Class 6 Playing with Numbers Exercise 2.8-1

\begin{array}{l} \text {Q1. Find the largest number which divides 615 and 963 leaving remainder 6 in each case. } \\ \\\text {Sol. To find the largest number which divides 615 and 963 leaving remainder 6 means } \\ \text {we have to find the largest number which divides } (615-6) =609 \text { and } (963-6)= 957 \text { exactly.} \\ \Rightarrow \text {The required number is HCF of 609 and 957} \\ \text {Prime factors of 609 } =3 \times 7 \times 29 \\ \text {Prime factors of 957} =3 \times 11 \times 29 \\ \Rightarrow \text {HCF of 609 and 957} = 29 \times 3=87 \\ \text {Hence, the largest number which divides 615 and 963 leaving remainder 6 in each case is 87.} \\ \\\text {Q2. Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively. } \\ \\\text {Sol. To find the largest number which divides 285 and 1249 leaving } \\ \text {remainder 9 and 7 respectively means we have to find the largest number which } \\ \text {divides } (285-9) =276 \text { and } (1249-7)= 1242 \text { exactly.} \\ \Rightarrow \text {The required number is HCF of 276 and 1242} \\ \text {Prime factors of 276 } =2 \times 2 \times 3 \times 23 \\ \text {Prime factors of 1242} =2 \times 3 \times 3 \times 3 \times 23 \\ \Rightarrow \text {HCF of 276 and 1242} = 2 \times 3 \times 23=138 \\ \text {Hence, the largest number which divides 285 and 1249 leaving remainders 9 and 7 is 138.} \\ \\\text {Q3. What is the largest number that divides 626,3127 and 15628 } \\ \text {and leaves remainders of 1,2 and 3 respectively? } \\ \\\text {Sol. To find the largest number which divides 626,3127 and 15628 leaving } \\ \text {remainder 1,2 and 3 respectively means we have to find the largest number } \\ \text {which divides } (626-1) =625, (3127-2) =3125 \text { and } (15628-3)= 15625 \text { exactly.} \\\Rightarrow \text {The required number is HCF of 625, 3125 and 15625} \\ \text {Prime factors of 625 } =5 \times 5 \times 5 \times 5 \\ \text {Prime factors of 3125} =5 \times 5 \times 5 \times 5 \times 5 \\ \text {Prime factors of 15625} =5 \times 5 \times 5 \times 5 \times 5 \times 5 \\\Rightarrow \text {HCF of 625, 3125 and 15625} = 5 \times 5 \times 5 \times 5 =625 \\ \text {Hence, the greatest number which divides 626,3127 and 15628 leaving remainders 1,2 and 3 is 625.} \\ \\\text {Q4. The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm } \\ \text { and 4 m 50 cm respectively. Determine the longest rod which can measure } \\ \text {the three dimensions of the room exactly.} \\ \\\text {Sol. Given that Length of the room =8 m 25 cm =825 cm } \quad \quad [\text { 1 m = 100 cms}]\\ \text {Breadth of the room =6 m 75 cm =675 cm} \quad \quad [\text { 1 m = 100 cms}]\\ \text {Height of the room =4 m 50 cm=450 cm} \quad \quad [\text { 1 m = 100 cms}]\\ \Rightarrow \text {The longest rod that can measure all 3 dimensions } \\ \text {will be the HCF of 825,675 and 450.} \\\text {Prime factors of 825} =3 \times 5 \times 5 \times 11 \\ \text {Prime factors of 675} =3 \times 3 \times 3 \times 5 \times 5\\ \text {Prime factors of 450} =2 \times 3 \times 3 \times 5 \times 5 \\ \Rightarrow \text {HCF of 825,675 and 450 }=3 \times 5 \times 5=75 \\ \text {Thus, the required length of the longest rod is 75 cm} \\ \\\text {Q5. A rectangular courtyard is 20 m 16 cm long and 15 m 60 cm broad. } \\ \text {It is to be paved with square stones of the same size. Find the } \\ \text {least possible number of such stones.} \\ \\\text {Sol. Given Length of the rectangular courtyard } \\ = \text {20 m 16 cm = 2016 cm} \quad \quad [\text { 1 m = 100 cms}]\\ \text {Breadth of the rectangular courtyard =15 m 60 cm=1560 cm} \\\text {Least possible no. of square stones used to pave the rectangular courtyard } \\ \text {means side of square should be maximum i.e. HCF of 2016 and 1560} \\\text {Prime factorization of 2016 }=2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \\\text {Prime factorization of 1560 }=2 \times 2 \times 2 \times 3 \times 5 \times 13 \\\text {HCF of 2016 and 1560}=2 \times 2 \times 2 \times 3=24 \\\Rightarrow \text {Side of square stones used to pave the rectangular courtyard is 24 cm} \\\Rightarrow \text {Number of square stones used to pave the rectangular courtyard } \\= {\large \frac {\text {Area of rectangular courtyard}}{\text {Area of square stone}} }\\ ={\large\frac {2016 \times 1560}{24 \times 24} }\\ =5460 \\ \text {Thus, the least number of square stones used to pave the rectangular courtyard is 5460.} \\ \\ \end{array}