1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution:
Let, 21y5 is a multiple of 9.
According to the divisibility rule of 9, we know the sum of all the digits should be a multiple of 9.
So, 2+1+y+5 = 8+y
Therefore, 8+y is a factor of 9.
This is possible when 8+y is any one of these numbers 0, 9, 18, 27, and so on
Since, y is a single digit number, so that the sum can be 9 only.
Hence, the value of y should be 1 only.
Hence, 8+y = 8+1 = 9.
2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Solution:
Given, 31z5 is a multiple of 9.
According to the divisibility rule of 9, we know that the sum of all the digits should be a multiple of 9.
So, 3+1+z+5 = 9+z
so, 9+z is a multiple of 9
This is only possible when 9+z is any one of these numbers: 0, 9, 18, 27, and so on.
9+0 = 9 and 9+9 = 18
Hence 0 and 9 are two possible answers.
3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6+x is a multiple of 3; so 6+x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … . But since x is a digit, it can only be that 6+x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.)
Solution:
Let, 24x is a multiple of 3.
According to the divisibility rule of 3, we know that the sum of all the digits should be a multiple of 3.
So, 2+4+x = 6+x
Therefore, 6+x is a multiple of 3,
And 6+x is one of these numbers: 0, 3, 6, 9, 12, 15, 18 and so on.
Since it is given that, x is a digit, then value of x will be either 0 or 3 or 6 or 9, and the sum of the digits can be 6 or 9 or 12 or 15 respectively.
Hence, x can have any of the four different values: 0 or 3 or 6 or 9.
4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution:
Given, 31z5 is a multiple of 3.
According to the divisibility rule of 3, we know that the sum of all the digits should be a multiple of 3.
so, 3+1+z+5 = 9+z
Therefore, 9+z is a multiple of 3.
This is possible when the value of 9+z is any of the values: 0, 3, 6, 9, 12, 15, and so on.
For z = 0, 9+z = 9+0 = 9
For z = 3, 9+z = 9+3 = 12
For z = 6, 9+z = 9+6 = 15
For z = 9, 9+z = 9+9 = 18
So, the value of 9+z can be 9 or 12 or 15 or 18.
Hence 0, 3, 6 or 9 are four possible answers for z.
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