Question

How many ordered pairs (a,b) exists such that hcf of a,b=1 and a*b=1386

Answer 1

answer is 7
use prime factorization
18,77
22,63
99,14
693,2
7,198
126,11
154,9

Answer 2

There are 7 ordered pairs such that their HCF 1 and product = 1386.

Given:

HCF (a, b)=1 and a*b=1386.

To Find:

The number of ordered pairs (a, b) satisfying the given conditions.

Solution:

HCF or the highest common factor of two numbers is the largest number that divides both the numbers.

Since prime numbers have only two divisors, 1 and the number itself, their HCF is always 1.

Those numbers that do not have any common factors other than 1 ( mutually prime) also have HCF = 1.

We are given that for two numbers a and b,

HCF (a, b)=1 and a*b=1386.

Let us find the prime factorization of 1386.

1386 = 2×3×3×7×11 = 2 x 9 x 7 x 11.

We need to find two numbers that a mutually prime and their product = 1386. The possible combinations are:

(a,b) = (7,198)

(a,b) = (126,11)

(a,b) = (154,9)

(a,b) = (18,77)

(a,b) = (22,63)

(a,b) = (99,14)

(a,b) = (693,2)

We obtain a total of 7 pairs of numbers that satisfies the given conditions.

∴ There are 7 ordered pairs such that their HCF 1 and product = 1386.

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