hcf for 285 and 1245
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Answered by
4
Hi!
Here is the answer to your question.
Since on dividing 285 by the required number, the remainder is 9.
∴ 285 – 9 = 276 will be exactly divisible by the required number.
Similarly, 1249 – 7 = 1242 will be also exactly divisible by the required number.
∴The required number is HCF of 276 and 1242.
Prime factorization of 276 = 2×2×3×23
Prime factorization of 1242 = 2×3×3×3×23
HCF of 276 and 1242 = 2×3×23 = 138
Thus, the required greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively is 138.
Here is the answer to your question.
Since on dividing 285 by the required number, the remainder is 9.
∴ 285 – 9 = 276 will be exactly divisible by the required number.
Similarly, 1249 – 7 = 1242 will be also exactly divisible by the required number.
∴The required number is HCF of 276 and 1242.
Prime factorization of 276 = 2×2×3×23
Prime factorization of 1242 = 2×3×3×3×23
HCF of 276 and 1242 = 2×3×23 = 138
Thus, the required greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively is 138.
Answered by
3
Here is your answer :
1245 = 285 × 4 + 105
285 = 105 × 2 + 75
105 = 75 × 1 + 30
75 = 30 × 2 + 15
30 = 15 × 2 + 0
So the H.C.F is 15
1245 = 285 × 4 + 105
285 = 105 × 2 + 75
105 = 75 × 1 + 30
75 = 30 × 2 + 15
30 = 15 × 2 + 0
So the H.C.F is 15
udit43:
Your welcome
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