To begin with, let’s determine the highest common factor (HCF) of the numbers 65 and 117.
First, we’ll find the prime factorization of both numbers:
- 65 can be factored into: 5 × 13.
- 117 can be factored into: 3 × 39 or equivalently 3 × 3 × 13.
From the factorizations above, we see that both 65 and 117 share the prime factor 13.
Thus, the HCF of 65 and 117 is 13.
Next, we are tasked with finding integral values of m and n such that HCF(65m, 117n) equals 13.
To achieve this, we can express both 65m and 117n in terms of their prime factors:
- Since 65 = 5 × 13, we can write: 65m = 5m × 13.
- And for 117 = 3 × 39, we can assert: 117n = 3n × 39.
For the HCF of 65m and 117n to equal 13, both m and n should be chosen such that the common factor remains only 13.
A straightforward choice is to set m = 1 and n = 1. In this case:
- HCF(65 × 1, 117 × 1) = HCF(65, 117) = 13.
Therefore, one pair of integral values for m and n that achieves our requirement is (m, n) = (1, 1).
However, there are other possible pairs as well, such as:
- m = 2 and n = 1
- m = 1 and n = 2
In each case, if one of the values is a multiple of the respective number that does not introduce additional factors of 13 or any other common prime factors, we ensure that the HCF remains 13. This shows that the selections for m and n can be somewhat flexible, as long as they are appropriately chosen.
In summary, the HCF of 65 and 117 is 13, and one pair of integral values for m and n satisfying HCF(65m, 117n) = HCF(65, 117) is (m, n) = (1, 1).