2008 Mathematics Game

Are you ready for a challenge? Join us for the 2008 Mathematics Game. Here are the rules:

Use the digits in the year 2008 and the operations +, -, x, ÷, sqrt (square root), ^ (raise to a power), and ! (factorial) — along with parentheses, brackets, or other grouping symbols — to write expressions for the counting numbers 1 through 100.

  • All four digits must be used in each expression.
  • Only the digits 2, 0, 0, 8 may be used.
  • Multi-digit numbers such as 20, 208, or .02 MAY be used this year.
  • The square function may NOT be used.
  • The integer function may NOT be used.

By definition:
0! = 1
[See Dr. Math's Why does 0 factorial equal 1?]

For this game we will accept the value:
{0}^{0} = 1
[See the Dr. Math FAQ 0 to the 0 power.]

According to the Math Forum 2007 game page:

For many years mathematicians, scientists, engineers and others interested in mathematics have played “year games” via e-mail and in newsgroups. We don’t always know whether it is possible to write expressions for all the numbers from 1 to 100 using only the digits in the current year, but it is fun to try to see how many you can find. This year may prove to be a challenge.

This year, also — those zeros make it tricky! Still, working together, we managed to find 76 answers last year.

Use the comments section below to post a running list of the numbers you have been able to calculate. We had quite a lively discussion last year. I will be out of town this weekend, visiting the in-laws for a belated Christmas celebration, but I hope to dig into this puzzle during the long drive. I look forward to seeing your comments when I get back….

Warning: Some teachers may be using this puzzle as a classroom assignment, and there will always be students looking for people to do their work for them. Please do not post any solutions!

Edited to Add

Many people have asked in the comments, so I will keep a running tally here. These are our results so far:

Percentage solved = 86%.
Numbers that remain unsolved =
43, 52, 53, 67, 68, 70, 72, 76, 87, 92, 93, 94, 98, 99.

Clarifying The Rules

The only digits that can be used to build 2-or-more-digit numerals or decimals are the standard base-10 digits 2, 0, 0, 8.

  • “0!” is not a digit, so it cannot used to create a base-10 numeral.
  • The decimal point is not an operation that can be applied to other mathematical expressions: “.0!” does not make sense.

No exponent may be used except that which is made from the digits 2, 0, 0, 8.

  • You may not use a square function, but you may use “^2.”
  • You may not use a cube function, but you may use “^(2+0!).”
  • You may not use a reciprocal function, but you may use “^(-0!).”

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150 comments on “2008 Mathematics Game

  1. Ok, on a first pass, I have everything up to 21, 23-26, 30, 32, 34-36, 38-42, 44-50, 56, 64, 74, 77-83, 86, 89-91.

    (If I counted correctly, that’s 56 answers so far.)

    Shall we try to predict whether it’s going to be easier or harder this year than last? I was thinking it might be harder since everything is a power of two, so we have fewer factors to play with. (By harder, I mean I think fewer of the numbers will be solvable.)

    Like

  2. I’ve gotten a whole bunch:

    1-30,32,35-43,45,47-50,56,58,60,62-66,69,71,
    73-74,77-84,86,89-92,95-97,100

    Mathmom has reported 34, 44, and 46 which I don’t have.

    So, if neither of us have made a mistake the missing ones are:

    31,33,51-55,57,59,61,67-68,70,72,75-76,
    85,87-88,93-94,98-99

    That means there are only 23 numbers missing.

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  3. Thanks to n0n4m3’s hint, I have 60. I was trying too hard to find ways around using that 2-digit rule. I’m still missing 37, 43, and 92, but I marked them for increased attention.

    From the missing list, I have a expressions for 31, 55, and 57 without using 2-digit numbers, and 75 and 85 with.

    Like mathmom, I thought this year would be harder than last, but my current score is 81% and there are still more to find…

    Like

  4. By the way, has anyone found expressions for any of the following which do NOT use 2-digit numbers:
    28, 29, 60, 74, 75, 77-79, or 83-86?

    Two-digit numbers are allowed in the rules, but somehow it still feels like “cheating.”

