Question

A is a digit and 3A18 is a multiple of 6. Which of the following can be value of A

Answer 1

Answer:

Step-by-step explanation:

3+1+8=12

So, for division by 6 it's needed to be divisible by 2 and 3

So, at the place of A

we can put  

either 0,3,6

Answer 2

Concept: By looking at the digits of an integer, a divisibility rule can be used to quickly and conveniently determine if it can be divided by a specific number.

Any number that can be divided by both 2 and 3 is also divisible by 6. In other words, if the provided number's last digit is even and its digits added together are a multiple of 3, then it is also a multiple of 6. For instance, the number 630 is divisible by two because the last digit is zero.

Given: 'A' is a digit

            3A18 is a multiple of 6.

To find: the value of A

Solution: As we know that a number is divisible by 6 if the last digit of the number is even and that the sum of the digits is divisible by 3. Here the number is 3A18. The last digit here is 8.

Now, substituting the digits one by one at the place of A, and adding the digits,

if A = 0, 3 + 0 + 1 + 8 = 12  (since the sum of the digits is 12 which is divisible by 3, so at the place of A we can put 0)

if A = 1, 3 + 1 + 1 + 8 = 13  (since the sum of the digits is 13 which is not divisible by 3, so at the place of A we cannot put 1)

if A = 2, 3 + 2 + 1 + 8 = 14  (since the sum of the digits is 14 which is not divisible by 3, so at the place of A we cannot put 2)

if A = 3, 3 + 3 + 1 + 8 = 15  (since the sum of the digits is 15 which is divisible by 3, so at the place of A we can put 3)

if A = 4, 3 + 4 + 1 + 8 = 16  (since the sum of the digits is 16 which is not divisible by 3, so at the place of A we cannot put 4)

if A = 5, 3 + 5 + 1 + 8 = 17  (since the sum of the digits is 17 which is not divisible by 3, so at the place of A we cannot put 5)

if A = 6, 3 + 6 + 1 + 8 = 18  (since the sum of the digits is 18 which is divisible by 3, so at the place of A we can put 6)

if A = 7, 3 + 7 + 1 + 8 = 19  (since the sum of the digits is 19 which is not divisible by 3, so at the place of A we cannot put 7)

if A = 8, 3 + 8 + 1 + 8 = 20  (since the sum of the digits is 20 which is not divisible by 3, so at the place of A we cannot put 8)

if A = 9, 3 + 9 + 1 + 8 = 21  (since the sum of the digits is 21 which is divisible by 3, so at the place of A we can put 9)

Therefore, at the place of A we can put 0,3,6 and 9.

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