# Find the greatest number which when divides $ 253 $ , $ 568 $ and $ 813 $ leaves the same remainder each time.

Answer

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**Hint:**First we have to define what the terms we need to solve the problem are.

Since in the given question they are asking to find the greatest number among the given set of three numbers, and it will need to leave some remainder each time, the concept of highest common factor and prime factorization is going to be used to solve further.

**Complete step by step answer:**

Since we know something about GCD, LCM and HCF in our schools, where GCD is the greatest common divisor if the GCD is one then it is relatively prime too and LCM is the refers us the least common multiply and hence we can also see the HCF is the highest common factor one among as;

Let from the given question is the greatest among some numbers which divides $ 253 $ , $ 568 $ and $ 813 $

And it will leave as the same remainder each time; hence we first need to find the HCF which is common difference of two terms so take $ 568 - 253,813 - 568,813 - 253 $ (needs to be positive so put the biggest values first)

Hence the HCF of the given three numbers is $ 315,245,560 $

Now we are going to find the prime factors of the $ 315,245,560 $ (taking common multiplies out)

Thus 315 can be written as $ 315 = 3 \times 3 \times 5 \times 7 $ , thus similarly for others too $ 245 = 7 \times 7 \times 5 $ also final term is $ 560 = 2 \times 2 \times 2 \times 2 \times 5 \times 7 $ now we check the highest common factors are five and seven only (two is lowest term as well as three)

Therefore, the greatest number which when divides $ 253 $ , $ 568 $ and $ 813 $ leaves the same remainder each time is $ 35 $

**Note:**Since in this question they are asking to find the highest common factor only, or else two and three also divides $ 253 $ , $ 568 $ and $ 813 $ leaves the same remainder. (If it is least common multiple question the answer will be two and three respectively, and for greatest common divisor question means the answer will be seven only)

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