问题1.如果21y5是9的倍数，其中y是一个数字，则y的值是多少?
解决方案：
According to the divisibility rule of 9,the sum of all digits should be a multiple of 9
Sum of the digits of 21y5 = 2 + 1 + y + 5 = 8 + y
(8 + y) ÷ 9 = 1
8 + y = 9
y = 9 – 8 = 1
Hence, the required value of y = 1.
问题2.如果31z5是9的倍数，其中z是一个数字，z的值是多少?您会发现最后一个问题有两个答案。为什么会这样呢?
解决方案：
According to the divisibility rule of 9, the sum of all digits should be a multiple of 9
Sum of the digits of 31z5 = 3 +1 + z + 5 = 9 + z
9 + z = 9
z = 0
9 + z = 18
z = 9
Hence, 0 and 9 are two possible answers.
问题3.如果24x是3的倍数，其中x是一个数字，那么x的值是多少?
(由于24x是3的倍数，所以它的数字6 + x是3的倍数；所以6 + x是以下数字之一：0、3、6、9、12、15、18，…。但是，由于x是一个数字，只能是6 + x = 6或9或12或15。因此，x = 0或3或6或9。因此，x可以具有四个不同值中的任何一个。
解决方案：
Let us assume that 24x is a multiple of 3
According to the divisibility rule of 3,the sum of all digits should be a multiple of 3
Sum of the digits of 24x = 2 + 4 + x = 6 + x
6 + x = 6 if x = 0
6 + x = 9 if x = 3
6 + x = 12 if x = 6
6 + x = 15 if x = 9
Hence, x can have any of the four values
问题4.如果31z5是3的倍数，其中z是一个数字，z的值可能是多少?
解决方案：
According to the divisibility rule of 3,the sum of all digits should be a multiple of 3
Sum of the digits of 31z5 = 3 + 1 + z + 5 = 9 + z
9 + z = 9 if z = 0
9 + z = 12 if z =3
9 + z = 15 if z = 6
9 + z = 18 if z = 9
Hence, 0, 3, 6 and 9 are four possible values