14. Find the greatest number which divides 285 and 1249 leaving remain

1.	14. Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively15. If the HCE f 657 and 063 is aynrocsibl in th fm6063x- 15 find the value of
Answer» SOLUTION :Given numbers are 285 and 1249 and remainders are 9 and 7 respectively. Then new numbers after subtracting remainders are : 285 – 9 = 276 1249 – 7 = 1242.The required number is HCF of 276 and 1242.HCF by prime factorization method : Prime factorization of 276 = 2×2×3×23 = 2² × 3¹ × 23¹ Prime factorization of 1242 = 2×3×3×3×23 = 2¹ × 3³ × 23¹HCF of 276 and 1242 = 2¹ ×3¹×23¹= 6 × 23 = 138[HCF of two or more numbers = product of the smallest power of each common prime factor involved in the numbers.] HCF of 276 and 1242 is 138.Hence, the required greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively is 138.

14. Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively15. If the HCE f 657 and 063 is aynrocsibl in th fm6063x- 15 find the value of

Answer»

SOLUTION :Given numbers are 285 and 1249 and remainders are 9 and 7 respectively. Then new numbers after subtracting remainders are : 285 – 9 = 276 1249 – 7 = 1242.The required number is HCF of 276 and 1242.HCF by prime factorization method : Prime factorization of 276 = 2×2×3×23 = 2² × 3¹ × 23¹

Prime factorization of 1242 = 2×3×3×3×23 = 2¹ × 3³ × 23¹HCF of 276 and 1242 = 2¹ ×3¹×23¹= 6 × 23 = 138[HCF of two or more numbers = product of the smallest power of each common prime factor involved in the numbers.]

HCF of 276 and 1242 is 138.Hence, the required greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively is 138.

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