Find the largest no. that will divide 20,57,85 leaving remainders 2.3,4 respectively

Hi!

Here is the answer to your question.

 

Since on dividing 20 by the required number, the remainder is 2

∴ 20 – 2 = 18 will be exactly divisible by the required number.

Similarly, 57 – 3 = 54 and 85 – 4 = 81 will be also exactly divisible by the required number.

∴The required number is HCF of 18, 54 and 81.

Prime factorization of 18 = 2 × 3 × 3

Prime factorization of 54 = 2 × 3 × 3 × 3

Prime factorization of 81 = 3 × 3 × 3 × 3

HCF of 18, 54 and 81 = 3 × 3 = 9

Thus, the largest number that will divide 20,57,85 leaving remainders 2,.3,4 respectively is 9.

 

Cheers!