一道迷惑题,谢谢!

zhqkai

请问各位:

三个数均是自然数, M<<Q,三个数平均数是10,中数是:m+6, 问Q最大是多少?

我算的是: 最大是27, 请各位帮看看,对否, 先谢了!

bestxuan

23-m 吧？

zhqkai

bestxuan:

其实m也有取值范围:所以m+6>0,m+6 should be natural number and m+6 shoulbe be greater than M;supposing M=1, then m+6 can be 2, so m= -4, then Q will be max= 23-(-4)=27; it seems that we both can get the same result for this question.

Thanks!

mymengming

还以为m是笔误M呢

P=1,M=2 MAX(Q)=27