Find the largest number which on dividing 1251

Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively Solution: The remainders of 1251, 9377 and 15628 are 1, 2 and 3 By subtracting these remainders from the numbers, we get 1251 - 1 = 1250, 9377 - 2 = 9375 15628 - 3 = 15625,which are divisible by the required number. Required number = HCF (1250, 9375, 15625). Using Euclid’s division algorithm, a = bq + r --- (i) We know that dividend = divisor x quotient + remainder Consider a = 15625 and b = 9375 15625 = 9375 × 1 + 6250 [from eq. (i)] 9375 = 6250 × 1 +3125 6250 = 3125 × 2 + 0 HCF (15625, 9375) = 3125. Taking c = 1250 and d = 3125, Again by using Euclid’s division algorithm, d = cq + r 3125 = 1250 × 2 + 625 1250 = 625 × 2 + 0 HCF (1250, 9375,15625) = 625 Therefore, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainders, 1, 2 and 3, respectively. ✦ Try This: Using Euclid’s division algorithm, find the largest number that divides 730, 245 and 2190 leaving remainders 1, 2 and 3, respectively ☛ Also Check: NCERT Exemplar Class 10 Maths Exercise 1.3 Problem 9 Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively Summary: Using Euclid’s division algorithm,625 is the largest number which divides 1251, 9377 and 15628 leaving remainders, 1, 2 and 3, respectively ☛ Related Questions: • • •

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It is given that 1, 2 and 3 are the remainders of 1251, 9377 and 15628, respectively. Subtracting these remainders from the respective numbers, we get 1251 − 1 = 1250 9377 − 2 = 9375 15628 − 3 = 15625 Now, 1250, 9375 and 15625 are divisible by the required number. Required number = HCF of 1250, 9375 and 15625 By Euclid's division algorithm a = bq + r, 0 ≤ r < b For largest number, puta= 15625 andb= 9375 15625 = 9375 × 1 + 6250 ⇒ 9375 = 6250 × 1 +6250 ⇒ 6250 = 3125 × 2 + 0 Since remainder is zero, therefore, HCF(15625 and 9375) = 3125 Further, takec= 1250 andd= 3125. Again using Euclid's division algorithm d = cq + r, 0 ≤ r < c ⇒ 3125 = 1250 × 2 +625 ...[∵ r ≠ 0] ⇒ 1250 = 625 × 2 + 0 Since remainder is zero, therefore, HCF(1250, 9375 and 15625) = 625 Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3, respectively.