1) Same Remainder

2) Different Remainders

3) Based on number of digits  

Concept 1:

LCM Model 1: (Same Remainder)

            The least number which when divided by x, y, z so as to leave the same remainder R in each case. Then, the

Required number = LCM (x, y, z) + R

Eg: Find the least number which when divided by 4, 9, 12 leaves the same remainder in each case.

Sol: Required least number = LCM (4, 9, 12) + 3 = 36 + 3 = 39.

Concept 2:

LCM Model 2: (Different Remainders)

     The least number which when divided by x, y, z so as to leave the same remainders R_1, R₂

and R₃ respectively, then in each case. Then, the

Required number = LCM (x, y, z) – Common Difference

Where, common difference = x – R₁ = y – R₂ = z – R₃

Note: Common difference value must be equal for all the values. If common difference values are not equal, then the answer is calculated through options.

Eg: Find the least number which when divided by 10, 15, 20 leaves the remainders 6, 11 and 16 respectively.

Sol: Here, common difference = 10 – 6 = 15 – 11 = 20 – 16 = 4 (Equal for all values)

Required least number = LCM (10, 15, 20) – Common difference = 60 – 4 = 56.

Concept 3:

LCM Model 3: (Based on number of digits)

            Again in LCM third model we have 3 cases

Case 1: The least number of n–digits which is exactly divisible by x, y, z. Then,

Required number = Least n–digit number + LCM – Remainder (R)

Where ‘R’ is the remainder obtained by dividing least n–digit number with LCM of x, y, z.

Eg: Find the least number of 4–digits which is exactly divisible by 6, 8, 12.

Sol: Required number = Least 4–digit number + LCM (6, 8, 12) – R

LCM of 6, 8, 12 = 24 & Least 4–digit number = 1000

After dividing 1000 with 24 we will get 16 as Remainder.

Required number = 1000 + 24 – 16 = 1008.

Case 2: The least number of n–digts which when divided by x, y, z so as to leave the same remainder R₁ in each case. Then,

Required number = (Least n–digit number + LCM – R) + R₁

Where ’R’ is the remainder obtained by dividing least n–digit number with LCM of x, y, z and ‘R₁’ is the remainder mentioned in the given problem.

Eg: Find the least number of 4 digits which when divided by 10, 14, 16 leaves the same remainder 7 in each case.

Sol: Required number = Least 4–digit number + LCM (10, 14, 16) – R + 7

LCM of 10, 14, 16 = 560 & Least 4–digit number = 1000

After dividing 1000 with 560 we will get 440 as Remainder.

Required number = 1000 + 560 – 440 + 7 = 1127.

Case 3: The least number of n–digits which when divided by x, y, z so as to leave the remainders R₁, R₂ and R₃ respectively. Then,

Required number = (Least n–digit number + LCM – R) – Common difference

Where ‘R’ is the remainder obtained by dividing Least n–digit number with LCM of   x, y, z and common difference = x – R₁ = y – R₂ = z – R₃

Eg: Find the least number of 4–digits which when divided by 4, 9, 12 leaves the remainders 2, 7, 10 respectively.

Sol: Required number = Least 4–digit number + LCM (4, 9, 12) – R – Common difference

LCM of 4, 9, 12 = 36, Least 4–digit number = 1000 & C. D = 4 – 2 = 9 – 7 = 12 – 10 = 2

After dividing 1000 with 36 we will get 28 as Remainder.

Required number = 1000 + 36 – 28 – 2 = 1006.

Concept 4:

HCF Model 1: Same Remainder

Again in HCF first model we have 2 cases.

Case 1: (Remainder mentioned)

The greatest number which will divide x, y, z so as to leave the same remainder R in each case. Then,

Required number = HCF [(x – R), (y – R), (z – R)]

Eg: Find the greatest number which will divide 14, 24, 34 so as to leave the same remainder 4 in each case.

Sol: Required number = HCF [(14 – 4), (24 – 4), (34 – 4)] = HCF [10, 20, 30] = 10

Case 2: (Remainder not mentioned)

The greatest number which on dividing x, y, z so as to leave the same remainder in each case. Then,

Required number = HCF [|x – y|, |y – z|, |z – x|]

Eg: Find the greatest number which on dividing 34, 52, 76 leves the same remainder in each case.

Sol: Required number = HCF [|34 – 52|, |52 – 76|, |76 – 34|] = HCF (18, 24, 42) = 6

Concept 5:

HCF Model 2: Different Remainders

The greatest number which will divide x, y, z so as to leave the remainders R₁, R₂ and R₃ respectively. Then,

Required number = HCF [(x – R₁), (y – R₂), (z – R₃)]

Eg: Find the greatest number which will divide 126 and 234 so as to leave the remainders 6 and 2 respectively.

Sol: Required number = HCF [(126 – 6), (234 – 2)] = HCF (120, 232) = 8.

Concept 6:

HCF Model 3: Based on number of digits

Again in HCF third model we have 3 cases.

Case 1: The greatest number of n–digits which when divided by x, y, z leaves no remainder. Then,

Required number = Greatest n–digit number ­­­­­– Remainder (R)

Where ‘R’ is the remainder obtained by dividing  greatest n–digit number with LCM.

Eg: Find the greatest 4–digit number which is exactly divisible by 8, 14, 26.

Sol: Required number = Greatest 4–digit number – Remainder (R)

Greatest ­4–digit number = 9999 & LCM of 8, 14, 26 = 728

After dividing 9999 with 728 we will get 535 as Remainder.

Required number = 9999 – 535 = 9464.

Case 2: The greatest number of n–digits which when divided by x, y, z leaves the same remainder R₁ in each case. Then,

Required number = (Greatest n–digit number ­­­­­– Remainder R) + R₁

Where ‘R’ is the remainder obtained by dividing greatest n–digit number with LCM and R₁ is the remainder mentioned in the given problem.

Eg: Find the greatest number of 4–digits which when divided by 12, 16 and 18 leaves the same remainder 9 in each case.

Sol: Required number = (Greatest 4–digit number – Remainder R) + R₁

Greatest 4–digit number = 9999, R₁ = 9 & LCM of 12, 16, 18 = 144

After dividing 9999 with 144 we will get 63 as Remainder.

Required number = 9999 – 63 + 9 = 9945.

Case 3: The gretest number of n–digits which when divided by x, y, z leaves the remainders R₁, R₂ and R₃ respectively. Then,

Required number = (Greatest n–digit number ­­­­­– Remainder R) – Common difference

Where ‘R’ is the remainder obtained by dividing greatest n–digit number with LCM.

Common difference = x – R₁ = y – R₂ = z – R₃

Eg: Find the greatest 4–digit number which when divided by 24, 32, 48 leaves the remainders 15, 23, 39 respectively.

Sol: Required number = (Greatest 4–digit number – R) – Common difference

Common difference = 24 – 15 = 32 – 23 = 48 – 39 = 9 & LCM of 24, 32, 48 = 96

After dividing 9999 with 96 we will get 15 as Remainder.

Required number = 9999 – 15 – 9 = 9975.

Find the LCM of 24, 36, 48 and 60 by factorisation method.

A) 480                           B) 720                          C) 360                          D) 240 

Sol:          LCM of  24, 36, 48 and 60 by long division method

                           24 = 2³ × 3

                           36 = 2² × 3²

                           48 = 2⁴ × 3

                           60 = 2² × 3 × 5

⸫ Required LCM =  2⁴ × 3² × 5 = 720. 

ANSWER : B                                     

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