LCM and HCF
 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 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
 Find the greatest number of five digits which when divided by 3, 5, 8, 12 have 2 as remainder :

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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.
The greatest number of five digits is 99999.
LCM of 3, 5, 8 and 12
∴ LCM = 2 × 2 × 3 × 5 × 2 = 120
After dividing 99999 by 120, we get 39 as remainder
99999 – 39 = 99960 = ( 833 × 120 )Correct Option: D
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.
The greatest number of five digits is 99999.
LCM of 3, 5, 8 and 12
∴ LCM = 2 × 2 × 3 × 5 × 2 = 120
After dividing 99999 by 120, we get 39 as remainder
99999 – 39 = 99960 = ( 833 × 120 )
99960 is the greatest five digit number divisible by the given divisors.
In order to get 2 as remainder in each case we will simply add 2 to 99960.
∴ Greatest number = 99960 + 2 = 99962
 The least perfect square, which is divisible by each of 21, 36 and 66 is

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LCM of 21, 36 and 66
∴ LCM = 3 × 2 × 7 × 6 × 11
⇒ LCM = 3 × 3 × 2 × 2 × 7 × 11
∴ Required number = 3^{2} × 2^{2} × 7^{2} × 11^{2}Correct Option: C
LCM of 21, 36 and 66
∴ LCM = 3 × 2 × 7 × 6 × 11
⇒ LCM = 3 × 3 × 2 × 2 × 7 × 11
∴ Required number = 3^{2} × 2^{2} × 7^{2} × 11^{2}
Required number = 213444
 Let the least number of six digits which when divided by 4, 6, 10, 15 leaves in each case same remainder 2 be N. The sum of digits in N is :

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LCM of 4, 6, 10, 15 = 60
Least number of 6 digits = 100000
The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
∴ Required number (N) = 100020 + 2 = 100022Correct Option: B
LCM of 4, 6, 10, 15 = 60
Least number of 6 digits = 100000
The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
∴ Required number (N) = 100020 + 2 = 100022
Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5