# find the least digit no which leaves remainder 3 in each case when divided by 5,10,12,15,18,25,30. find the least natural no which when divided by 989and 1892 leaves a remainder 1. find the greast 4-digit novwhich when divided by 36,30,24and 16 leaves a remainder 13 in each case

(1) The given numbers are 5, 10, 12, 15, 18, 25, 30.

The smallest number which when divided by the given numbers leaves the remainder 3 can be obtained by adding 3 to the LCM of given numbers.

Prime factorization of 5 = 5

Prime factorization of 10 = 2 × 5

Prime factorization of 12 = 2 × 2 × 3

Prime factorization of 15 = 3 × 5

Prime factorization of 18 = 2 × 3 × 3

Prime factorization of 25 = 5 × 5

Prime factorization of 30 = 2 × 3 × 5

LCM of 5, 10, 12, 15, 18, 25, 30 = 2 × 2 × 3 × 3 × 5 × 5 = 900

∴ Least / smallest number which when divided by 5, 10, 12, 15, 18, 25, and 30 leaves the remainder 3 is 900 + 3 = 903.

(2) The required number can be obtained by adding 1 to the LCM of given numbers 989 and 1892.

Try it yourself and if you face any problem, please do get back to us.

(3) LCM of 36, 30, 24 and 16 can be calculated as:

36 = 2 × 2 × 3 × 3

30 = 2 × 3 × 5

24 = 2 × 2 × 2 × 3

16 = 2 × 2 × 2 × 2

Thus, LCM of 36, 30, 24 and 16 is  2 × 2 × 2 × 2 × 3 × 3 × 5 = 720.

Greatest four digit number is 9999

Now,

∴ Greatest 4 digit number which is divisible by 36, 30, 24 and 16 = 9999 – 639 = 9360

Hence, the greatest 4 digit number which when divided by 36, 30, 24 and 16 leaves a remainder 13 in each case is 9360 + 13 = 9373.

• -5
What are you looking for?