Q1. Find the H.C.F. of the following numbers using prime factorization method:
(i) 144,198
(ii) 81,117
(iii) 84,98
(iv) 225,450
(v) 170,238
(vi) 504,980
(vii) 150,140,210
(viii) 84,120,138
(ix) 106,159,265
Sol.
\begin{array}{l|l} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline & 3 \\ \end{array}$$ \Rightarrow \text {The prime factorization of 144 is } 2 \times 2 \times 2 \times 3 \times 3 $$\begin{array}{l|l} 2 & 198 \\ \hline 3 & 99 \\ \hline 3 & 33 \\ \hline & 11 \\ \end{array}$$ \Rightarrow \text {The prime factorization of 198 is } 2 \times 3 \times 3 \times 11 $$We can clearly see that common prime factors are 2 and 3.
The minimum number of times the prime factor 2 occurs is one time.
Similarly, The minimum number of times the prime factor 3 occurs is two times.
$$ \text {Hence, HCF of 144, 198 is } 2 \times 3 \times 3 = 18 $$$$ \text {(ii) Prime factorization of 81 } =3 \times 3 \times 3 \times 3 $$ $$ \text {Prime factorization of 117} =3 \times 3 \times 13 $$ $$ \text {Hence, HCF of 81,117 is } 3 \times 3=9 $$$$ \text {(iii) Prime factorization of 84 }=2 \times 2 \times 3 \times 7 $$ $$ \text {Prime factorization of 98 } =2 \times 7 \times 7 $$ $$ \text {Hence, HCF of 84, 98 } =2 \times 7=14 $$$$ \text {(iv) Prime factorization of 225} =3 \times 3 \times 5 \times 5 $$ $$ \text {Prime factorization of 450 }=2 \times 3 \times 3 \times 5 \times 5 $$$$ \text {Hence, HCF of 225, 450 }=3 \times 3 \times 5 \times 5 = 225 $$$$ \text {(v) Prime factorization of 170} =2 \times 5 \times 17 $$ $$ \text {Prime factorization of 238}=2 \times 7 \times 17 $$ $$ \text {Hence, HCF of 170,238 }=2 \times 17=34 $$$$ \text {(vi) Prime factorization of 504} =2 \times 2 \times 2 \times 3 \times 3 \times 7 $$$$ \text {Prime factorization of 980}=2 \times 2 \times 5 \times 7 \times 7 $$ $$ \text {Hence, HCF of 504,980 }=2 \times 2 \times 7=28 $$$$ \text {(vii) Prime factorization of 150 }=2 \times 3 \times 5 \times 5 $$ $$ \text {Prime factorization of 140}=2 \times 2 \times 5 \times 7 $$ $$ \text {Prime factorization of 210 }=2 \times 3 \times 5 \times 7 $$ $$ \text {Hence, HCF of 150,140 and 210 }=2 \times 5=10 $$$$ \text {(viii) Prime factorization of 84}=2 \times 2 \times 3 \times 7 $$ $$ \text {Prime factorization of 120}=2 \times 2 \times 2 \times 3 \times 5 $$ $$ \text {Prime factorization of 138} =2 \times 3 \times 23 $$ $$ \text {Hence, HCF of 84,120 and 138}=2 \times 3=6 $$$$ \text {(ix) Prime factorization of 106}=2 \times 53 $$ $$ \text {Prime factorization of 159}=3 \times 53 $$ $$ \text {Prime factorization of 265}=5 \times 53 $$ $$ \text {Hence, HCF of 106,159 and 265}=53 $$Q2. What is the H.C.F. of two consecutive
(i) Numbers
(ii) even numbers
(iii) odd numbers
Sol. (i) The common factor of two consecutive numbers is always 1.
Therefore, HCF of two consecutive numbers =1
(ii) The common factors of two consecutive even numbers are 1 and 2.
Therefore, HCF of two consecutive even numbers = 2
(iii) The common factor of two consecutive odd numbers is 1.
Therefore, HCF of two consecutive odd numbers =1
Q3. H.C.F. of co-prime numbers 4 and 15 was found as follows:
$$ 4=2 \times 2 \text { and } 15= 3 \times 5 $$ Since there is no common prime factor.
So, H.C.F. of 4 and 15 is 0. Is the answer correct?
If not, what is the correct H.C.F.?
Sol. No, it’s not correct.
We know that HCF of two co-prime number is 1.
And, 4 and 15 are co-prime numbers because 1 is the only common factor to them.
