anonymous
  • anonymous
how many 3-digit whole #s can be formed from the digits 3-9 if: the result number is a multiple of 3?
Mathematics
chestercat
  • chestercat
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yuki
  • yuki
good one
yuki
  • yuki
what you first need to do is to know how a number is can be a multiple of 3
yuki
  • 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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yuki
  • yuki
for example, 222 is divisible by 3 because 2+2+2 =6
yuki
  • 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
  • yuki
I would suggest this strategy
yuki
  • 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
  • yuki
so if you have exactly one number with a multiple of 3, you have exactly 18 numbers
anonymous
  • 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
  • anonymous
so would it be 7*6 as the answer?
yuki
  • 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
  • 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
  • yuki
this one is tough to count because there is no easy formula for it
anonymous
  • 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
  • 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
  • yuki
gian franco, 0 is not allowed to use
anonymous
  • anonymous
you guys have a final answer
yuki
  • yuki
anyway, now let's count the exactly 3 digits being divisible by 3
anonymous
  • anonymous
im going to do it this way 7x7x - ( how many with at least one 7, one 4,,5,8
anonymous
  • anonymous
because (3,3,7) and permutations thereof dont add up to multiply of 3, (3,7,7) doent add up, etc
anonymous
  • anonymous
so this seems quicker
yuki
  • 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
  • yuki
now we need to think if there is any 3 digit number that has no multiple of 3s in its digits.
yuki
  • 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
  • yuki
I'm trying to see if there is anything else...
anonymous
  • anonymous
we have 3,2^2, 5, 7, 2^3, 3^2
yuki
  • yuki
I don't think there is anything else
yuki
  • yuki
so the total is,,, 4 + 27 + 18 = 59 of them
anonymous
  • 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
  • anonymous
yuki, theres a mistake
anonymous
  • 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
  • anonymous
3+5+7 = 15
anonymous
  • anonymous
yuki?
anonymous
  • anonymous
oh that's true!
anonymous
  • anonymous
and 6 + 5 + 7 = 18 a multiple of three
anonymous
  • anonymous
9 + 5 + 7 = 21, a multiple of 3
anonymous
  • anonymous
so we'd have to take into account (4,8) and (5,7) solutions?
anonymous
  • anonymous
right
anonymous
  • 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
  • anonymous
oh wait
anonymous
  • anonymous
well the 444 and etc
anonymous
  • anonymous
right, im wrong
anonymous
  • 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
  • anonymous
how would one do that then?
anonymous
  • anonymous
so lets look at this way you have the numbers 3,4,5,6,7,8,9 , correct
anonymous
  • anonymous
you get how i have total cases , 7x7x7 , right ?
anonymous
  • anonymous
i want to find the number of NON multiples of 3, subtract from the total, will give you multiples of 3
anonymous
  • anonymous
ok
anonymous
  • 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
  • anonymous
you can have 1*3 + 2*9 , for instance
anonymous
  • anonymous
the coefficients here have to add to 3 ,
anonymous
  • 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
  • anonymous
ok so far?
anonymous
  • anonymous
hmm i'm not understanding the multiplication tables at the top
anonymous
  • anonymous
so clearly, , well i get 3*9 from (9,9,9)
anonymous
  • anonymous
and we are adding them
anonymous
  • anonymous
ohhhhhhh OK
anonymous
  • anonymous
so if you randomly picked 1*3, and say 2*7, you get (3,7,7)
anonymous
  • anonymous
yeah i get it now XD
anonymous
  • anonymous
so automatically you rule out all the 3's cases, and you rule out 6's, 9's
anonymous
  • anonymous
i take that back
anonymous
  • anonymous
you rule out the last column, of 3*3, 3*4, 3*5,... 3*9
anonymous
  • anonymous
ok
anonymous
  • 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
  • anonymous
right those are left
anonymous
  • anonymous
alright
anonymous
  • anonymous
wait, you left in 3x3, that goes away
anonymous
  • 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
  • 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
  • anonymous
right :)
anonymous
  • 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
  • anonymous
clearly 1*3 , 2*3 is rule out, since thats just 3,3,3
anonymous
  • anonymous
1*4, 2*4 1*5,2*5, 1*7, 2*7, 1*8,2*8 1*9,2*9
anonymous
  • 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
  • anonymous
not the 9's or the 6's
anonymous
  • anonymous
oh good point
anonymous
  • 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
  • anonymous
ok i see, so there is the case of one 3, one 4 , one 5, etc
anonymous
  • anonymous
yeah cause 1*3,2*4 works
anonymous
  • 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
  • anonymous
right ,
anonymous
  • anonymous
theres also a strange case 4+8+9 = 21
anonymous
  • 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
  • anonymous
4,5,7 4,5,8, 5,6,8 5,6,9
anonymous
  • anonymous
6,7,9
anonymous
  • anonymous
3,4,6 3,4,7 3,4,9
anonymous
  • anonymous
right , so the ones cases you can do by hand
anonymous
  • anonymous
those are all the ones with non multiples of 3 so 8 * 6 = 48 ways that aren't possible
anonymous
  • anonymous
and then you have permutations of them
anonymous
  • anonymous
right
anonymous
  • anonymous
ill double check, but that looks right
anonymous
  • anonymous
so for the one and 2's cases i have 1*3 - ( 2*4, 2*5, 2*7, 2*8, )
anonymous
  • 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
  • anonymous
and you systematically eliminate
anonymous
  • anonymous
4 + (2*3, 2*7, 2*6, 2*9)
anonymous
  • anonymous
5 + (2*3, 2*4, 2*6, 2*7, 2*9)
anonymous
  • anonymous
6 + (2*4, 2*5, 2*7, 2*8)
anonymous
  • anonymous
7 + (2*3, 2*5, 2*6, 2*8, 2*9)
anonymous
  • anonymous
8 + (2*3, 2*4, 2*6, 2*7, 2*9)
anonymous
  • anonymous
9 + (2*4, 2*5, 2*7, 2*8)
anonymous
  • anonymous
looks good
anonymous
  • anonymous
so then you'd take the permutations of the singles + the permutations of the doubles and subtract from 7 *7*7?
anonymous
  • anonymous
right
anonymous
  • anonymous
do you have the answer in the book,
anonymous
  • anonymous
that would be nice to check
anonymous
  • anonymous
even # :(

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