let 13 and 273 are the hcf and lcm of two numbers respectively, and if one of them is less than 140 and greater than 60. then what will be that number?

2 months ago 20

Given:

Let the two numbers be aaa and bbb. Since the HCF is 13, we can express the numbers as:

a=13x,b=13ya=13x,\quad b=13ya=13x,b=13y

where xxx and yyy are coprime integers (because the HCF is factored out). From the relation between HCF and LCM:

a×b=HCF×LCM ⟹ (13x)(13y)=13×273a\times b=\text{HCF}\times \text{LCM}\implies (13x)(13y)=13\times 273a×b=HCF×LCM⟹(13x)(13y)=13×273

169xy=3549169xy=3549169xy=3549

xy=3549169=21xy=\frac{3549}{169}=21xy=1693549=21

Now, xxx and yyy are factors of 21, and both must be greater than 1 (since aaa and bbb are greater than 13):

Factors of 21: 1, 3, 7, 21
Possible pairs for (x,y)(x,y)(x,y) are (3,7) or (7,3) (excluding 1 and 21 because x,y>1x,y>1x,y>1).

Therefore:

a=13×3=39,b=13×7=91a=13\times 3=39,\quad b=13\times 7=91a=13×3=39,b=13×7=91

Among these, the number that is greater than 60 and less than 140 is 91. Answer: The number that lies between 60 and 140 is 91