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

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.

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.