Problem on HCF and LCM

Question 1: Find the HCF by long division method of two no’s the sequence of quotient from top to bottom is 9, 8, 5 and the last divisor is 16. Find the two no’s. Solution: Start with the divisor and last quotient. Divisor x quotient + remainder = Dividend 16 x 5 + 0 = 80 80 x 8 + 16 = 656 656 x 9 + 80 = 5984 Hence, two numbers are 656 and 5984. Question 2: The LCM and HCF of two numbers is 210 and 5. Find the possible number of pairs. Solution: HCF = 5 so it should be multiple of both numbers. So both numbers 5x : 5y LCM = 5 * x * y = 210 x * y = 42 {1 x 42}, { 2 x 21}, {3 x 14}, { 6 x 7 } . Four pairs are possible. Question 3: The sum of two numbers is 132 and their LCM is 216. Find both the numbers. Solution: Note: HCF of Sum & LCM is also same as actual HCF of two numbers. Factorize both 132 and 216 and find the HCF. 132= 2² x 3 x 11 216= 2³x 3³ HCF= 2² x 3 =12 Now, 12x + 12y = 132 x + y = 11 And 12 * x * y = 216 x * y = 18 Solve for x and y, we get y = 9 and x = 2. Hence both numbers are 12*2 = 24 and 12*9 = 108 Question 4: The LCM of two numbers is 15 times of HCF. The sum of HCF and LCM is 480. If both number are smaller than LCM. Find both the numbers. Solution: LCM = 15 * HCF We know that LCM + HCF = 480 16 * HCF = 480 HCF = 30 Then LCM = 450 LCM = 15 HCF 30 * x * y = 15 * 30 x * y = 15 Factors are {1 x 15} and { 3 x 5} Both numbers less than LCM so take {3 x 5} Hence numbers are 3 * 30 = 90 and 5 * 30 = 150 Question 5: Find the least perfect square number which when divided by 4, 6, 7, 9 gives remainder zero. Solution: Find the LCM for 4, 6, 7, 9 LCM= 2² * 3² * 7 = 252 To become perfect square all factors should be in power of 2. So, multiply it by 7 LCM = 2² * 3² * 7² = 1764 And it is perfect square of 42.