Each of the digits $6,7,8,9$ is used no more than once to form an integer $N.$ How many possible $N$s are multiples of $3$?
Every $N$ possible is divisible by 3 bc every number that has digits adding to a multiple of three is divisible by 3. so the answer is 24 bc 24 is four factroial which is the number of possiblities there are.
It would have been easier for me to answer if you had just typed it in.
I cannot just copy it and I like the question to be in my answer as sometimes kids change the original question.
Each of the digits 6,7,8,9, is used no more than once to form an integer N.
How many possible N's are multiples of 3 ?
To be a multiple of 3 the digits have to add to a multiple of 3
6+7+8+9=31 that is no good, not all digits can be used.
31-1 =30 but there is no 1s
31-1-6 = 31-7=24=3*8 so a three digit number could contain 6,8 and 9 there are 3!=6 of those
31-1-9 = 31-10 but there are no 10s
So there are 6 3-digit possibilities.
If I add pairs I get
13, 14, 15, 15, 16, 17
Only 15 is a multiple of 3
so the number could be 69, 96, 78, 87 there are 4 of them
There are 2 one digit ones , 6 and 9
so we have 6+4+2 = 12