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

- anonymous

- chestercat

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- yuki

good one

- yuki

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

- yuki

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

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## More answers

- yuki

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

- yuki

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

- yuki

I would suggest this strategy

- yuki

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

- yuki

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

- anonymous

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

- anonymous

so would it be 7*6 as the answer?

- yuki

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

- yuki

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

- yuki

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

- anonymous

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.

- anonymous

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

- yuki

gian franco, 0 is not allowed to use

- anonymous

you guys have a final answer

- yuki

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

- anonymous

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

- anonymous

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

- anonymous

so this seems quicker

- yuki

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

- yuki

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

- yuki

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

- yuki

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

- anonymous

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

- yuki

I don't think there is anything else

- yuki

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

- anonymous

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?

- anonymous

yuki, theres a mistake

- anonymous

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

- anonymous

3+5+7 = 15

- anonymous

yuki?

- anonymous

oh that's true!

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

right

- anonymous

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

- anonymous

oh wait

- anonymous

well the 444 and etc

- anonymous

right, im wrong

- anonymous

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

- anonymous

how would one do that then?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

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

- anonymous

the coefficients here have to add to 3 ,

- anonymous

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

- anonymous

ok so far?

- anonymous

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

- anonymous

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

- anonymous

and we are adding them

- anonymous

ohhhhhhh OK

- anonymous

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

- anonymous

yeah i get it now XD

- anonymous

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

- anonymous

i take that back

- anonymous

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

- anonymous

ok

- anonymous

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,

- anonymous

right those are left

- anonymous

alright

- anonymous

wait, you left in 3x3, that goes away

- anonymous

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

- anonymous

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,

- anonymous

right :)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

not the 9's or the 6's

- anonymous

oh good point

- anonymous

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,

- anonymous

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

- anonymous

yeah cause 1*3,2*4 works

- anonymous

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

- anonymous

right ,

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

6,7,9

- anonymous

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

- anonymous

right , so the ones cases you can do by hand

- anonymous

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

- anonymous

and then you have permutations of them

- anonymous

right

- anonymous

ill double check, but that looks right

- anonymous

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

- anonymous

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

- anonymous

and you systematically eliminate

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

8 + (2*3, 2*4, 2*6, 2*7, 2*9)

- anonymous

9 + (2*4, 2*5, 2*7, 2*8)

- anonymous

looks good

- anonymous

so then you'd take the permutations of the singles
+ the permutations of the doubles and subtract from 7 *7*7?

- anonymous

right

- anonymous

do you have the answer in the book,

- anonymous

that would be nice to check

- anonymous

even # :(

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