Given:
- HCF = 13
- LCM = 273
- One of the two numbers is less than 140 and greater than 60.
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