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Ok, I give my solution. Welcome others.
following liangqi's
6m=8n+2
then m=(4n+1)/3 we want to make sure m is an integre, then not every n could satisfy this eqution.
we could get m=(n+1)/3 so if we want get m is integre n should be n=3k-1
subsitutes this into x=8n+4
we get x=24k-4
if k=2Y x=48y-4=48y+44
if k=2Y+1 x=48y+20
So it is 2 possible numbers. every x,y k m n is integer.
Hi lianqi if you still think it not very 爽. you can try that way like you said, just always in mind, because 6 and 8's minimal mulitiple is 24, so if divied by 48 there would be two possible values. |
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