陳那遜的一題上課題

[himorgan]

If r and s are positive integers, each greater than 1, and if 11(s-1)=13(r-1), what is the least possible value of r+s?
(A)2
(B)11
(C)22
(D)24
(E)26

答案為(E)
誰可以幫忙解釋一下呢

[DARY]

If r and s are positive integers, each greater than 1, and if 11(s-1)=13(r-1), what is the least possible value of r+s?
(A)2
(B)11
(C)22
(D)24
(E)26

答案為(E)
誰可以幫忙解釋一下呢

整理一下 13r-11s=2
很直覺得看得出r=1, s=1時成立，但r與s都比1大
所以設r=1+a, s=1+b 帶入上面的式子(此時a,b皆是大於0的整數)
13r-11s=(13+13a)-(11+11b)=2+(13a-11b)=2
可知13a-11b=0, 得a,b的最小值a=11, b=13
則r=1+11=12
s=1+13=14
r+s=12+14=26

選(E)