Question

what is the HCF of 485 and 1029​

Answer 1

Answer:

It is given that the required number when divides 245 and 1029, the remainder is 5 in each case, This means that 245−5=240 and 1029−5=1024 are completely divisible by the required number.

It follows from this that the required number is a common factor of 240 and 1024. It is also given that the required number is the largest number satisfying the given property. Therefore, it is the HCF of 240 and 1024.

Let us now find the HCF of 240 and 1024 by Prime factorization method(refer above image).

Clearly, HCF of 240 and 1024 is common divisor i.e., 2×2×2×2=16.

Hence, required number =16.