Monday Math Madness #12: The Martian Problem
04 Aug 2008 Daniel Sutoyo 38 comments 4,564 views
The Problem Statement
Tired of the Earth obstructing his view of Venus, Marvin the Martian desires to destroy the Earth with his Illudium Q-36 Explosive Space Modulator. The only thing stopping him is his distant cousin Jeffrey’s belief that there are intelligent life forms on Earth. In order to show Marvin that there are indeed intelligent lifeforms, Jeffrey and Marvin travel to earth and approach two mathematicians with the following problem:
Jeffrey: I am thinking of two numbers X and Y in my head. Both X and Y are positive integers greater than 2 and less than 99. I am going to tell Dan the sum of these two number; similarly, I will tell Quan the product of these two numbers. Without telling each other what I have told you, can you tell me what the two numbers X and Y are?
Dan and Quan stare at each other for a bit. Shortly thereafter, Dan and Quan have the following conversation between them:
Dan: Quan, you don’t know what X and Y are.
Quan: That was true, Dan, but now I do!
Dan: Hmmmmmmmmmmmm. And now I do, too!
Both: X and Y are _ and _!
What are X and Y?
The Bounty
The winner will receive a 10 dollar gift certificate to Amazon.com!
The Rules
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Email your answers to mondaymathmadness at gmail dot com.
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Depending on how many submissions we receive, we will be giving out additional prizes, giving you yet more chances to win. We’d also like to give a prize out to the “best” or “most creative” answer.
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Inelgible for any one person to win more than once per year. But you should still submit your answer!
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Answer must be explained. You must show your work! We will accept answers in the form of a MATLAB script as well. We will be the final judge on whether an answer was properly explained or not.
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The deadline to submit answers is August 11th 2008, Monday 11:59 PM Pacific Standard Time
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The winners will be chosen randomly from all the submittals using a random number generator.
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The winner will be announced at 9:00 AM PST August 15th, 2008.
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Comments for this post should only be used to clarify the problem. Please do not discuss ANY potential solutions.
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Please spread the word about our contest by stumbling this webpage!
38 Responses to “Monday Math Madness #12: The Martian Problem”
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Include MATLAB code in your comment by doing the following:
<pre lang="MATLAB">
%insert code here
</pre>
This problem seems harder than usual. xD
Yes it is harder than usual, but we got some complaints of past problems being too easy : )
Is X and Y distinct?
I know there are different variations of the problem, but in this case I believe it does not have to be distinct.
Erm.. is that a yes or a “it might be”?
X <= Y
Are we supposed to come up with exact values, or values relative to the sum and product?
the exact values of X and Y
yep, doesn’t change the result.
So, the max product can be 98*98.. or.. 98? =(
Maybe I should just give up already, haha.
Done the problem.
The hardest problem for MMM yet, Martians do have some intelligence.
Can be hand solved… but I suggest for this problem, use a computer to do some filtering. Least amount of time expected for hand solving is 15 minutes, and that’s if you know how to solve it already and found a smart starting point.
Yeah, I solved it too.. nevermind my previous comment. :p (or I think I did)
[...] and Multiplication Monday Math Madness #12 is up at Blinkdagger, and features Marvin the Martian (who turned 60 years old this past July 24. [...]
Sorry guys, I won’t be sending a solution to this one - I’ve seen it far too many times (seven or eight times at least, in the last 3 or 4 years).
Further, many solutions are available online by direct googling on information in the question (I found four or five on the first page of a search I just did); even if I hadn’t already seen the problem, I don’t think I’d be so keen to compete with people’s ability to just google for the solution.
Hello efrique,
While you may have seen this problem many times over the years, I know that many other people will be seeing it for the first time. Perhaps you would like to offer ideas for the next MMM? Any ideas? Contact us via the Contact Form if you have suggestions!
Quan
efrique, totally understand where you are coming from and I respect you for not participating.
I first saw this problem in my discrete class 7 years ago. It is a good, challenging, and fun problem especially for those who have not seen it. Several people have found this problem fun and challenging (based on incoming links).
You are right that with google you can almost find anything (even wrong solutions). It is possible that a contestant gets lucky and win the contest by googling (we do read the submissions thoroughly for posers). Having said that, ultimately our goal is to let engineers/mathematicians have some fun doing neat problems. Hope you understand.
