### Brain Teasers

# Divisible?

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

The digits one through nine are randomly arranged to make a number. What is the probability that the resulting number is divisible by eighteen?

### Hint

Use your divisibility rules!### Answer

Answer: 4/9If the digits one through nine are arranged, the resulting number will always be divisible by nine (a number is divisible by nine if the sum of its digits is divisible by nine). The question can be simplified to ask for the probability that the resulting number is even. Now, there are four options for the last digit (2, 4, 6, 8). There are then eight options for the digit before the last (one option is already taken). There are seven options for next, and so on. The total number of combinations divisible by eighteen is 4*8*7*6*5*4*3*2*1. The total number of combinations is 9*8*7*6*5*4*3*2*1. The probability a combination is divisible by eighteen is 4*8*7*6*5*4*3*2*1/9*8*7*6*5*4*3*2*1 or 4/9.

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## Comments

Your calculations are correct, but this problem is much easier than that.

For a number to be divisible by 18 it need only be even and divisible by nine. A number is divisible by nine if the sum of its digits is divisible by nine (or, if you continue to sum the digits, eventually equals nine). A nine digit number comprised of the digits 1-9 with none of the digits repeating digit sums to 45 (which then sums again to 9) and thus is divisible by nine. All we have to do is determine which combinations are even, so all we care about is the final digit. Of the numbers 1-9, only four are even (2,4,6,, so there is simply a 4 in 9 chance that the final digit will be even, which leaves the number divisible by 18. Discrete mathematics are cool, but here they are entirely unnecessary.

For a number to be divisible by 18 it need only be even and divisible by nine. A number is divisible by nine if the sum of its digits is divisible by nine (or, if you continue to sum the digits, eventually equals nine). A nine digit number comprised of the digits 1-9 with none of the digits repeating digit sums to 45 (which then sums again to 9) and thus is divisible by nine. All we have to do is determine which combinations are even, so all we care about is the final digit. Of the numbers 1-9, only four are even (2,4,6,, so there is simply a 4 in 9 chance that the final digit will be even, which leaves the number divisible by 18. Discrete mathematics are cool, but here they are entirely unnecessary.

That is the procedure I took. I think we are saying the same thing in different ways.

I glanced at it again. Your way is actually easier. (I could not edit my previous post). Thank you for the alternate solution!

Would things be made more interesting if you were asking for it to be divisible by 36 instead of 18? Because it isn't a 2/9 probability then, is it?

That way you need it obviously divisible by 4 instead as well as 9. You look at the last two digits then: the probability isn't just 2/9 because you obviously can't repeat the digits, and 0 isn't there to add to the complications.

Just a thought

That way you need it obviously divisible by 4 instead as well as 9. You look at the last two digits then: the probability isn't just 2/9 because you obviously can't repeat the digits, and 0 isn't there to add to the complications.

Just a thought

It seems you can't change the comments.

Anyway, thinking it through, trying to get it divisible by 36 isn't more interesting than the original problem after all. A quick calculation and I make it 16/72 = 2/9.

Sorry for embarking on a dead-end journey!

Anyway, thinking it through, trying to get it divisible by 36 isn't more interesting than the original problem after all. A quick calculation and I make it 16/72 = 2/9.

Sorry for embarking on a dead-end journey!

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