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anonymous
 5 years ago
how many 3digit whole #s can be formed from the digits 39 if:
the result number is a multiple of 3?
anonymous
 5 years ago
how many 3digit whole #s can be formed from the digits 39 if: the result number is a multiple of 3?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what you first need to do is to know how a number is can be a multiple of 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if the numbers digits add up to a multiple of 3, then we know that the number itself is divisible by 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for example, 222 is divisible by 3 because 2+2+2 =6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would suggest this strategy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if you have exactly one number with a multiple of 3, you have exactly 18 numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0so would it be 7*6 as the answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this one is tough to count because there is no easy formula for it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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 12 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
 5 years ago
Best ResponseYou've already chosen the best response.0there are 7x7x7 possible 3 digit numbers from that list 39, of those subtract how many dont add up to a muiltiple of 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gian franco, 0 is not allowed to use

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you guys have a final answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anyway, now let's count the exactly 3 digits being divisible by 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im going to do it this way 7x7x  ( how many with at least one 7, one 4,,5,8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because (3,3,7) and permutations thereof dont add up to multiply of 3, (3,7,7) doent add up, etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so this seems quicker

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0333, 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we need to think if there is any 3 digit number that has no multiple of 3s in its digits.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm trying to see if there is anything else...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have 3,2^2, 5, 7, 2^3, 3^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think there is anything else

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the total is,,, 4 + 27 + 18 = 59 of them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow 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
 5 years ago
Best ResponseYou've already chosen the best response.0yuki, theres a mistake

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0and 6 + 5 + 7 = 18 a multiple of three

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.09 + 5 + 7 = 21, a multiple of 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we'd have to take into account (4,8) and (5,7) solutions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it 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
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0how would one do that then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so lets look at this way you have the numbers 3,4,5,6,7,8,9 , correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you get how i have total cases , 7x7x7 , right ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i want to find the number of NON multiples of 3, subtract from the total, will give you multiples of 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youre 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
 5 years ago
Best ResponseYou've already chosen the best response.0you can have 1*3 + 2*9 , for instance

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the coefficients here have to add to 3 ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01*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
 5 years ago
Best ResponseYou've already chosen the best response.0hmm i'm not understanding the multiplication tables at the top

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so clearly, , well i get 3*9 from (9,9,9)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and we are adding them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if you randomly picked 1*3, and say 2*7, you get (3,7,7)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so automatically you rule out all the 3's cases, and you rule out 6's, 9's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you rule out the last column, of 3*3, 3*4, 3*5,... 3*9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0only 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
 5 years ago
Best ResponseYou've already chosen the best response.0wait, you left in 3x3, that goes away

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01*3, 2*3, 1*4, 2*4 1*5,2*5, 1*7, 2*7, 1*8,2*8 1*9,2*9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0now 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
 5 years ago
Best ResponseYou've already chosen the best response.0clearly 1*3 , 2*3 is rule out, since thats just 3,3,3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01*4, 2*4 1*5,2*5, 1*7, 2*7, 1*8,2*8 1*9,2*9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh 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
 5 years ago
Best ResponseYou've already chosen the best response.0not the 9's or the 6's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01*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
 5 years ago
Best ResponseYou've already chosen the best response.0ok i see, so there is the case of one 3, one 4 , one 5, etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah cause 1*3,2*4 works

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so would we have to find the individual sums for all of the outcomes for each 1*{3,4,5,6,7,8,9}

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0theres also a strange case 4+8+9 = 21

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.04,5,7 4,5,8, 5,6,8 5,6,9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right , so the ones cases you can do by hand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0those are all the ones with non multiples of 3 so 8 * 6 = 48 ways that aren't possible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then you have permutations of them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill double check, but that looks right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so for the one and 2's cases i have 1*3  ( 2*4, 2*5, 2*7, 2*8, )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for 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
 5 years ago
Best ResponseYou've already chosen the best response.0and you systematically eliminate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04 + (2*3, 2*7, 2*6, 2*9)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.05 + (2*3, 2*4, 2*6, 2*7, 2*9)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.06 + (2*4, 2*5, 2*7, 2*8)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.07 + (2*3, 2*5, 2*6, 2*8, 2*9)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.08 + (2*3, 2*4, 2*6, 2*7, 2*9)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.09 + (2*4, 2*5, 2*7, 2*8)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then you'd take the permutations of the singles + the permutations of the doubles and subtract from 7 *7*7?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you have the answer in the book,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that would be nice to check
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