Question

The sum of two numbers is 1215 and their HCF is 81. How many pairs of such numbers can be formed?

(a) 2

(b) 6

(c) 4

(d) None of these

Answer 1

Correct option : (c) 4

Since, HCF of both number is 81.

\(\therefore\) Both numbers are multiple of 81.

Let numbers are 81a & 81b when a & b are coprimes.

\(\therefore 81a +81b = 1215\)

\(\Rightarrow a+b = \frac{1215}{81} =15.\)

Possible pair of a & b are (1,14), (2,13), (4,11) & (7,8).

\(\therefore\) 4 pairs of such numbers are possible.