LCM and HCF

LCM and HCF

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

1. 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

Correct 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

1. The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is :

1. 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

Correct 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

1. 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

1. 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

Correct 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

1. The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?

1. Given that , The L.C.M. of three different numbers is 120.
LCM = 2 × 2 × 2 × 3 × 5
Hence, HCF = 4, 8, 12 or 24

Correct 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.

1. 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 :

1. Let the numbers be 23p and 23q .where p and q are co-prime.
∴ LCM = 23pq
According to question ,
23pq = 23 × 13 × 14
∴ p = 13, q = 14

Correct Option: E

Let the numbers be 23p and 23q .where p and q are co-prime.
∴ LCM = 23pq
According to question ,
23pq = 23 × 13 × 14
∴ p = 13, q = 14
∴ The larger number = 23q = 23 × 14 = 322