Question

find the smallest number which when divide by 28,32 and 48 leaves remainder 24,4 and 20 respectively .
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Answer 1

Answer:

Therefore, the required number will be 20 less than the LCM of 28 and 32. LCM(28, 32) = 2 x 2 x 2 x 2 x 2 x 7 = 224. Therefore, the required the smallest number = 224 – 20 = 204.

Answer 2

Step-by-step explanation:

LCM of 28 and 32. LCM(28, 32) = 2 x 2 x 2 x 2 x 2 x 7 = 224. Therefore, the required the smallest number = 224 – 20 = 204.