Question

The h.c.f of two numbers is 17 and their product is 4624 .what is the sum of their reciprocal?

Answer 1

Answer:

$\dfrac{1}{16}$

Step-by-step explanation:

It is given that the h.c.f of two numbers is 17.

Let the two number are 17a and 17b, where a and b are positive integers and the HCF(a,b)=1.

The product of these two numbers is 4624 .

$17a\times 17b=4624$

$289ab=4624$

Divide both sides by 289.

$ab=16$

The factor pairs of 16 are (1,16), (2,8) and (4,4).

Only (1,16) factor pair has HCF 1. So,

First number = 17 × 1 = 17

second number = 17 × 16 = 272

We need to find the sum of their reciprocal.

$\dfrac{1}{17}+\dfrac{1}{272}=\dfrac{16+1}{272}\Rightarrow \dfrac{17}{272}=\dfrac{1}{16}$

Therefore the the sum of their reciprocal is $\dfrac{1}{16}$.