Class 6 Playing with Numbers Exercise 2.11-1

\begin{array}{l} \text {Q1. For each of the following pairs of numbers, verity the property:} \\ \text {Product of the number = Product of their H.C.F. and L.C.M.} \\ (i) \quad 25,65 \\ (ii) \quad 117,221 \\ (iii) \quad 35,40 \\ (iv) \quad 87,145 \\ (v) \quad 490,1155 \\ \\\text {Sol. (i) Prime factorization of 25 }=5 \times 5 \\ \text {Prime factorization of 65} =5 \times 13 \\\therefore \text {HCF of 25 and 65} =5 \\ \text {LCM of 25 and 65} =5 \times 5 \times 13=325 \\ \text {Product of the given numbers }=25 \times 65=1625 \\ \text {Product of their HCF and LCM}=5 \times 325=1625 \\ \Rightarrow \text {Product of the given numbers} = \text {Product of their HCF and LCM} \\ \text {Hence verified} \\ \\\text {(ii) Prime factorization of 117 }=3 \times 3 \times 13 \\ \text {Prime factorization of 221} =13 \times 17 \\\therefore \text {HCF of 117 and 221} =13 \\ \text {LCM of 117 and 221} =3 \times 3 \times 13 \times 17=1989 \\ \text {Product of the given numbers }=117 \times 221=12857 \\ \text {Product of their HCF and LCM}=13 \times 1989=12857 \\ \Rightarrow \text {Product of the given numbers} = \text {Product of their HCF and LCM} \\ \text {Hence verified} \\ \\\text {(iii) Prime factorization of 35 }=5 \times 7 \\ \text {Prime factorization of 40} =2 \times 2 \times 2 \times 5 \\\therefore \text {HCF of 35 and 40} =5 \\ \text {LCM of 35 and 40} =2 \times 2 \times 2 \times 5 \times 7=280 \\ \text {Product of the given numbers }=35 \times 40=1400 \\ \text {Product of their HCF and LCM}=5 \times 280=1400 \\ \Rightarrow \text {Product of the given numbers} = \text {Product of their HCF and LCM} \\ \text {Hence verified} \\ \\\text {(iv) Prime factorization of 87 }=3 \times 29 \\ \text {Prime factorization of 145} =5 \times 29 \\\therefore \text {HCF of 87 and 145} =29 \\ \text {LCM of 87 and 145} =3 \times 5 \times 29=435 \\ \text {Product of the given numbers }=87 \times 145=12615 \\ \text {Product of their HCF and LCM}=29 \times 435=12615 \\ \Rightarrow \text {Product of the given numbers} = \text {Product of their HCF and LCM} \\ \text {Hence verified} \\ \\\text {(v) Prime factorization of 490 }=2 \times 5 \times 7 \times 7 \\ \text {Prime factorization of 1155} =3 \times 5 \times 7 \times 11 \\\therefore \text {HCF of 490 and 1155 } =35 \\ \text {LCM of 490 and 1155} =2 \times 3 \times 3 \times 5 \times 7 \times 7 \times 11=16710 \\ \text {Product of the given numbers }=490 \times 1155=565950 \\ \text {Product of their HCF and LCM}=490 \times 1155=565950 \\ \Rightarrow \text {Product of the given numbers} = \text {Product of their HCF and LCM} \\ \text {Hence verified} \\ \\\text {Q2. Find the H.C.F. and L.C.M. of the following pairs of numbers: } \\ (i) \quad 117,221 (ii) \quad 234,572 \\ (iii) \quad 145,232 \\ (iv) \quad 861,1353 \\\text {Sol. (i) Prime factorization of 117}=3 \times 3 \times 13 \\ \text {Prime factorization of 221}=13 \times 17 \\ \therefore \text {HCF of 117 and 221} =13 \\ \text {And, LCM of 117 and 221}=3 \times 3 \times 13 \times 17=1989 \\ \\\text {(ii) Prime factorization of 234 } =2 \times 3 \times 3 \times 13 \\ \text {Prime factorization of 572}=2 \times 2 \times 11 \times 13 \\ \therefore \text {HCF of 234 and 572}=226 \\ \text {And, LCM of 117 and 221}=2 \times 2 \times 3 \times 3 \times 11 \times 13=5148 \\ \\\text {(iii) Prime factorization of 145} =5 \times 29 \\ \text {Prime factorization of 232} =2 \times 2 \times 2 \times 29 \\ \therefore \text {HCF of 145 and 232} =289 \\ \text {And, LCM of 145 and 232} =2 \times 2 \times 2 \times 5 \times 29=1160 \\ \\\text {(iv) Prime factorization of 861 }=3 \times 7 \times 41 \\ \text {Prime factorization of 1353}=3 \times 11 \times 41 \\ \therefore \text {HCF of 861 and 1353} =123 \\ \text {And, LCM of 861 and 1353} =3 \times 7 \times 11 \times 41=9471 \\ \\\end{array}