2
$\begingroup$

Edit #1: Remember that a, b, and c are positive digits, so their values are between 1 & 9, inclusive

I was thinking casework for this problem, but I got stuck on the last part. I can't figure out whether there are four cases or three.

$20 = 2^2 * 5$, so you need two 2's and a 5.

Case 1: One five, either a four or an eight, and an odd number that is not five

Case 2: Two fives and either a four or an eight


Here's where I get confused. I'm not sure whether it should be:

Case 3: One five and two even numbers

or:

Case 3: One five and two even numbers that are different

Case 4: One five and two even numbers that are the same


Here are the calculations that I have done so far:

Case 1: $2* 4 * 3!$, as the five is fixed, you can choose between the 4 & 8, and four choices for the odd number that is not five. The $3!$ accounts for the various permutations of the three that are chosen.

Case 2: $2 * 3$. The fives are fixed, you choose between the 4 & 8, and there are three different ways to arrange them after you choose your numbers.

Edit #2: I also think that the answer is $102$. I 'cheated' (used Python and wrote a script) to figure that part out, but I want to figure out the answer with math, not programming.

Thank you in advance!

$\endgroup$
  • 2
    $\begingroup$ There are infinitely many. Do you mean the product is exactly equal 20? $\endgroup$ – fleablood Mar 21 '17 at 2:56
  • $\begingroup$ $(20,1,1),(40,1,1),(60,1,1),\dots,(20n,1,1),\dots$ all have their product divisible by twenty, and that is just one class of solutions... or perhaps you are restricting each of $a,b,c$ to be integers in $\{0,1,\dots,9\}$? $\endgroup$ – JMoravitz Mar 21 '17 at 2:57
  • $\begingroup$ @YiyuanLee The constraints are stated in the question: a, b, and c must be positive digits, so they have to be less than 10. $\endgroup$ – S. Kim Mar 21 '17 at 3:08
  • 1
    $\begingroup$ digits... you really should have said that. "digits" meaning less than ten is not always standard.. $\endgroup$ – fleablood Mar 21 '17 at 3:11
  • $\begingroup$ @fleablood I assumed that digits was a standard term...guess not. Anyway, I edited the question to make that clearer. $\endgroup$ – S. Kim Mar 21 '17 at 3:13
0
$\begingroup$

Okay, If any of them are $0$ the product is $0$ which is divisible by $20$.

There are $1$ way to have all three being $0$. $3*9*9$ ways for there to be exactly $1$ zero (3 places to put the zero and $9*9$ choices for the other 2). There $3*9$ ways for there to be $2$ zeros.

Otherwise you need 5. You can have 5, 4 or 8, and an odd. There are $3$ places to put the $5$, there are two remaining places to put the $4$ or $8$. So that is a total of $5*2*2 = 20$ ways.

Or you can have a $5$ and two different evens. There are $3$ places to put the $5$ and there are $4$ options for the first even and $3$ for the second even. So there are $3*4*3 = 24$ ways to do that.

Or you can have a $5$ and two equal evens. There are $3$ places to put the $5$ and there $4$ options for the two equal evens. That is $5*4 = 20$ ways.

So in total there are $20 + 24+20 + 3*9*9 + 3*9 + 1$ ways to do this.

$\endgroup$
0
$\begingroup$

This is assuming that by positive digit you mean a number in {0, 1, 2,...,9} since that wouldn't give an infinite answer.

Fix 3 digits - 1 unique unordered triple
$(2, 2, 5)$

Fix 2 digits - 9 + 8 unique unordered triples
$(4, 5, x) \text{ for }x={1,...,9}\\(5, 8, x)\text{ for }x={1,2,3,5,...,9}$
One less in the second to avoid double counting $\because(5,8,4) = (4,5,8)$ in my construct

Fix 1 digit
I claim these are already counted
--> Case 1: choose digit in 2, 4, 5, 8. Then we must choose another digit to ensure divisibility by 20. If divisible, then we are in the second section. If not divisible, we must choose the third digit to ensure divisibility. Then we must be in the first section.
--> Case 2: choose digit not in 2, 4, 5, 8. Then we must fix the other two to ensure divisibility by 20 as in the second section and thus is already counted.