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  5. Hmm. I just got 60 without using 2-digit numbers, but I didn’t have any pieces left over to get the surrounding ones. Guess I need a different solution… I did figure out 37 — what a bear! I was able to use the same trick to find an alternate expression for 25, for whatever that might be worth.

    I keep thinking I have found one of the missing numbers, only to realize that I used a digit twice. Of the numbers you all have reported, I still don’t have 43, 59, 61, and 92.

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  6. Okay, so what you are telling me is there’s a way to make 60 using only 0, 2, and 8, so I’ll have something left over to make 59 & 61, right? But it’s “complicated.” Worse than getting 71 or 37 or *gasp* 31?

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  7. Are you using a “cube root” function? I see a way to get it if I take the cube root of 8, but I don’t think that’s allowed. I’ll keep playing…

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  8. Nope, I use a: sum, sqrt, 3 factorials a division (not in this order, to get 60) and the last operations: a factorial and sum/sub to get the 61 or 59 respectively.

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  9. It sounds like non4m3 did the same thing as I did for 60. (I also have a decimal in there) — I don’t have 31 yet — are you telling me there’s something similar I can do there?

    I don’t have 43 or 92 yet, so I can’t help you on those. Hopefully I’ll have time to play some more tonight and see if I can fill some of my gaps. I have 22 numbers to go, several of which have been reported as found by some of you.

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  10. Ok, now I’m down to 16 to go. Of those, two (43 and 92) have been solved by others (Sol — or so he claims. ;-) )

    Having gotten 71, I’ll say that 59/60/61 is worse than that. It’s kind of similar to 31 but uses a different internal trick that I haven’t used much elsewhere. Let me know if you want a little more hint on that.

    The one that drives me nuts is 72 — it just seems like it should be doable! ;-)

    I think we may be getting down to the bottom of the barrel. I’m not ready to give up quite yet, but maybe soon. ;-)

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  11. Unfortunately I don’t really have 43 or 92. I thought I did but I screwed up. Now, in the year 2028 I’ll have a head start :)

    Sorry.

    It’d be interesting to discuss approaches to doing this puzzle.

    I spent some time looking at last year’s results for tricks. That gave me ideas about dividing by decimals to get integers, about using factorials of factorials, and about taking a particular factorial, adding a particular digit to it, then taking the square root of the whole thing. The last one, in particular, would not have occurred to me without this “research”.

    I wonder if there’s a finite bag of tricks used for the problem every year and if one could apply all the tricks to the four digits each year and quickly crank out all the possible solutions.

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  12. Aha! I finally got the “trick” for 59-61 while tossing in bed about 11:37 last night. I was so stuck on doing square roots as the final or nearly-final step that it hadn’t occurred to me to try one so early in a calculation.

    Past experience certainly makes the list go faster. Last year we were into February before we had all the solutions. This year, with help from all of you, I have a score of 82% after only 4 days. I will keep playing at it to see if any other numbers are possible, but since no one else has reported new discoveries, I think we might be done.

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  13. There are usually several ways to make any particular number. The easiest way I can think of to make 15 would be to take advantage of how close it is to 16 (which is a basic times-table number). You would need to subtract one, but the rules give you a way to do that without using any factorials, if you wish.

    For any number that gives you trouble, it can help to look at the neighboring numbers and ask yourself: “Can I make one of these, and then add or subtract something to get the number I want?”

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  14. my teacher changed the rules a little. He made us keep the numbers in order, but in order to subtract one you would have to do that after multiplying the 2 by 8????????

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  15. I can do it with the numbers in order, but I need three factorials.

    Whatever set of rules you’re using, there will be numbers that are not solvable. Have you been given reason to believe that it is possible to make 15 using the numbers in order, and at most two factorials? What other rules do you have.

    Your post is interesting to me because I was thinking about trying to make as many as possible using the numbers in order (but with no other restrictions other than the usual rules) but haven’t tried it yet.

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  16. Sol, what I do is figure out everything I can make using certain subsets of the numbers. What can I make using only (2,0), (2,0,0), (8,0), (8,0,0) — those are my main sets, but I also looked at (2,8) and (2,0,8) as well (but those aren’t so helpful, because there’s only so far you can get from anything there.

