假設m,n皆為自然數,則恰有2個m值滿足不等式8/15<m/(m+n)<7/13的最大n值及最小n值為何?

Sol
8/15<m/(m+n)<7/13
15/8>(m+n)/m>13/7
7/8>n/m>6/7
8/7<m/n<7/6
設m兩解k，k+1
(k－1)/n<=8/7<k/n<(k+1)/n<7/6<=(k+2)/n
(k－1)/n<=8/7
K－1<=8n/7
k<=1+8n/7
7/6<=(k+2)/n
7n<=6k+12
7n－12<=6k
7n/6－2<=k
7n/6－2<=k<=1+8n/7
7n/6－2<=1+8n/7
(7n/6－8n/7)<=3
(49n－48n)<=126
n<=126

8/15<m/(m+n) --> 8(m+n)<15m --> 8n<7m
m/(m+n)<7/13 --> 13m<7(m+n) --> 6m<7n

48n<42m<49n
(8/7)n<m<(7/6)n
2<[(7/6)-(8/7)]n<3
2<(1/42)n<3
84<n<126
n=85 時, m=98,99
n=125 時, m=123,124,125
...
n=121 時, m=139,140,141
n=120 時, m=138,139

Ans:n 最大值 120, 最小值 85