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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 $3$?

 May 13, 2021
 #1
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There are 35 possible values of N.

 May 13, 2021
 #2
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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.

 May 14, 2021
 #3
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Thanks for your effort but both are wrong.

 May 15, 2021
 #4
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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

 May 15, 2021
 #5
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its wrong ;-;

Guest May 20, 2021

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