The mechanical computer is a very sensitive device, in consists of four sequential cog wheels which are in constant mesh. The largest cog has 63 teeth and the others have 42, 35 and 27 respectively. By accident, Daniel started to rotate the largest cog. How many revolutions must he rotate the largest cog make before the computer is back in its starting position with all of the cogs where they started?
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The given number of teeth are : 63, 42, 35, 27
3 * 3* 7 , 2* 3*7 , 5 * 7, 3 * 3 * 3
the LCM = 2 * 3 * 3 * 3 * 5 * 7 = 1890
So the number of teeth to turn = 1890, before all the cogs will be in the starting state.
If we are rotating the first wheel, then: it has to turn: 1890/63 = 30 full rotations to come back to the initial state.
3 * 3* 7 , 2* 3*7 , 5 * 7, 3 * 3 * 3
the LCM = 2 * 3 * 3 * 3 * 5 * 7 = 1890
So the number of teeth to turn = 1890, before all the cogs will be in the starting state.
If we are rotating the first wheel, then: it has to turn: 1890/63 = 30 full rotations to come back to the initial state.
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