What is the largest number which divides 747 and 638 leaving remainder 6 and 11 respectively

Step 1: Since 747 > 638, we apply the division lemma to 747 and 638, to get

747 = 638 x 1 + 109

Step 2: Since the reminder 638 ≠ 0, we apply division lemma to 109 and 638, to get

638 = 109 x 5 + 93

Step 3: We consider the new divisor 109 and the new remainder 93, and apply the division lemma to get

109 = 93 x 1 + 16

We consider the new divisor 93 and the new remainder 16,and apply the division lemma to get

93 = 16 x 5 + 13

We consider the new divisor 16 and the new remainder 13,and apply the division lemma to get

16 = 13 x 1 + 3

We consider the new divisor 13 and the new remainder 3,and apply the division lemma to get

13 = 3 x 4 + 1

We consider the new divisor 3 and the new remainder 1,and apply the division lemma to get

3 = 1 x 3 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 1, the HCF of 638 and 747 is 1

Notice that 1 = HCF(3,1) = HCF(13,3) = HCF(16,13) = HCF(93,16) = HCF(109,93) = HCF(638,109) = HCF(747,638) .

We can take hcf of as 1st numbers and next number as another number to apply in Euclidean lemma

Step 1: Since 891 > 1, we apply the division lemma to 891 and 1, to get

891 = 1 x 891 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 1, the HCF of 1 and 891 is 1

Notice that 1 = HCF(891,1) .

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Find the largest number which divides 438 and 606 leaving remainder 6 in each case.

Answer:

Largest number which divides 438 and 606, leaving remainder 6

438 – 6 = 432

606 – 6 = 600 [remainder – 0]

HCF of 432 and 600 gives the largest number.

Prime factors of 432 and 600:

432 = 24 × 33

600 = 23 × 3 × 52

HCF = 23 × 3

HCF = 24

The largest number which divides 438 and 606, leaving remainder 6 is 24.

Find the largest number which divides 438 and 606 leaving remainder 6 in each case.

Largest number which divides 438 and 606, leaving remainder 6 is actually the largest number which divides 438 – 6 = 432 and 606 – 6 = 600, leaving remainder 0.Therefore, HCF of 432 and 600 gives the largest number.Now, prime factors of 432 and 600 are:

432 = 24 × 33

Concept: Euclid’s Division Lemma

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Page 2

Find the largest number which divides 320 and 457 leaving remainders 5 and 7 respectively.

We know that the required number divides 315 (320 – 5) and 450 (457 – 7).∴ Required number = HCF (315, 450)

On applying Euclid’s lemma, we get: 

Therefore, the HCF of 315 and 450 is 45.
Hence, the required number is 45.

Concept: Euclid’s Division Lemma

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Page 3

Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91.Prime factorization of 35, 56 and 91 is:35 = 5 × 7

56 = 23 × 7

LCM = product of greatest power of each prime factor involved in the numbers = 23 × 5 × 7 × 13 = 3640

Thus, the required number is 3647.