# If no repetitions are allowed, how many 9-digit numbers can be formed from the digits 1, 2, 3, …, 9? How many of these are divisible by 4 ?

If no repetitions are allowed, how many 9-digit numbers can be formed from the digits 1, 2, 3, …, 9? How many of these are divisible by 4 ?

My Attempt:

the first part is obviously 9!

Now the 2nd part is the tricky one. I saw a similar question where the person who answered sums up all the numbers, in which the sum = 45, but I don't quite understand what to do next. Even if the unit digit is 2,4 or 8, there's no guarantee that it's divisible by 4

A number $$n$$ is divisible by $$4$$ if and only if $$n\mod 100$$ is divisible by $$4$$ (the last two digits).

You can find other divisibility rules here: http://www.softschools.com/math/topics/divisibility_rules_2_4_8_5_10/

Now notice that $$12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96$$ are the only possible two digit combinations that can be made which are divisible by $$4$$. Thus our answer comes out to be $$16\cdot7!$$

• Thanks for the tip as well! Oct 1, 2018 at 12:29

HINT

A number is divisible by four when the number made of the last two digits is divisible by four or both equal zero. You just missed the point about two digits by only refering to the unit digit. Therefore you only have to focus on numbers with i.e. $$24$$ in the end. The rest is irrelevant and can be computed again by using the factorial.

• Alright thanks for the tip! Oct 1, 2018 at 12:28