Question

HCF of two integers is 4 and one of the integers is 24. Write down 2 possible values of the other integer.

Answer 1

If the Highest Common Factor (HCF) of two integers is 4 and one of the integers is 24, we can denote the two integers as \( a = 24 \) and \( b \) (the integer we are looking for).

Since the HCF is 4, both integers must be multiples of 4. We can express \( a \) and \( b \) as follows:

1. The integer \( a = 24 \) can be factored as \( 24 = 4 \times 6 \).
2. Thus, we can say that the integer \( b \) must be of the form \( b = 4k \) for some integer \( k \).

To ensure the HCF of \( a \) and \( b \) is 4, \( b \) must not contain any higher factors of 4 aside from the one already given. Therefore, \( k \) must be an integer that is **not** a multiple of 6 (because including 6 would make the HCF greater than 4).

Here are two possible values for \( k \):
- Let’s choose \( k = 1 \):
\[
b = 4 \times 1 = 4
\]

- Let’s choose \( k = 2 \):
\[
b = 4 \times 2 = 8
\]

Now we check HCF(24, 4) and HCF(24, 8):
- HCF(24, 4) = 4
- HCF(24, 8) = 8 (not valid, we need another).

Let's choose \( k = 3 \):
\[
b = 4 \times 3 = 12
\]
HCF(24, 12) = 12 (not valid).

Let’s choose \( k = 5 \):
\[
b = 4 \times 5 = 20
\]
HCF(24, 20) = 4 (valid).

Let’s choose \( k = 7 \):
\[
b = 4 \times 7 = 28
\]
HCF(24, 28) = 4 (valid).

Thus, two possible values of the other integer \( b \) could be **20** and **28**.