Questions (with answers) on HCF and LCM

1. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
2. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
3. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
4. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:  
5. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: 
6. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is
   
   answers
7. Answers
   1.      Let the remainder be R
   So, 43 – R, 91 – R, and 183 – R are all divisible by the said divisor.
   Since we have to find the greatest number that divides the above numbers, we have to find the HCF.
   
   
   Now if 2 numbers are multiples of  a given number (HCF), then their difference is also a multiple of the HCF.
   
   
   So, (91 – R) – (43 – R) = multiple of HCF
   Or, 48 = multiple of HCF
   So, (183 – R) – (91 – R) = multiple of HCF
   Or, 92 = multiple of HCF
   So, (183 – R) – (43 – R) = multiple of HCF
   Or, 140 = multiple of HCF
   
   
   Now, HCF of 48, 92 and 140 is 4.
   So, 4 is the answer.
   
   
   2.      Let n be the least multiple
   So, 7n + 4 is divisible by 6, 9, 15 and 18
   Which means the number must be LCM
   LCM of 6,9,15 and 18 is 90
   So, 7n + 4 = 90
   So, n = 12
   
   
   3.      Let the remainder be R
   So, 1305 – R, 4665 – R, and 6905 – R are all divisible by the said divisor.
   Since we have to find the greatest number that divides the above numbers, we have to find the HCF.
   
   
   Now if 2 numbers are multiples of  a given number (HCF), then their difference is also a multiple of the HCF.
   
   
   So, (4665 – R) – (1305 – R) = multiple of HCF
   Or, 3360 = multiple of HCF
   So, (6905 – R) – (4665 – R) = multiple of HCF
   Or, 2240 = multiple of HCF
   So, (6905 – R) – (1305 – R) = multiple of HCF
   Or, 5600 = multiple of HCF
   
   
   3360  = 2 x 2 x 2 x 2 x 2 x 7 x 5 x 3
   2240  = 2 x 2 x 2 x 2 x 2 x 7 x 5 x 2
   5600 =  2 x 2 x 2 x 2 x 2 x 7 x 5 x 5
   
   
   HCF = 1120
   Therefore, the sum of digits of HCF = 4
   
   
   4.      The smallest number which is divisible by 15, 25, 40 and 75 is the LCM
   LCM of 15, 25, 40, 75 is 600
   
   
   To find the greatest number of 4 digit that is divisible by 600, we will take the smallest number of 5 digit, i.e 10000 and divide by 600
   We get 16 ⁴⁰⁰/₆₀₀
   So, the greatest number of 4 digit is 10000 – 400 = 9600
   
   
   5.      Product of 2 numbers = LCM x HCF
   4107 = LCM x 37
   Or, LCM = 111
   The factors of LCM are 3 x 37
   Therefore the greater number is 3 x 37 =  111
   
   
   6.      Product of 2 numbers = LCM x HCF
   2028= LCM x 13
   Or, LCM = 156
   The factors of LCM are 2 x 2 x 3 x 13x 1
   The possible pair of numbers will be, (13 x 2 x 2, 13 x 3), (13 x 2 x 2 x 3 , 1 x 13)
   (52, 39), (156, 13)