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Daily topic primary school mathematics
China remainder theorem

The fixed routine is as follows:

1) Find the smallest number that can be divisible by 7,8 and 9,3, that is, 56×6=336.

2) Find the smallest number that can be divisible by 7,9 and 8 to get 4, that is, 63×4=252.

3) Find the smallest number divisible by 8 and 9, which is divisible by 7 and 2, and it is 72.

4) Add the three found numbers, that is, 336+252+72=660.

5) Find the least common multiple of 7, 8 and 9, that is, 7×8×9=504.

6) Compare the sum with the least common multiple, and if it is greater than the least common multiple, subtract it (it can be repeated).

660>504

660-504= 156

156 is the minimum number that meets the requirements.