The smallest three digit number which leaves reminders 8 and 12 when divided by 28 and 32 respectively is.......
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The answer can be expressed as 28y+8=32x+12
28y=32x+4
7y=8x+1
y=(8x+1)/7
8x+1 is divisible by 7 where y is an integer. The smallest possible x is x=6
32x+12= 32(6)+12= 192+12= 204