edit: I screwed up the combinations for the final total, whoops

$\endgroup$
  • $\begingroup$ You missed $2,5,6$ as well as the ones that have zero $\endgroup$ – Ross Millikan Mar 21 '17 at 3:48
  • $\begingroup$ @RossMillikan 0 is not a positive digit. $\endgroup$ – S. Kim Mar 21 '17 at 3:50
  • $\begingroup$ @S.Kim yay I can go back to forgetting about 0 haha! $\endgroup$ – Charlie Mar 21 '17 at 3:51
  • $\begingroup$ @RossMillikan oh $(2, 5, 6)$. Yes. Yes I did. The even/odd approach might work better $\endgroup$ – Charlie Mar 21 '17 at 4:06
0
$\begingroup$

Starting over since @fleablood's answer doesn't add up when disregarding the 0's...

The correct answer is 102 as confirmed by my and OP's scripts.

Let $(a, b, c)$ be our triple with $a,b,c\in{1,...,9}$
$20 = 2^2(5)$. So $5$ must be one element of the triple.
WOLOG set $a=5$. Then $20\mid abc \implies 4\mid bc$.
Thus $bc$ is even, and so at least one of $b$ or $c$ is even.
WOLOG say $b$ is even:

Cases
1. $(5, b, b)$
2. $(5, b, c)\text{ where }c\text{ is even and } c \ne b$
3. $(5, b, c)\text{ where }c\text{ is odd}$
$\rightarrow$However, $c$ odd means $4\not\mid c$ and together with $4\mid bc$ from above, have $4\mid b$. So $b=4$ or $b=8$.

Counts
1. 3 positions for $5$. $b\in \lbrace 2,4,6,8\rbrace$ means 4 choices for $b$, will fill other 2 slots at once.
$\implies 3*4 = 12$
2. 3 positions for $5$. Fix $b=2$, 3 choices for $c$, 2 perms. Fix $b=4$, 2 choices for $c$, 2 perms. Fix $b=6, c=8$, 2 perms. Can't fix $b=8$ as it's already counted.
$\implies 3*2*(3+2+1+0) = 42$
3. 1 choice for $5$. 2 choices for $b$. 5 choices for $c$.
$\implies 6*(1*2*5) = 60$

But these add to $112 > 102$, which is the correct answer. I'm lost.

Other thing I know: If $a,b,c$ are all distinct, then there are $3!=6$ permutations of (a, b, c).

I feel like I should be asking the question, not answering. Please someone help, I got sniped hard by this and my combinatorics is weak.

$\endgroup$
  • $\begingroup$ @fleabood can you help? $\endgroup$ – Charlie Mar 21 '17 at 9:54
  • $\begingroup$ @RossMillikan can you help? $\endgroup$ – Charlie Mar 21 '17 at 9:54
  • 1
    $\begingroup$ 2 cases for $b,c$. Case 1 [Both even]. 2 subcases: $(5,b,b), (5, b, c)$ with $c$ even. Case $(5, b, b)$ has $\binom{3}{1} \times \binom{4}{1} = 12$. Case $(5, b, c)$ has $ 3! \times \binom{4}{2} = \mathbb{36}$. Yes 36. $\endgroup$ – mdave16 Mar 24 '17 at 21:20
  • 1
    $\begingroup$ Case 2 [Even, Odd]. 2 subcases: $(5, b, 5), (5, b, c)$ with $c$ odd. Case $(5, b, 5)$ has $ \binom{3}{1} \times \binom{2}{1} = 6$. Case $(5, b, c)$ has $ 3! \times \binom{2}{1} \times \binom{4}{1} = 48$. Note that $48 + 6 = 54 \neq 60$. $\endgroup$ – mdave16 Mar 24 '17 at 21:27
  • 1
    $\begingroup$ Also $12 + 36 + 6 + 48 = 102$, I think this question took me longer to type out in the comments without using enter, than if I had just answered this question properly. Alas. $\endgroup$ – mdave16 Mar 24 '17 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.