Question

If m = dn+r, where m, n are positive integers and d and r are integers, then n is HCF(m,n) if
[a] r = 1
[b] $0 < r\le 1$
[c] r = 0
[d] r is a real number.

Answer 1

Hint: Assume that n is the HCF(m,n). Use the fact that if g is the HCF of a and b, then g divides a and g divides b. Use the fact that if a is divisible by b, then the remainder obtained on dividing a by b is 0. Hence find the value of r.

Complete step-by-step solution -
We have m = dn+r.
Let HCF(m,n) = n.
We know that if g is the HCF of a and b, then g divides a and g divides b.
Hence, we have n divides m.
We know that if a is divisible by b, then the remainder obtained on dividing a by b is 0
Hence, we have r = 0.
Hence option [c] is correct.

Note: Alternative Solution: By Euclid’s division algorithm:
For finding HCF of two numbers m and n, we set the smaller of the two numbers as b and the larger of the two numbers as a. Then we apply Euclid’s division lemma on a and b, i.e.
a = bq+r .
If r = 0, then HCF(m,n) is b. Otherwise we repeat the above process with a = b and b = r.
We set a = m and b = n.
Now, we apply Euclid’s division lemma on a, b, we get
a = bq+r
i.e. m = nq+r
Since n is the HCF(m,n), we get r = 0, which is the same as obtained above.
Hence option [c] is correct.