    But yes, there are certain tricks that I learned by doing this last year that certainly made this easier this year!

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  17. jesuschick: well, you don’t need to subtract it after multiplying 2 by 8.. just “subtract” it before using a – before the number.

    This was the first time I played this game and I used technics similar to what mathmom and Denise described, i.e., check how can I get a number and then try getting the surrounding ones, and what numbers could I get from the combinations of the digits.
    Other thing I used was working backwards: if I want a number x, what is the: sqrt(x), x^2, x +- 2, x * 2, etc. so I can get it using the numbers I have available.

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  18. jesuschick,

    As mathmom said, keeping the numbers in order, 15 can be made with exactly 3 factorials. This is actually the way I originally did it, not because I was trying to keep the numbers in order, but because I was working my way up the number line. I made 14 as 3!+8, so to make 15 I simply added 1 to that.

    Without using more than two factorials AND keeping the numbers in order… hmmm. I’ll have to puzzle over that one. Are you sure it is possible?

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  19. I guess I misunderstood the problem then… I thought what jesuschick wanted was the digits of the calculation in order: 0 … 2 … 8 not the actual intermediate numbers that each operation create.

    In that case the only solutions I see are getting 14 (there are at least two different ways both using a factorial) and then the other factorial to get 15. I can’t find one without using two factorials.

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  20. My interpretation of “in order” was she had to use them in the order 2 … 0 … 0 … 8, but since she hasn’t been back to clarify, we may never know.

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  21. Emily — I’ll add another hint that is key for getting 61, which is that it’s possible to get 3 using only 8 and 0. I didn’t use that for 31. For 31 I used a total of three factorials, a decimal, a square root, addition, multiplication and division. Let us know if you want more of a hint.

    non4m3 — I think two factorials are allowed, so if you can do it with two, you’ve got it. Best I could do was 15 = (2 + 0!)! +0! +8 which uses 3 factorials.

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  22. I still i can’t manage to get 60. As said in a previous post i need, 2 sqrt’s, 3 factorials, 1 division and 1 addition. But this doesn’t seem right, because i’l need to use one square root with 3600 to get 60, then then one factorial and one addition to get 3 by doing 2 x o!, and since i will later need to divide either 3 or 8 by a number, the number that i will need to divide must be made on its own using 2 factorials and 2 sqrt’s. And i can’t manage to get this number.
    Sorry if it’s a bit confusing

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  23. Sorry, I thought I have two ways (maybe I had) but now I only have one… Tip: I solved it by using a method that I explained (or tried to) before, that is starting from the end. What are the possible numbers from where we can get 15?
    225, 10 + 5, 3 * 5, 20 – 5, etc..
    Try getting one of the ones above and you’ll only need to use 2 factorials ;)

    Emily: How can you 3600 from 720 and how can you get the 720 you need using the remaining digits? ;)

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  24. n0n4m3: Of course! I had that method written in my notes, but I had put the numbers in a different order. I wasn’t paying enough attention to notice they could be rearranged.

    Emily: That’s the same place I was stuck for a long time and almost conceded defeat. What finally cleared it up for me was finding a different way to make 3.

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  25. There are two ways I know to get 31. One method gets it directly, by first finding 31^2 and then taking the square root. The other method is to make 30 using only 3 of the digits, and then add 1 to get 31.

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  26. Hey guys how many have % have we figured out are possible, I have about 79 of the 100. Of the ones I’m missing, (31, 33, 37, 43, 51-53, 59, 61, 67-68, 70, 72, 76, 85, 87, 92-94, 98-99) how many are possible?

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  27. I’m still stuck on the question of making 15 using the digits in order and only two factorials. I could do it if I could make 5 from 0,8 but I don’t think I can!

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  28. sully: as Denise I also have 84%.

    mathmom: “I could do it if I could make 5 from 0,8 but I don’t think I can!” – maybe if you try 5 * 3 instead of 3 * 5 it will help ;)

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  29. Hi
    I’ve nearly got 60, i know how to do the equation just that i need a 5 and my only remaining digit is a 2 . I got the three using 8 and 0!. But im stuck in the middle of the equation i can get up to 720 doing 3!! and i can go backwards from 60 by doing a sqrt but i can’t get the step in the middle.
    Can someone please help me

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  30. non4m3: I can do 5 with 2 and 0, but the 0 has to be first. I can’t think of a way to do 5 with 2 then a 0.

