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5 May, 07:04

Find the largest number which divides 224,250 and 302 and leaves remainder 3 in each case.

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Answers (2)
  1. 5 May, 07:08
    0
    Required number = HCF of (224 - 3), (250 - 3) and (302 - 3)

    Required number is HCF of 221, 247 and 299. Required number = HCF of 221 and 247 and then HCF of (221 & 247) and 299.

    Step (1) HCF of 221 and 247 = 13 [by using Euclid’s division algorithm]

    Step (2) HCF of 13 and 299 = 13 [By using Euclid’s division algorithm]

    Hence HCF of 221, 247 and 299 is 13. Hence, required number is 13.
  2. 5 May, 07:15
    0
    To have a remainder 3, we take away 3 from each number first.

    Take away 3:

    224 - 3 = 221

    250 - 3 = 247

    302 - 3 = 299

    Prime factorisation of each of the numbers:

    221 = 13 x 17

    247 = 13 x 19

    299 = 13 x 23

    Find HCF:

    HCF = 13

    Answer: The largest number that can be divided is 13.
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