Question

There are five odd numbers \[1,3,5,7\] and \[9\]. What is the HCF of these odd numbers?
A. \[2\]
B. \[1\]
C. \[6\]
D. \[3\]

Answer 1

Hint: In general factors of a number are that numbers which when multiplied together gives the original number. For example \[2\times 3=6\]. and $2,3$ are the factors of \[6\] here. The highest common factor or H.C.F. states finding a common factor in which all the asked numbers in the given question are divisible.

Complete step-by-step solution -
Let us calculate and check what is the highest common factor of \[1,3,5,7\] and \[9\].
Step 1: Find out all the possible factors of the given number]
Step 2: Find the common factor of the number that has been produced.
Step 3: the longest number that you have in the previous step is the highest common factor.
H.C.F. of \[1,3,5,7\] and \[9\]
Factors of \[1=1\]
Factors of \[3=1\] and \[3\]
Factors of \[5=1\] and \[5\]
Factors of \[7=1\] and \[7\]
Factors of \[9=1,3\] and \[9\]
We can see that \[1\] is the common factor with which all the given \[5\] numbers are divisible.
Hence H.C.F. of \[1,3,5,7\]and \[9\] is \[1\].

Note: A number can have many factors. The highest common factor shows that the common factor in which the number’s asked is divisible. Though you can also check the H.C.F. with the long divisibility method. In this method first, we take two numbers of the given set. Then we divide the largest number by the small number until we get a remainder equal to zero. Then the last divisor from which we get the remainder equal to zero is H.C.F of the first two numbers. Now we can find H.C.F of third given number and H.C.F of the first two numbers. This process continues till the last number.