高一數學題目〈急〉

題目是：「設n為二位之整數，若n^3/6+n/2－5/3為整數，則n有幾個解？
Sol
n^3/6+n/2－5/3=(n^3+3n－10)/6
p為整數
(1) n=3p(不合)
(2) n=6p+1
n^3+3n－10
=>1^3+3－10
=>－6----------------------
(3) n=6p+2
n^3+3n－10
=>2^3+6－10
=>4(不合)
(4) n=6p+3
n^3+3n－10
=>3^3+9－10
=>26(不合)
(5) n=6p+4
n^3+6n－10
=>4^3+4*6－10
=>78-----------------------
(6) n=6p+5
n^3+6n－10
=>5^3+6*5－10
=>145(不合)
So
n=6p+1 or n=6p+4
10<=6p+1<=99
9<=6p<=98
1.5<=p<=16.33
2<=p<=16
16－2+1=15
10<=6p+4<=99
6<=6p<=95
1<=p<=15.83
1<=p<=15
15－1+1=15
15+15=30

月下的解法不錯耶!

消去3次方, 只剩n

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