Find the largest number which divides 245 and 1029 leaving remainder 5 in each case.

It is given that required number which divides 245 and 1029, the remainder is 5 in each case.

⇒ 245 - 5 = 240 and 1029 - 5 = 1024 are completely divisible by the required number.

Since, it is given that the required number is the largest number.

Therefore, it is the HCF of 240 and 1024.

Now, finding HCF by Euclid’s division algorithm.

Given integers are 240 and 1024.

Clearly 1024 > 240.

Therefore, it is the HCF of 240 and 1024 and 240, we get



II. Since, the remainder 64 ≠ 0, we apply division lemma to get



III. We consider the new divisor 64 and remainder 48 and apply division lemma to get



IV. We consider the new divisor 48 and new remainder 16 to get


V. The remainder at this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e., 16 is the HCF of 245 and 1029.

 Given integers are 4052 and 12576, clearly 12576 > 4052.

Therefore, by applying Euclid's division lemma to 4052 and 12576, we get

I. 12576 = 4052 × 3 + 420



II. Since the remainder 420 ≠ 0, we apply division lemma to 4052 and 420 to get



III. We consider the new divisor 420 and new remainder 272 and apply division lemma to gel   



IV. We consider the new divisor 272 and new remainder 148 and apply division lemma to get




V. We consider the new divisor 148 and new remainder 124 and apply division lemma to get



VI. We consider the new divisor 124 and new remainder 24 and apply division lemma to get

VII.We consider the new divisor 24 and new remainder 4 and apply division lemma to get



The remainder at this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e. 4 is the HCF of 4052 and 12576.

It is given that required number when divides 615 and 963, the remainder is 6 in each case.

⇒ 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the required number.

Since, it is given that the required number is the largest number. Given integers are 957 and 609 clearly 957 > 609.

Therefore, it is the HCF of 609 and 957.

Now, finding HCF by using Euclid’s division lemma to 609 and 957, we get



II. Since the remainder 348 ≠ 0, we apply division lemma to 348 and 609 to get



III. We consider the new divisor 348 and new remainder 261 and apply division lemma to get



IV. We consider the new divisor 261 and new remainder 87 and apply division lemma to get



The remainder at this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e., 87 is the HCF of 615 and 963.

Given integers are 867 and 255.

Clearly 867 > 255

Therefore, by applying Euclid’s division lemma to 867 and 255, we get



II.  Since the remainder 102 ≠ 0, we apply division lemma to get,

III. We consider the new divisor 102 and new remainder 51 and apply division lemma to get,

The remainder at this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e., 51 is the HCF of 867 and 255.