## Friday, June 6, 2014

### Finding Missing Digit - Divisibility Rule, 14

Category: Arithmetic

"Published in Vacaville, California, USA"

Find the missing digit so that it becomes divisible by 15 for

a. 13?2
b. 337?

Solution:

a. Consider the given number

A number is divisible by 15 if it is both divisible by 3 and 5. The given number is not divisible by 5 because the last digit is not 5 or 0. Because of this, the given number is not divisible by 15. The multiples of 15 ends with 5 or 0 always.There's nothing that we can do in order to become divisible by 15 since the last digit of a given number is not 5 or 0. You can assign any number to the missing digit but still, the given number will never become divisible by 15.

b. Consider the given number

A number is divisible by 15 if it is both divisible by 3 and 5. Since the last digit of a given number is missing, then we can assign 5 or 0. A number is divisible by 5 if the last digit ends with 5 or 0 and a number is divisible by 3 if the sum of the digits is a multiple of 3. Let's do the test for 5 and 0 as their last digit.

If the last digit is 0, then the sum of the digits is

Since 4 is not a multiple of 3, then the given number is not divisible by 3. 0 is not the last digit of a given number. Let's do the test again for the other last digit which is 5.

If the last digit is 5, then the sum of the digits is

Since 9 is a multiple of 3, then the given number is divisible by 3. 5 is the last digit of a given number. Because of this, the given number is divisible by 15. Therefore, the possible number is only 3375.