Well, you definitely did a good job of coming up with such a problem. I had fun figuring it out.. even if I turn out to be wrong. I’ve never done anything like this before.
Is it 2 and 99 inclusive or exclusive?
Hi Anneleen,
2 and 99 are not included.
Quan
Also uhm, I’m guessing we don’t get to know whether or not we’re right until Monday, do we? =(
I completely agree that it’s a fun problem, well worth the time if you haven’t done it before.
I certainly wasn’t objecting to you guys asking any question you see fit - it’s not up to you to offer only problems that I haven’t seen; if you keep asking interesting questions you’re bound to hit questions that some of us have seen before.
I may sometimes point out that I’m choosing not to enter, though. No slight is intended.
[...] a contest from blinkdagger (9rules member). Good [...]
See
Lets say x+y = m
and x*y = n
then x = (4n-m^2 +m)/2
and y = m - (4n-m^2 +m)/2
So it is something like both the sum and products are same for the numbers
so the possible solutions is 2 and 2
If only it were that simple.
It seems like the answer to this problem is no. We have to know what the product and sum is of X and Y.
No you don’t :p
And nevermind, my solution is entirely whacked.
4 & 13 . ..
EZZZ
4 * 13 = 52 giving factors of
1,52 2,26 4,13
Factors 1,52 and 2,26 are eliminated, leaving 4 and 13. Quan would be able to figure out what X and Y are with his product alone. It clearly states he does not know initially.
“EZZZ” eh?
Do note I’m only posting this because the deadline is long since passed (and who knows, maybe I’m wrong).
[...] results are in, and the winner for MMM #12 is Anneleen Van Geenhoven! She will be receiving a $10 Amazon gift certificate. We only had 9 [...]
Kinks,
You are correct on one part, you are wrong.
Where did you get 4 & 13?
If Quan knew his product was 52 then he could have easily known X & Y were 4 & 13 for the very reason you stated.
Dan on the other hand could not know if his X & Y were 3 & 14 or
4 & 13 or
5 & 12 or
6 & 11 or
7 & 10 or
8 & 9.
I was replying to the comment above.. you know, the guy who said it WAS 4 and 13.
its simple…
First off, Dan and Quan solve the problem, not the reader. So it is fair game to assume the following:
Quan hears: 52
Dan hears: 17
the first statement:
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Dan: Quan, you don’t know what X and Y are.
—-
is just that, a statement. He isn’t asking Quan. He is telling Quan “I KNOW that you don’t know what X and Y are”.
This gives Quan additional information.
If Quan knows the product, what are the possibilities that given the product, he can know what X and Y are: There are only 2 factors > 2 and 2 is always even, this means that Dan’s Sum was Odd. So now we know that BOTH numbers CANNOT be prime and that 1 number is even and 1 number is odd.
The second statement
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“Quan: That was true, Dan, but now I do!”
—-
lets Dan know that the Product of X and Y is the Only product that could allow Quan to determine the values of X and Y correctly
For example, some possibilities:
7 * 10 = 70 but also: 5 * 14 = 70
5 * 12 = 60 but also: 3 * 20 = 60
6 * 11 = 66 but also: 3 * 22 = 66
So it can’t be those combinations.
Dan knows his Sum is 17
and since Quan says that he now knows the number, this lets Dan know the number as well since 4 + 13 = 17 is the only combination of X & Y where x * y presents a unique value given the constraints:
3 * 14 = 42 (but so does 6*7)
4 * 13 = 52 (no other factors where both are > 2 & < 99)
5 * 12 = 60 (so does 3 * 20)
6 * 11 = 66 (and 3 * 11)
7 * 10 = 70 (and 5 * 14)
8 * 9 = 72 ( and 3 * 24)
So now that Dan knows that Quan knows, Dan can narrow his choices down to 4 and 13
The secret is to extract every ounce of information you can. Sometimes its easier to work backwards.
Go here to find out the winner and answer!
http://www.blinkdagger.com/monday-math-madness/monday-math-madness-12-the-winner-is
Yes, please do click the URL. I don’t know why you’re arguing about it now after the answer has been posted..
[...] who submitted (23 this time) got it correct! Maybe we need more difficult problems like the Martian Problem or our most popular problem Mac Madness. Give us some feedback on ideas and promotions. And thanks [...]
how is it fair game to assume they heard 17 and 52?