LCM and HCF
 The least number, which when divided by 12, 15, 20 or 54 leaves a remainder of 4 in each case, is :

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Here , Remainder ( r ) = 4
As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
So , LCM of 15, 12, 20, 54 ( k ) = 540Correct Option: D
Here , Remainder ( r ) = 4
As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
So , LCM of 15, 12, 20, 54 ( k ) = 540
Hence , Then number = k + r = 540 + 4 = 544
 The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is :

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Here , Remainder ( r ) = 2
As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
So , L.C.M. of 4, 6, 8, 12 and 16 = k = 48Correct Option: C
Here , Remainder ( r ) = 2
As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
So , L.C.M. of 4, 6, 8, 12 and 16 = k = 48
∴ Required number = k + r = 48 + 2 = 50
 The H.C.F. and L.C.M. of two numbers are 44 and 264 respectively. If the first number is divided by 2, the quotient is 44. The other number is

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Given , The H.C.F. and L.C.M. of two numbers are 44 and 264 .
According to question ,
First number = 2 × 44 = 88
As we know that ,
∴ First number × Second number = H.C.F. × L.C.M.
⇒ 88 × Second number = 44 × 264Correct Option: C
Given , The H.C.F. and L.C.M. of two numbers are 44 and 264 .
According to question ,
First number = 2 × 44 = 88
As we know that ,
∴ First number × Second number = H.C.F. × L.C.M.
⇒ 88 × Second number = 44 × 264⇒ Second number = 44 × 264 = 132 88
 The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?

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Given that , The L.C.M. of three different numbers is 120.
LCM = 2 × 2 × 2 × 3 × 5
Hence, HCF = 4, 8, 12 or 24Correct Option: D
LCM = 2 × 2 × 2 × 3 × 5
Hence, HCF = 4, 8, 12 or 24
According to question ,
35 cannot be H.C.F. of 120.
Hence , required answer is 35.
 The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. The larger of the two numbers is :

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Let the numbers be 23p and 23q .where p and q are coprime.
∴ LCM = 23pq
According to question ,
23pq = 23 × 13 × 14
∴ p = 13, q = 14Correct Option: E
Let the numbers be 23p and 23q .where p and q are coprime.
∴ LCM = 23pq
According to question ,
23pq = 23 × 13 × 14
∴ p = 13, q = 14
∴ The larger number = 23q = 23 × 14 = 322