Question

Find the smallest number of three digits
which when divided by 16, 24 and 40
leaves a remainder of 6 in each case.​

Answer 1

Answer:

246

Step-by-step explanation:

First, we should find the LCM of 16,24, 40

That is 240

Then, 240 + 6 = 246

246 is the answer.

thus, the number which us divided by 16, 24 and 40 leaves 8 as remainder in each case....

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Answer 2

we need to find LCM

16 = 2^4

24 = 2^2×7

40 = 2^3×5

LCM = 2^4×7×5( taking highest power of nos.)

lcm =560

so 560 is the smallest number which the above 3 numbers divide. but what is requested is that the smallest number which leaves the remainder 5.

so if we increase 5 from 560 everytime these 3 numbers will leave a remainder as 5.

hence the number will be 5+560 = 565