GG 51

rainbowmanutd

N=a+b+c=XYZ, abc连续，xyz连续，问N除以5的余数
（1）a除以5余1  

（2）x除以5余1



（提供者ID：PsycheCC原文：余数题 N=a+b+c=XYZ, abc连续，xyz连续，问N除以5的余数
（1）a除以5余1  （2）x除以5余1

思路：因为a+b+c=XYZ，abc连续，所以必有xyz=3b，则xyz肯定是3的倍数，当（1）a除以5余1时，a的尾数可能为1或者6，这时b的尾数为2或者7，则xyz的尾数为6或者4，则xyz除以5的余数为1，可以求到（2）x除以5余1，则x可能为1或者6，此时可以算出xyz为123或者678，此时xyz除以5的余数为3。选D
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xyz的尾数为6或者4，则xyz除以5的余数为1,为什么? 234/5, 余数就等于4... NOT 1..

rainbowmanutd

求助??

sailing198961

That's how I think :
first consider (1) . Let's set a =5x+1 according to the question. Because a.b .c are Consecutive numbers。 b=5x+2  c=5x+3
Add a.b and c together ,we get N, which is 15x+6 . When N is divided by 5 , the reminder  is always 1 .

Consider(2) because XYC is a three-digit number ,and X divided by 5 has a reminder of 1. The only possible number of X either 1 or 6 . 123 and 678  both have a reminder of 3 .

Hence, both are sufficient (D).

keven_chen

1. a, b, c 的尾数可以为1,2,3或者6,7,8. 相加之和均为6。所以N除5的余数是12. x,y,z 的尾数也同样是1,2,3或者6,7,8. 相乘尾数也是6。所以N除5的余数是1

rainbowmanutd

That's how I think :
first consider (1) . Let's set a =5x+1 according to the question. Because a.b .c are Consecutive numbers。 b=5x+2  c=5x+3
Add a.b and c together ,we get N, which is 15x+6 . When N is divided by 5 , the reminder  is always 1 .

Consider(2) because XYC is a three-digit number ,and X divided by 5 has a reminder of 1. The only possible number of X either 1 or 6 . 123 and 678  both have a reminder of 3 .

Hence, both are sufficient (D).

-- by 会员 sailing198961 (2012/4/16 5:52:20)

Thanks man!