    Emily: you just need one more trick to get from 720 to 3600 and you can do it using the 2. You need a decimal point. Hope that helps!

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  31. Emily, it sounds like you are almost there! You can use the 2 in place of a 5, which is a useful trick for many numbers. Look again at this part of the rules: “Multi-digit numbers such as 20, 208, or .02 MAY be used this year.” Decimals are allowed…

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  32. Mathmom, reading your comment, I just realized I don’t have 15 in order either. I guess my coffee hadn’t kicked in yet when I thought I could rearrange those numbers this morning. Division isn’t commutative! Back to the drawing board…

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  33. But now I really do have it. My antique dial-up Internet connection crashed my computer (again!), and while I was waiting for it to reboot, I realized that there’s more than one way to skin a reciprocal.

    I like your cool 4-factorial method, and I’ve added it to my list, too.

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  34. mathmom: as Denise said, to make the reciprocal you don’t need to change the order of 2 and 0 ;)

    I guess I didn’t have 2 ways of getting 15, I’ve been trying but couldn’t get it so probably I was wrong and there’s only this way.

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  35. Oh, finally! I figured out how to do 5 “backward” using 2 and 0 in that order. Assuming we can use a unary – to create a negative, which it doesn’t explicitly say we can do, but seems within the spirit of things.

    With only one factorial!? hmm….

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  36. Pingback: Ars Mathematica » Blog Archive » 24th Carnival of Mathematics

  37. mmm… i have a question, if you put 0.2 does the zero count as if u were using one of the zeros in 2008 or does it count as a separate number?

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  38. You can just use “.2″ without the leading zero. (I think it would be kosher to use 0.2 also if you wanted to use up the zero, but there are other ways of getting rid of a zero you don’t need anyhow.)

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  39. I found a second way to get 15 (not using 5×3) with the numbers in order and only two factorials. But now there is a way with only one factorial? And n0n4m3 says it is “tricky”? That will probably keep me busy until we are ready for next year’s game…

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  40. I think I just figured the way Denise is talking and it happens to be the one I talked some posts ago but have forgotten… That’s what happens when you don’t write the things :\

    About the method with one factorial… I think it’s not as tricky as the 59/61 way but it’s tricky in the sense that it’s calculated in a different way from most of the numbers I have.

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  41. i got 15 with 2 factorials by taking the square root of 20! four times ; i then added that to 0!^8 ; i came out with 15.079 Is that close enough?

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  42. jesuschick — I don’t think that’s close enough. I think you should only use methods that result in whole numbers. But really whether or not it is close enough is up to your teacher!

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  43. I don’t think we’re allowed repeating decimals, though that would be a fun addition!

    You have 70? Hmm, I’ll have to go back and look at that one again…

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  44. @ Qi Yanjun — We are trying not to post specific answers, because some teachers are using this as a homework assignment. I made an exception for mathmom’s 4-factorial version of 15 above only because there is a much easier way to get 15 that most students would already have figured out.

    @ jd2718 — It seemed like 70 and 72 ought to be possible, but I hadn’t found a solution for either of them. I’ll have to give 70 another look. Here is the list of numbers I am still missing:
    33, 43, 51-53, 67-68, 70, 72, 76, 87, 92-94, 98-99

    And then there is the challenge of seeing how many numbers you can do with the digits in order. I haven’t started that (except to play with jesuschick’s 15 problem), but perhaps when I get caught up on bills and email…

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  45. I got 33 but only by using -0! is that allowed? also, can someone help me get 37, I dont know where to start, and it’s the only one i have left

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  46. For 37 I used the fact that I can make the number 3 using only an 8 and a 0. Let me know if you want more of a hint than that.

    I’m not sure if a unary negative is allowed or not. The rules don’t explicitly allow it, but it seems “within the spirit” of things. Still, I’m having a hard time thinking of how you would need it if you didn’t care about the order of the digits. I’ll have to look at 33 again in light of that new hint.

