If the HCF of two numbers is 23 and their sum is 161, the possible numbers CANNOT be:

Correct option is D

Solution:

Option A: (46, 115)

The prime factorization of 46 is 46 = 2 × 23

The prime factorization of 115 is 115 = 5 × 23

The HCF of 46 and 115 is 23, which matches the given condition.

Sum: 46 + 115 = 161, which also satisfies the sum condition.

So, (46, 115) is a valid pair.

Option B: (23, 138)

The prime factorization of 23 is 23 (a prime number).

The prime factorization of 138 is 138 = 2 × 3 × 23

The HCF of 23 and 138 is 23, which matches the given condition.

Sum: 23 + 138 = 161, which satisfies the sum condition.

So, (23, 138) is a valid pair.

Option C: (69, 92)

The prime factorization of 69 is 69 = 3 × 23

The prime factorization of 92 is 92 = $$2^2 \times 23.$$​​

The HCF of 69 and 92 is 23, which matches the given condition.

Sum: 69 + 92 = 161, which satisfies the sum condition.

So, (69, 92) is a valid pair.

Option D: (61, 100)

The prime factorization of 61 is 61 (a prime number).

The prime factorization of 100 is 100 = 2$$^2 \times 5^2.$$​​

The HCF of 61 and 100 is 1 (since 61 and 100 do not have any common factors except 1),

which does not match the given condition that the HCF should be 23.

Sum: 61 + 100 = 161, but the HCF is not 23.

So, (61, 100) is not a valid pair.
The pair of numbers that cannot satisfy the conditions is D) (61, 100).