2016年全美数学竞赛（AMC）竞赛
数字$$N$$是个两位数．  $$N$$除以$$9$$，余数为$$1$$．  $$N$$除以$$10$$，余数为$$3$$．  则$$N$$除以$$11$$时，余数为几？
课内体系->知识点->式->因式分解->十字相乘法->二次项系数为±1的十字相乘 美国AMC8->Knowledge Point->Number Theory->Remainder Problems->Remainder Characteristics
数字$$N$$是个两位数．  $$N$$除以$$9$$，余数为$$1$$．  $$N$$除以$$10$$，余数为$$3$$．  则$$N$$除以$$11$$时，余数为几？  从第二个子弹开始，我们知道第二个数字必须是$$3$$，因为除以$$9$$还有一个余数$$1$$，$$9$$的倍数必须以$$2$$结束，我们找到：  $$9\left( 1 \right)=9$$  $$9\left( 2 \right)=18$$  $$9\left( 3 \right)=27$$  $$9\left( 4 \right)=36$$  $$9\left( 5 \right)=45$$  $$9\left( 6 \right)=54$$  $$9\left( 7 \right)=63$$  $$9\left( 8 \right)=72$$  这个上数字$$71+1=73$$都满足条件，我们减去$$11$$的小于$$73$$的最大倍数得到剩余的，也就是$$73-11\left( 6 \right)=73-66=7$$．  故选$$\text{E}$$．  From the second bullet point, we know that the second digit must be $$3$$. Because there is a remainder of $$1$$ when it is divided by $$9$$ the multiple of $$9$$ must end in a $$2$$. We now look for this one:  $$9(1) = 9$$  $$9(2) =18$$  $$9(3) = 27$$  $$9(4) = 36$$  $$9(5) = 45$$  $$9(6) =54$$  $$9(8) = 72$$  The number $$72 +1= 73$$ satisfies both conditions. We subtract the biggest multiple of $$11$$ less than $$73$$ to get the remainder. Thus, $$73- 11(6) = 73 - 66 =\boxed{(\rm E) 7 }$$. 