which when divided by 16, 20 and 40 leaves the same remainder 4
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Let’s see straight through it…. All three numbers will only divide a number when all of them are factors of it. So it’s obvious we need to find the LCM.
16 = 2^4
28 = 2^2 X 7
40 = 2^3 X 5
LCM = 2^4 X 7 X 5 (Taking the highest power of the nos.)
LCM = 560
So 560 Is the smallest number which the above three nos divide . But what is requested is that the smallest number which leaves the remainder 5.
So if we increase 5 from 560 , everytime these three nos will leave a remainder as 5. (Isn’t that obvious) :)
Hence the no. will be 560+ 5 = 565
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