Question

how many 3-digit whole #s can be formed from the digits 3-9 if:
the result number is a multiple of 3?

Answer 1

good one

Answer 2

what you first need to do is to know how a number is can be a multiple of 3

Answer 3

if the numbers digits add up to a multiple of 3, then we know that the number itself is divisible by 3

Answer 4

for example, 222 is divisible by 3 because 2+2+2 =6

Answer 5

so you are going to use the numbers 3,4,5,6,7,8,9 in order to find out all combinations of the sum of the digits that are a multiple of 3

Answer 6

I would suggest this strategy

Answer 7

if the exactly one of the digits is a multiple of 3, namely 3,6,9
the other two has to add up to a multiple of 3,
so the only pair that does that is (4,5) (5,4) because you are not
allowed to use other multiples of 3s.

so you have 3,4,5 and all the permutations of that, 6 of them
6,4,5 and it's permutations 6 of them
9,4,5 and its permutations 6 of them

Answer 8

so if you have exactly one number with a multiple of 3,
you have exactly 18 numbers

Answer 9

ok i found 7 combinations
3,4,5/3,4,8/4,5,6/4,5,9/5,6,7/6,7,8/7,8,9
but the order doesn't matter because the sum of the digits will be the same

Answer 10

so would it be 7*6 as the answer?

Answer 11

now you will think about having exactly two digits that are multiple of 3,
If you see this carefully, the other digit MUST be a multiple of 3 as well,
so there are no 3 digit number with exactly two of them being a multiple of 3

Answer 12

now we think of a way to count how many 3 digit numbers are divisible by 3 when it has exactly 3 digits that are multiple of 3s

Answer 13

this one is tough to count because there is no easy formula for it

Answer 14

If the digits you choose total 3, 6 or 9 then they are divisible by 3.
Start with 300.
You now just add 3 each time...
300, 303, 306, 309, ( but you cannot have 312, 315, etc because digits 1-2 cannot be used so next numbers are...) 330. 333, 336, 339,

What have you found so far?
300, 303, 306, 309, 330. 333, 336, 339, It is tempting to look for a pttern already. You will see the numbers end, 0,3,6,9 but this could be a trap!
Your next choise of starter digits should systematically be 34... but they must total 3, 6 or 9 so that means 34 2 can be the only one but it is illeagal as 2 is not allowed, so move on to 35...
352 (not leagal) 354, (allowed as 3+5+4+= 12 and 1+2=3) 357 also allowed
now we have
300, 303, 306, 309, 330. 333, 336, 339, 354, 357,..
te next numbers using this approach yield 360, 363, 366, 369, 375, 378, 384, 387 390, 393, 396, 399.
Put them all together and start patern spotting
300, 303, 306, 309, 330. 333, 336, 339, 354, 357,360, 363, 366, 369, 375, 378, 384, 387 390, 393, 396, 399.

End digits are 0, 3, 6, 9, 0, 3, 6, 9, 4, 7, 0, 3, 6, 9, 5, 8, 4, 7, 0, 3, 6, 9
This represents all possible 300 type numbers, do the same for the 400s and 500s and see if you can spot a useful pattern.

Answer 15

there are 7x7x7 possible 3 digit numbers from that list 3-9, of those subtract how many dont add up to a muiltiple of 3

Answer 16

gian franco, 0 is not allowed to use

Answer 17

you guys have a final answer

Answer 18

anyway, now let's count the exactly 3 digits being divisible by 3

Answer 19

im going to do it this way 7x7x - ( how many with at least one 7, one 4,,5,8

Answer 20

because (3,3,7) and permutations thereof dont add up to multiply of 3,
(3,7,7) doent add up,
etc

Answer 21

so this seems quicker

Answer 22

333, 666,999 3 ways

3,3,6 can be ordered in 3 ways
as well as 339,663,669,993,996 so 18 ways in total

3,6,9 can be ordered in 6 ways,

so we have 27 ways in total

Answer 23

now we need to think if there is any 3 digit number that has no multiple of 3s in its digits.

Answer 24

this is the toughest, you need to count them carefully

for example, 555 is a multiple of 3 because the sum is 3 * 5
so using these types of digits will be

444,555,777,888 4 of them

Answer 25

I'm trying to see if there is anything else...

Answer 26

we have 3,2^2, 5, 7, 2^3, 3^2

Answer 27

I don't think there is anything else

Answer 28

so the total is,,, 4 + 27 + 18 = 59 of them

Answer 29

wow that was a ton of math.... i had no idea it was that hard. thanks so much yuki
could you explain why you went with the strategy of how many possibilities there were based on the number of digits divisble by 3?

Answer 30

yuki, theres a mistake

Answer 31

you said if you have one of 3,6,9, the only other numbers that add up to a multipl of 3 is 4,5. what about 5,7

Answer 32

3+5+7 = 15

Answer 33

yuki?

Answer 34

oh that's true!

Answer 35

and 6 + 5 + 7 = 18 a multiple of three

Answer 36

9 + 5 + 7 = 21, a multiple of 3

Answer 37

so we'd have to take into account (4,8) and (5,7) solutions?

