Question

which is the largest number which divides 328 and 1032 leaving 8 as a remainder in each case?​

Answer 1

Answer:

the answer is 16

Step-by-step explanation:

248 - 8 = 240 and 1032 - 8 = 1024 are completely divisible by the required number.

- Therefore, it is the HCF of 240 and 1024.

- Prime factorization of 240 = 2 * 2 * 2 * 2 * 3 * 5

- Prime factorization of 1024 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2.

- HCF(240,1024) = 2 * 2 * 2 * 2   = 16.

Therefore the largest number which divides 248 and 1032 leaving remainder 8 in each case = 16.

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