Find the greatest number that will divide 285 and 1249 by leaving the remainder 4 and 3 respectivly
Answers
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 × 23¹
HCF 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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Answer:
1
Step-by-step explanation:
When 285 is divided by the required number, 4 is left as remainder. Thus, 285 - 4 = 281 must be completely divisible by the number.
When 1249 is divided by the required number, 3 is left as remainder. Thus, 1249 - 3 = 1246 must be completely divisible by the number.
Thus, the required number = HCF of 281 and 1246.
(i) Prime factorization of 281 = 281
(ii) Prime factorization of 1246 = 2 * 7 * 89
HCF(281,1246) = 1
Hence, the greatest number is 1.
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