Chapter 16 Playing with Numbers Exercise 16.2
Question 1: If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Answer:
Let 21y5 be a multiple of 9.
Divisibility rule of 9 is that the sum of all the digits must be a multiple of 9.
= 2 + 1 + y + 5 = 9
= 8 + y = 9
= y = 9 - 8
= y = 1
Therefore, the value of y is 1. The number becomes 2115, which is divisible by 9.
Question 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?
Answer:
Let 31z5 be a multiple of 9.
Divisibility rule of 9 is that the sum of all the digits must be a multiple of 9.
= 3 + 1 + z + 5 = 9
= 9 + z = 9
= y = 9 - 9
= y = 0 (Case 1)
= 3 + 1 + z + 5 = 18
= 9 + z = 18
= y = 18 - 9
= y = 9 (Case 2)
Therefore, the value of z may be either 0 or 9. The number becomes 3105 or 3195, which is divisible by 9.
Question 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.)
Answer:
Let 24x be a multiple of 3.
Divisibility rule of 3 is that the sum of all the digits must be a multiple of 3.
= 2 + 4 + x = 6 + x (6 + x is the multiple of 3, 6 + x may be 0, 3, 6, 9, 12, 15, 18…)
But x is only 1 digit so the value of x can be either 0, 3, 6, 9 and the sum of the digits can be 6, 9, 12, 15 which is multiple of 3.
Therefore, the value of x may be either 0 or 3 or 6 or 9.
Question 4: If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Answer:
Let 31z5 be a multiple of 3.
Divisibility rule of 3 is that the sum of all the digits must be a multiple of 3.
= 3 + 1 + z + 5 = 9 + z (9 + z is the multiple of 3, 9 + x may be 0, 3, 6, 9, 12, 15, 18…)
But x is only 1 digit so the value of z can be either 0, 3, 6, 9 and the sum of digits can be 9, 12, 15, 18 which is multiple of 3.
If z = 0, 3 + 1 + 0 + 5 = 9 which is divisible by 3.
If z = 3, 3 + 1 + 3 + 5 = 12 which is divisible by 3.
If z = 6, 3 + 1 + 6 + 5 = 15 which is divisible by 3.
If z = 9, 3 + 1 + 9 + 5 = 15 which is divisible by 3.
Therefore, 0, 3, 6, 9 might be the values of z.
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