Answer 38

right

Answer 39

it does seem to be the case that you need at least one multiple of 3 to get a sum of three, like 4,5,7 doesnt work, etc. but 3,5,7 does

Answer 40

oh wait

Answer 41

well the 444 and etc

Answer 42

right, im wrong

Answer 43

but the negation case seems to be easier to count,
find the number of cases which dont add up to a multiple of 3, subtract by the total 7x7x7

Answer 44

how would one do that then?

Answer 45

so lets look at this way
you have the numbers 3,4,5,6,7,8,9 , correct

Answer 46

you get how i have total cases , 7x7x7 , right ?

Answer 47

i want to find the number of NON multiples of 3, subtract from the total, will give you multiples of 3

Answer 48

ok

Answer 49

youre cases are 1*3, 2*3, 3*3
1*4, 2*4,3*3
1*5,2*5, 3*5
1*7, 2*7,3*7
1*8,2*8,3*8
1*9,2*9,3*9

Answer 50

you can have 1*3 + 2*9 , for instance

Answer 51

the coefficients here have to add to 3 ,

Answer 52

1*3, 2*3, 3*3
1*4, 2*4,3*4
1*5,2*5, 3*5
1*6, 2*6,3*6
1*7,2*7,3*7
1*8,2*8,3*8
1*9,2*9,3*9

Answer 53

ok so far?

Answer 54

hmm i'm not understanding the multiplication tables at the top

Answer 55

so clearly, ,
well i get 3*9 from (9,9,9)

Answer 56

and we are adding them

Answer 57

ohhhhhhh OK

Answer 58

so if you randomly picked 1*3, and say 2*7, you get
(3,7,7)

Answer 59

yeah i get it now XD

Answer 60

so automatically you rule out all the 3's cases, and you rule out 6's, 9's

Answer 61

i take that back

Answer 62

you rule out the last column, of 3*3, 3*4, 3*5,... 3*9

Answer 63

ok

Answer 64

only the last column right?
so it becomes

1*3, 2*3, 3*3
1*4, 2*4,
1*5,2*5,
1*7, 2*7
1*8,2*8,
1*9,2*9,

Answer 65

right those are left

Answer 66

alright

Answer 67

wait, you left in 3x3, that goes away

Answer 68

1*3, 2*3,
1*4, 2*4
1*5,2*5,
1*7, 2*7,
1*8,2*8
1*9,2*9

Answer 69

so we have left
1*3, 2*3,
1*4, 2*4,
1*5,2*5,
1*6, 2*6
1*7, 2*7
1*8,2*8,
1*9,2*9,

Answer 70

right :)

Answer 71

now start with 1*3 in the first column, and how many can it go with in the second column, so that its NOT a multiple of 3

Answer 72

clearly 1*3 , 2*3 is rule out, since thats just 3,3,3

Answer 73

1*4, 2*4
1*5,2*5,
1*7, 2*7,
1*8,2*8
1*9,2*9

Answer 74

oh wait no cause 1*3 can go with others


1*4, 2*4
1*5,2*5,
1*7, 2*7,
1*8,2*8
1*9,2*9

Answer 75

not the 9's or the 6's

Answer 76

oh good point

Answer 77

1*3, 2*3,
1*4, 2*4,
1*5,2*5,
1*6, 2*6
1*7, 2*7
1*8,2*8,
1*9,2*9,

Answer 78

ok i see, so there is the case of one 3, one 4 , one 5, etc

Answer 79

yeah cause 1*3,2*4 works

Answer 80

so would we have to find the individual sums for all of the outcomes for each 1*{3,4,5,6,7,8,9}

Answer 81

right ,

Answer 82

theres also a strange case
4+8+9 = 21

Answer 83

ok so lets rethink here. we have
1*3, 2*3,
1*4, 2*4,
1*5,2*5,
1*6, 2*6
1*7, 2*7
1*8,2*8,
1*9,2*9,

the cases of ones we can do seperately

3,4,5 3,4,6 3 , 4 , 7 3, 4 8 3 , 4 , 9

Answer 84

4,5,7 4,5,8, 5,6,8 5,6,9

Answer 85

6,7,9

Answer 86

3,4,6 3,4,7 3,4,9

Answer 87

right , so the ones cases you can do by hand

Answer 88

those are all the ones with non multiples of 3
so 8 * 6 = 48 ways that aren't possible

Answer 89

and then you have permutations of them

Answer 90

right

Answer 91

ill double check, but that looks right

Answer 92

so for the one and 2's cases i have
1*3 - ( 2*4, 2*5, 2*7, 2*8, )

Answer 93

for the one and twos cases.
so there are 3 possibilities.
either all of them are the same (3 of em)
2 of them are the same, and 1 different.
or all of them are different

Answer 94

and you systematically eliminate

Answer 95

4 + (2*3, 2*7, 2*6, 2*9)

Answer 96

5 + (2*3, 2*4, 2*6, 2*7, 2*9)

Answer 97

6 + (2*4, 2*5, 2*7, 2*8)

Answer 98

7 + (2*3, 2*5, 2*6, 2*8, 2*9)