maths challenge

Hi,

I'm struggling with this question - would anyone be able to help?

Consider all three-digit numbers formed by using different digits from 0,1,2,3 and 5. How many of these numbers are divisible by 6?

possible answers: (it's a multiple choice question) A 4 B 7 C 10 D15 E20

well I'm a little confused by "different digits from" to be honest! does that mean using 0123 and 5 or not using them and just using 4678 and 9?

ahh okay - that makes sense. so are the only possibilities 150, 120, 312 and 102 or are there any more?

ah okay thanks 😀

so...
If you recall your 6 times tables:
6, 12, 18, 24, 30, 36...... It is evident that the last digit of these numbers is a repeating cycle that goes 6, 2, 8, 4, 0 and then repeats.
since you only have the numbers 0 1 2 3 5, you already know that the numbers can only end with two of these digits - 0 and 2, making it far easier to work out multiples.

So, treating each case independently, lets start with 0.
The number ends in 0 so is in the form XX0 where X is an unknown unit.
Know consider your multiples of 6 ending in 0..... 60, 510, 360..... by inspection and intuitively the two digits preceding the 0 are a multiple of three in all cases, supported by the fact that 5x6=30 meaning your multiples of 6 ending in 0 should go up in indentations of 30. So, excluding 0 from your set leaves you with 1, 2, 3 and 5 and all you need do now is consider all the arrangements of these numbers that forms a two digit number divisible by 3 (A quick way to do this would be to acknowledge that the digits of multiples of 3 sum to a multiple of 3 eg. 51 -> 5+1=6 and 6 is a multiple of 3 therefore so is 51). You will find the possible arrangements are as such: 12, 15, 21, 51. This means 4 different three digit numbers an be formed that end in 0, namely 120, 150, 210 and 510.

Now you have to consider those that end in 2, in the form XX2. Again by inspection it is evident that the multiples of 6 that end in two have preceding digits that could be written in the form 3n+1 where n is positive real and whole (slightly different to the 3n form from just before). So, in a similar fashion to before, you have to consider the possible arrangements of the remaining numbers - 0, 1, 3 and 5 - that form two digit numbers in the form 3n+1. The solutions to this are 01, 10, 13 and 31, giving the three digit numbers 102, 132, 312 - note '012' has been omitted as it is a two digit number as you discount the 0, and the question specifies three digit numbers. Therefore there are 3 three digit numbers divisible by 6, from the set, that end in 2.

This means there are in total 3+4 three digit numbers divisible by 6, possible from that set of numbers; so there are 7.
I think this is all correct.

This problem was just solved, and you shouldnt really be posting full solutions @RDKGames

I was showing them how to solveproperly without just guessing, and why no full solutions?

Ask @RDKGames he has been known to care