OG 余数问题

sunshinehanUID：3762313

When the positive integer n is divided by 3, the remainder is 2 and when n is divided by 5, the remainder is 1. What is the least possible value of n?

请问这种余数问题，只能通过“硬”算来解题吗？

引用自：
https://gre.magoosh.com/forum/36 ... nteger-n-is-divided

We have,
3x + 2 = n
5y + 1 = n (x & y being their respective quotient)

Therefore, 3x + 2 = 5y + 1
3x = 5y - 1
x = (5y - 1)/3

Now we know that, 5y - 1 is divisible by 3, so lets substitute y = 0, 1, or 2

Trying these values we find y = 2

3x = 5(2) - 1
3x = 9
x = 3

n = 3x + 2 = 9 + 2 = 11
n = 5y + 1 = 10 + 1 = 11

Thus 11 is the least possible value of n

这种方式也是逐个来试验的

阿之UID：3452972

我的解法的话：
因为是5的倍数+1：第一个是5*1+1 = 6，但是6/3=2，不满足第一条件；
第二个是5*2+1 = 11，11/3=3余2，满足第一条件
所以是11.

也可以用3再算一次检查。

sunshinehanUID：3762313

阿之 发表于 2017-7-19 06:24
我的解法的话：
因为是5的倍数+1：第一个是5*1+1 = 6，但是6/3=2，不满足第一条件；
第二个是5*2+1 = 11， ...

谢谢，看来还是要一次次的去尝试验算