    Like

  47. Pingback: Teachers share, nice example « JD2718

  48. Can we get an updated list of missing values? I’ve got 22 left to find and want to know where to focus my attention (and if I’ve found anything no one else has…)

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  49. Aww i just found 69,71,73 and i was so excited because i thought i mighta been the first. lol. oh well. anyways it would be nice to get an updated missing list.

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  50. Missing values: 33, 43, 51-53, 67-68, 70, 72, 76, 87, 92-94, 98-99.
    Except: jd2718 says he found 70, and sully says he found 33. I haven’t found either of these, yet. Care to give us a small hint?

    I also have not found n0n4m3’s one-factorial, numbers-in-order way to make 15.

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  51. I was able to get 35 using only 3 of the digits, which left me an extra 0! to subtract for 34 (or add for 36). It is similar to the way I got 44-46, if that helps.

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  52. i also have 33, but im not 100% sure whether or not it is legal. I used a unary sign in the expression.

    i cant seem to get 3 from 8 and 0 and i have spent forever on it. can anyone leave a hint for what signs that uses or something

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  53. wow it just came to me duuuh

    i now have all except

    Missing:
    37, 43, 51, 52, 53, 67, 68, 70, 72, 76, 85, 87, 92, 93, 94, 98, 99

    that is 17 numbers and since people have said that 85 and 37 are possible, doesnt that mean that the best so far should be 85 total.

    i was wondering if anyone had some hints for 85 or 37

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  54. For 37, I used the trick that involved getting a 3 from 8 and 0 again. For 85, I used a 2-digit number and then added something to it. Hope that helps. I’ll see if I can come up with your 33, which I don’t have.

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  55. When I got 37, I hadn’t yet figured out that “Duh!” method for getting 3 without using up my 2, so I had to take the lo-o-o-ong way around. But I managed to square a number without using the 2 as an exponent, and still hold back the extra digit I needed to add to get 37.

    I was pretty impressed by the trick, so I used it to get alternate expressions for 4, 16, and 25, too. Unfortunately, I can’t think of a way to use it for anything on the missing list.

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  56. Rules clarification: No exponent may be used except that which is made from the digits 2, 0, 0, 8.

    You may not use a square function, but you may use “^2.”

    You may not use a cube function, but you may use “^(2+0!).”

    You may not use a reciprocal function, but you may use “^(-0!).”

    With that in mind, should I still be looking for 33? So far, I have been able to come up with an expression that would be useful in the year 20008, if anyone wants to play the game then. But for this year, I am stumped.

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  57. I have 33, in a way that doesn’t involve a single “-0!” (Is that a hint?) It must be one of Murphy’s many laws: Get on the computer at some ridiculous hour of the early morning and tell the whole world that you’re stumped, and then go back to bed — next thing you know, the answer is bound to pop into your head.

    Current percentage: 85%.
    Missing values: 43, 51, 52, 53, 67, 68, 70, 72, 76, 87, 92, 93, 94, 98, 99.

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  58. i must have the same way

    i thought i remembered someone saying that they got 43, was that actually possible, or were they wrong

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  59. I still need a hint for 33. I know there’s already a hint out there in that people used -0!. It seems to me that if you don’t care about the order of the numbers, you don’t need to ever use -0! (at least not as an exponent) because you could just do 0! / whatever. So… maybe there is a different use of -1 that I’m not thinking of. But anyhow, a little hint?

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  60. aballerinagirl,

    I don’t have 92, and I don’t think anyone else does either (at one point Sol thought he did, but he has retracted that claim).

    For 89 through 91 I used a method of getting 90 using only 3 of the numbers, leaving me free to adjust slightly with the remaining one to get 89 and 91. The trick to those is lots of factorials, and division.

    I think the only one I have left that anyone has solved is 33. I have 16 left, and am beginning to believe that 15 of them are impossible. I still think 72 should be doable though. :)

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  61. mathmom,

    I have the number 72, but it took a long time to find it. Subtract something from 82. Do the same for 92 but add something to 82.

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  62. if you happened to do
    82+0!0 for 92 or subtracted it for 72, then you got 92 and 72 illegally. 0!0 does not equal ten, it equals 1, because to use two digit numbers they have to be numbers before they are connected.

