \times 11 \), and \( 23 \times r \). Step 2: Use the LCM property to find \( r \).
The LCM of three numbers \( k \times a, k \times b, k \times c \) is \( k \times \text{LCM}(a, b, c) \).
Here, \( k = 23 \), \( a = 13 \), \( b = 11 \), and \( c = r \). Given LCM = 16445. \[ \text{LCM}(23 \times 13, 23 \times 11, 23 \times r) = 23 \times \text{LCM}(13, 11, r) \] \[ 16445 = 23 \times \text{LCM}(13, 11, r) \] Divide 16445 by 23 to find \( \text{LCM}(13, 11, r) \): \[ \text{LCM}(13, 11, r) = \frac{16445}{23} = 715 \] Now, we need to find \( r \) such that \( \text{LCM}(13, 11, r) = 715 \). First, find the prime factorization of 715: \[ 715 = 5 \times 143 = 5 \times 11 \times 13 \] We know that 11 and 13 are prime numbers. Since \( \text{LCM}(13, 11, r) = 5 \times 11 \times 13 \), and 11 and 13 are already factors of the first two numbers, \( r \) must contribute the factor 5.
For the HCF to be 23, \( r \) must not share any common factors with 11 or 13 (other than 1), which is true if \( r=5 \). So, \( r = 5 \). Step 3: Calculate the number 'A'.
We defined \( A = 23 \times r \).
Substitute the value of \( r \): \[ A = 23 \times 5 \] \[ A = 115 \] Thus, the third number 'A' is 115.