如果正整数n使得[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69,则n=

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如果正整数n使得[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69,则n=

如果正整数n使得[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69,则n=
如果正整数n使得[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69,则n=

如果正整数n使得[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69,则n=
由题意知,取[]里所有分母的最小公倍数60
60[n/2]+60[n/3]+60[n/4]+60[n/5]+60[n/6]=60*69
30n+20n+15n+12n+10n=87n=4140
n≈47.59,n不是整数,所以n>47,
取n=48时,等式成立;
取n=49时,等式成立;
取n=50时,等式左边>69
所以,n=48或n=49

高斯函数:x-1≤[x]≤x+1
∴n/2-1≤[n/2]<n/2+1
...
n/6-1≤[n/6]<n/6+1
相加,有n/2+n/3+n/4+n/5+n/6-5≤[n/2]+[n/3]+[n/4]+[n/5]+[n/6]≤n/2+n/3+n/4+n/5+n/6+5
∵[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69
∴n/2+n...

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高斯函数:x-1≤[x]≤x+1
∴n/2-1≤[n/2]<n/2+1
...
n/6-1≤[n/6]<n/6+1
相加,有n/2+n/3+n/4+n/5+n/6-5≤[n/2]+[n/3]+[n/4]+[n/5]+[n/6]≤n/2+n/3+n/4+n/5+n/6+5
∵[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69
∴n/2+n/3+n/4+n/5+n/6-5≤69,n/2+n/3+n/4+n/5+n/6+5≥69
不等式解得42≤n≤51
∴n=42,43,44,45,46...51

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