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  63. I don’t want to write the answer out, but i had a decimal to a negative power got 25 and added 8

    I also got 51 using a negative 8 I hope its allowed.

    Is .0!=.1=allowed if not I cant get 61 or 59 and need some help thanks

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  64. I cant find 37, not sure what you all see that I dont, I think I’m trying to solve it like past numbers instead of finding a special equation just for it… if anyone can help I’d appreciate it

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  65. Would you like me to tell you how I got 10 with two zeroes?

    I am still stumped as can be on how to get sixty without two zeroes.

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  66. sully,

    59 and 61 are the messiest ones I got. To get 60 using only one of the zeros, you need to use factorials, square roots, decimals, and division.

    There is a certain factorial, that when divided by a certain decimal, yields a useful square number. Getting to the certain factorial is a little messy, and uses a trick that also came in handy for 37.

    thanks for the hint on 33 — I finally got it with that!

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  67. I don’t think so. You can’t just shove a decimal point in front of an arbitrary mathematical expression. You’d never see .(54/9) used to represent .6, for example. You can only place decimal points in front of, between, or after decimal digits.

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  68. Rules clarification II: The only digits that can be used to build 2-or-more-digit numbers or decimals are the standard base-10 digits 2, 0, 0, 8. “0!” is not a digit.

    @ sully: 51 is cool! It reminds my of my long work-around for 37, but I didn’t make the connection before you mentioned it. If you are still looking for 37, you might try applying that 8 trick to some other expressions.

    Updated percentage: 86%.
    Still missing: 43, 52, 53, 67, 68, 70, 72, 76, 87, 92, 93, 94, 98, 99.

    Wow! I never expected to find so many.

    Like

  69. .0! is not .1, because .0 is equal to 0.0, which is equal to 0 which when factorialed(it should be a word) is 1

    i cant get 51, can anyone give anymore hints

    i assume that there is some way to get a 50 without using one of the o’s

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  70. after playing around for a while i realized that i could get 25 from just a 2 and an 8, but that only leaves the two zeroes which dont allow me to get 51. am i on the right track or am i going about it wrong

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  71. The method for getting 25 from a 2 and an 8 is close, except that you need to also use the 0, and your goal is to get 50. And it takes one more step at the end, but that step should be obvious when you get to it. Then you will still have the extra digit to make 51.

    Can you give me a hint about how you are getting a 5 from the 0 and 8?

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  72. i just got how to get 51, and that is a tricky one.

    i made a mistake in the previos post, i meant to say that i got 5 out of 8 and 2,

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  73. I am 11 years old, and can’t seem to find answers to the 2008 mathematics game. I have 37 answers, and have done everything you could imagine (square roots, factorial, and basic equations). Could you please help me with suggestions to the answers you have gotten?

    Thanks!

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  74. Hi, Callen!

    37 answers is pretty good. I gave this to my class, and I don’t think they got that many. (Well, one boy might have more by now. He took it home, and his older brother got interested in the puzzle.)

    The main thing to do is to try lots of things in various combinations. For instance, many of my answers required taking 2 or 3 factorials in a row, or maybe 2 square roots. At least one answer needs 3 square roots in a row. Also, remember that you are allowed to use decimals. Have you tried doing things with .2 or .08 or anything like that? Decimals can be helpful. Keep experimenting!

    Here is some advice mathmom gave somewhere earlier in this thread:

    What I do is figure out everything I can make using certain subsets of the numbers. What can I make using only (2,0), (2,0,0), (8,0), (8,0,0) — those are my main sets, but I also looked at (2,8 ) and (2,0,8 ) as well, but those aren’t so helpful, because there’s only so far you can get from anything there.

    And n0n4m3 offered this advice:

    This was the first time I played this game and I used techniques similar to what mathmom and Denise described, i.e., check how can I get a number and then try getting the surrounding ones, and what numbers could I get from the combinations of the digits.

    Other thing I used was working backwards: if I want a number x, what is the: sqrt(x), x^2, x +- 2, x * 2, etc. so I can get it using the numbers I have available.

