What is the test of divisibility of 18?

2 Answers
Bio
Mar 29, 2016

A number that is divisible by 18 must be divisible by both 2 and 9.

The inverse is also true:
A number that is divisible by both 2 and 9 must be divisible by 18.

Therefore, we just have to test for both divisibility by 2 and 9.

  • If a number is divisible by 2, its last digit must be even.
  • If a number is divisible by 9, the sum of all its digits must be a multiple of 9

If a number passes both tests, then it will surely be divisible by 18.

Shwetank Mauria
Mar 29, 2016

Test of divisibility of 1818 is that units digit is even i.e. it is either 0,2,4,60,2,4,6 or 88 and simultaneously sum of the digits is divisible by 99.

Explanation:

Factors of 1818 are 22 and 99 and hence divisibility by 1818 means divisibility both by 22 and 99.

Divisibility test of 22 being that units digit is divisible by 22 i.e. it is either 0,2,4,60,2,4,6 or 88.

Divisibility test of 99 is that sum of all the digits is divisible by 99.

Hence test of divisibility of 1818 is that units digit is even i.e. it is either 0,2,4,60,2,4,6 or 88 and simultaneously sum of the digits is divisible by 99.

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