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?

June 28, 2026

The required number is 91.

Step-by-step reasoning

We’re told:

HCF of two numbers = 13
LCM of the two numbers = 273
One of the numbers is between 60 and 140

For any two numbers aaa and bbb, the relation holds:
HCF×LCM=a×b\text{HCF}\times \text{LCM}=a\times bHCF×LCM=a×b.

So here:

a×b=13×273=3549a\times b=13\times 273=3549a×b=13×273=3549

Since HCF = 13, both numbers must be multiples of 13.
Let:

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

Then:

a×b=13x×13y=169xy=3549a\times b=13x\times 13y=169xy=3549a×b=13x×13y=169xy=3549

So:

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

Now we need factor pairs of 21:

1×211\times 211×21
3×73\times 73×7

So possible (x,y)(x,y)(x,y) pairs are:

(1,21)(1,21)(1,21) → numbers 13×1=1313\times 1=1313×1=13 and 13×21=27313\times 21=27313×21=273
(3,7)(3,7)(3,7) → numbers 13×3=3913\times 3=3913×3=39 and 13×7=9113\times 7=9113×7=91

Next, check which numbers lie between 60 and 140:

13 → not in range
273 → not in range
39 → not in range
91 → in range (60 < 91 < 140)

So the number that satisfies the condition is 91.

Therefore, if 13 and 273 are the HCF and LCM of two numbers respectively, and one number is less than 140 and greater than 60, that number is 91.

TL;DR: Use the formula HCF×LCM=product of the numbers\text{HCF}\times \text{LCM}=\text{product of the numbers}HCF×LCM=product of the numbers, express both numbers as multiples of 13, factor 21, and test which resulting number lies between 60 and 140 → 91.

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