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  75. Hello, I want to ask about if combinations are possible in this game? If so, then I can get 70:

    “(2+0!+0!)C8″
    =(8*7*6*5)/(4*3*2*1)

    Like

  76. Hi, Qi Yanjun!

    I am afraid your formatting did not come through on that at first. (If you know how to use LaTex, it works here.) I edited your comments to leave only the good parts.

    Unfortunately, combinations and permutations are not listed in the allowed functions, so we have to assume they are illegal, unless you can construct them straight from the digits. For example, one expression I have for 56 is the same as 8P6, but in factorial form.

    Edited to add: Oops! I meant 8P2, of course.

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  77. Hi, I teach G&T Math for grades K-6. My 5th and 6th graders have been working on this challenge and want to send in some of their answers to be posted, as the 2007 student solutions are posted now. What’s the procedure?
    Thanks!

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  78. The Math Forum at Drexel University has “hosted” the game and student answers in past years, but they don’t seem to have a page up for it this year, so unfortunately, I don’t know of anywhere where your students can submit their answers. But maybe you could make some posters for the hallway of your school, or create your own local website for their solutions?

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  79. I forgot to mention that the 15 with one factorial without changing the order of the digits uses a 2 digit number, so.. I don’t know if this is allowed or not but since I only saw a mention to the digit order I guess we can use 2 digit numbers.

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  80. No, or at least not a way that I know of. Take 2 decimals (one of them with two digit numbers) and a division to get 15. The tricky part is getting the first decimal which took me a while.

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  81. Yes, I think it’s late enough in the year that we don’t have to worry about giving away homework answers. Give me a few minutes to figure out how to format the LaTex, and I’ll post them…

    You can make 50 with a triple square root, and then 51 is easy:

    50 = \sqrt{\sqrt{\sqrt{.02^{-8}}}}

    59 and 61 rely on a trick that may come in handy next year:

    60 = \sqrt{ \left[ \left( \sqrt{ 8 + 0! } \right) ! \right] ! \div .2 }

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  82. I just handed this out to my middle schoolers on Friday. I told them they have until the end of 2008 to get as many answers as they can. It is meant to be filler for those spare moments that seem to crop up from time to time, or for when their main math teacher is reviewing a concept they’ve “got” and they are given permission to work on something else for a bit. I even listed on the worksheet the ones which are (or which we collectively agreed here were) impossible. But, I’m not too concerned about them googling for it and ending up here.

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  83. I do still get 1-2 searches per day for the 2008 game, on average. But I suppose that any student who reads through all 140+ comments has worked hard enough to earn the hints. And any student who dared to turn in these calculations without getting all the other, simpler ones would give himself away.

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  84. Since my middle schoolers currently don’t really know negative exponents (we did discuss it once in the context of scientific notation, in a previous year), anyone who turns in a solution using them is going to have to convince me that they actually know what it means. ;-) Perhaps some of them will be motivated to learn negative exponents in this context, however. :D

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  85. My son has this assignment for the year 1954. WE have been working on it for weeks. There are 13 numbers that we just can’t seem to find an equation to fit. If you can help us, please do. The numbers are: 51, 58, 66, 67, 75, 76, 78, 79, 82, 83, 84, 85, and 95. We are very frustrated at this point. He is so nervous and he is only in the 5th grade. ANY help would be much appreciated.

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  86. Wow! If your son is in 5th grade and can make that many numbers, he should be very proud. That is a LOT of creative calculating!

    If you do not have to keep the digits in order, then I can see a way to make 82 as 9×9+1. And 58 is 9×6+4. You may be able to make some of the others with similar multi-step calculations. Or it is possible that some of the numbers cannot be made — I know we have found several impossible numbers each year I’ve played the game.

    One thing that I have found helpful in working with the year game is to look at the digits individually (in particular, the factorials of that number), and then start pairing the numbers to see what I can do with two of them together. I don’t try to work with all of them at once or to go for specific numbers right away, but just trying to figure out all the possible combinations.

    But I think the most important thing would be to keep the attitude of playing with the numbers, which is why it’s called the year game. I hate to hear of a 5th grader getting stressed out over math homework.

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