Hence, HCF of 4 and 15 is 1.
(i) 144,198
(ii) 81,117
(iii) 84,98
(iv) 225,450
(v) 170,238
(vi) 504,980
(vii) 150,140,210
(viii) 84,120,138
(ix) 106,159,265
Sol.
\begin{array}{l|l} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline & 3 \\ \end{array}$$ \Rightarrow \text {The prime factorization of 144 is } 2 \times 2 \times 2 \times 3 \times 3 $$\begin{array}{l|l} 2 & 198 \\ \hline 3 & 99 \\ \hline 3 & 33 \\ \hline & 11 \\ \end{array}$$ \Rightarrow \text {The prime factorization of 198 is } 2 \times 3 \times 3 \times 11 $$We can clearly see that common prime factors are 2 and 3.
The minimum number of times the prime factor 2 occurs is one time.
Similarly, The minimum number of times the prime factor 3 occurs is two times.
$$ \text {Hence, HCF of 144, 198 is } 2 \times 3 \times 3 = 18 $$$$ \text {(ii) Prime factorization of 81 } =3 \times 3 \times 3 \times 3 $$ $$ \text {Prime factorization of 117} =3 \times 3 \times 13 $$ $$ \text {Hence, HCF of 81,117 is } 3 \times 3=9 $$$$ \text {(iii) Prime factorization of 84 }=2 \times 2 \times 3 \times 7 $$ $$ \text {Prime factorization of 98 } =2 \times 7 \times 7 $$ $$ \text {Hence, HCF of 84, 98 } =2 \times 7=14 $$$$ \text {(iv) Prime factorization of 225} =3 \times 3 \times 5 \times 5 $$ $$ \text {Prime factorization of 450 }=2 \times 3 \times 3 \times 5 \times 5 $$$$ \text {Hence, HCF of 225, 450 }=3 \times 3 \times 5 \times 5 = 225 $$$$ \text {(v) Prime factorization of 170} =2 \times 5 \times 17 $$ $$ \text {Prime factorization of 238}=2 \times 7 \times 17 $$ $$ \text {Hence, HCF of 170,238 }=2 \times 17=34 $$$$ \text {(vi) Prime factorization of 504} =2 \times 2 \times 2 \times 3 \times 3 \times 7 $$$$ \text {Prime factorization of 980}=2 \times 2 \times 5 \times 7 \times 7 $$ $$ \text {Hence, HCF of 504,980 }=2 \times 2 \times 7=28 $$$$ \text {(vii) Prime factorization of 150 }=2 \times 3 \times 5 \times 5 $$ $$ \text {Prime factorization of 140}=2 \times 2 \times 5 \times 7 $$ $$ \text {Prime factorization of 210 }=2 \times 3 \times 5 \times 7 $$ $$ \text {Hence, HCF of 150,140 and 210 }=2 \times 5=10 $$$$ \text {(viii) Prime factorization of 84}=2 \times 2 \times 3 \times 7 $$ $$ \text {Prime factorization of 120}=2 \times 2 \times 2 \times 3 \times 5 $$ $$ \text {Prime factorization of 138} =2 \times 3 \times 23 $$ $$ \text {Hence, HCF of 84,120 and 138}=2 \times 3=6 $$$$ \text {(ix) Prime factorization of 106}=2 \times 53 $$ $$ \text {Prime factorization of 159}=3 \times 53 $$ $$ \text {Prime factorization of 265}=5 \times 53 $$ $$ \text {Hence, HCF of 106,159 and 265}=53 $$Q2. What is the H.C.F. of two consecutive
(i) Numbers
(ii) even numbers
(iii) odd numbers
Sol. (i) The common factor of two consecutive numbers is always 1.
Therefore, HCF of two consecutive numbers =1
(ii) The common factors of two consecutive even numbers are 1 and 2.
Therefore, HCF of two consecutive even numbers = 2
(iii) The common factor of two consecutive odd numbers is 1.
Therefore, HCF of two consecutive odd numbers =1
Q3. H.C.F. of co-prime numbers 4 and 15 was found as follows:
$$ 4=2 \times 2 \text { and } 15= 3 \times 5 $$ Since there is no common prime factor.
So, H.C.F. of 4 and 15 is 0. Is the answer correct?
If not, what is the correct H.C.F.?
Sol. No, it’s not correct.
We know that HCF of two co-prime number is 1.
And, 4 and 15 are co-prime numbers because 1 is the only common factor to them.
Hence, HCF of 4 and 15 is 1.