An important and conceptual question from real numbers chapter in which we have given a number 161 and we have been asked to find the other number if their HCF is 23 and their LCM is 1449.
Kindly provide me a detailed solution of this question
RS Aggarwal, Class 10, chapter 1B, question 3
Given, HCF=23
LCM=1449
Let the two numbers be a and b. Let a=161. We have to find b.
We know that
HCF(a,b)×LCM(a,b)=a×b
⇒23×1449=161×b
⇒23×1449/161=b
⇒b=23×9
∴b=207
Thus, the other number is 207.