2009 Mathematics Game

Posted on by Denise Gaskins

new-year
[Photo by Amanda M Hatfield.]

Have you made a resolution to exercise your mental muscles this year? Then please join us for the 2009 Mathematics Game. Here are the rules:

Use the digits in the year 2009 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, 9 may be used.
  • The decimal point may be used, as in .9, .02, etc.
  • Multi-digit numbers such as 29 or 902 may be used, but preference is given to solutions that avoid them.

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

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

How Does It Work?

Use the comments section below to post a running list of the numbers you have been able to calculate. You may also share relatively cryptic tips and hints, but be warned: Some teachers use this puzzle as a classroom assignment, and there will always be students looking for people to do their work for them.

Do not post your solutions. I will delete them.

I know of no authoritative list of numbers that can be made with each year’s digits, so we will rely on our collective wisdom to decide when the game is done. We had a lively discussion the last two years. I’m looking forward to the fun!

Clarifying the Rules

Unary negatives are allowed. That is, you may use a “-” sign to create a negative number. This is particularly helpful if you are trying to keep the digits in order, using the 2 first and the 9 last.

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, 9.

  • “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, 9.

  • 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!).”

Also, you have to “hit” each number from 1 to 100 exactly — no rounding off or truncating decimals allowed. You may not use the integer function.

For more hints, check out this comment from the 2008 game.

Keeping Track

As the game results are reported below, I will keep a running tally of confirmed results (that is, numbers reported by two or more players) here:

Percent solved = 90%. Wow!
Numbers that remain unsolved =
52, 67-68, 74, 76-78, 86, 97-98.

Depending on how busy life gets, the tally may lag a few days behind the results posted in the comments, so be sure to scroll down and read the latest news.

And if you would like to join me in the “extended edition” game…

  • Still looking for an expression that does NOT use multi-digit numbers:
    57, 84-85, 87-89, and 91-95.
  • Found a way to keep the digits in order: (62%)
    1-36, 38, 40, 42, 44-46, 48, 50, 54, 57, 60, 62-64, 66, 69, 71-73, 75, 80-81, 90, 96, 99-100.

Related Number Puzzles

Check out Happy 41*7^2, What is interesting about the number 2009?, and The number 2009.

You may also enjoy the Famous Four 4’s. Or have a look at All the King’s Digits, or try Jonathan’s 4, 3, 2, 1 Puzzle.

This post is featured in the Carnival of Homeschooling: Week 158.

135 thoughts on “2009 Mathematics Game”

  1. So far I’ve got 1 – 30, 32, 33, 35, 36, 38, 40, 43 – 49, 53 – 55, 59 – 61, 63 – 65, 70 – 72, 79 – 83, 87 – 93, and 96.

    Reply

  2. 1 – 22 were fast. 23, 24. Tricks for 25. Different tricks for 26 – 29. 30.

    33 – 41. 43 – 51. 53 – 55. 58 – 62
    63 – 65. 72
    79 – 83. 84. 87 – 93. 96. 99. 100.

    Remains:
    31, 42, 52, 56-7, 66-69, 73-8, 85-6, 94-5, 97-8 (both)
    32, 70-1 (me)
    34, 37, 39, 50, 51, 58, 62, 84, 99, 100 (MH)

    Jonathan

    Reply

  3. It looks like 2009 is going to be a bit easier than the last two years. (I hate to think about next year’s game!) I can confirm all the numbers you guys found, making our percentage so far = 79%.

    I also found 57, 85, 94, and 95, but only by using 2-digit numbers. Since you guys didn’t report 94 and 95, can I assume you found a different way to do 91-93 that didn’t use the 2-digit trick?

    99 was a real bear for me, but I was able to use the same trick to get 66.

    Oh, and I was able to get 69 and 73, too. I used a slight variation of the trick for 70-72.

    Reply

  4. I am up to
    1 – 51, 53 – 55, 57 – 61, 63 – 65,
    72,
    79 – 84, 87 – 94, 96, 99 – 100

    Of those, I used double digits for
    23, 24, 26, 27
    57,58, 61, 84, 87 – 94, 96

    Reply

  6. Hello! I’m back again.

    Well, I do agree that the 2009 game is easier than last year, since we can make better use of the digit 9.

    I have managed to get 75 and 58-62, using techniques mentioned last year (without using multiple digits). 56 was easy too. Have anyone else gotten them too?

    Can I ask how is it possible to get 84-89 and 91-95 without using multiple digits?

    Reply

  7. I “played” with McRib this afternoon. We are around 84%.

    Add 75, (sort of a doubled trick, likely unique) 56, 85, 95.

    I am missing 52, 62, 66 – 78 (exc 72 and 75), 86, 97, 98.

    Have any of these?

    Jonathan

    Reply

  8. Why let 0^0 =1? Seems like you could use 0!+0=1, and not mess with something that really ought to stay undefined (since x^0=1 and 0^x=0 for all x except 0).

    Reply

  9. Hi, Sue! We allow 0^0 to stay consistent with the Math Forum rules, for the sake of students who may be playing. As far as I know, however, no one actually uses it. A zero is too valuable to waste that way, when 0! can give the same value with only one digit.

    The last of our holiday company just drove up, so I’ll be out of commission until sometime next week. But before I go, I’ll update the list here:
    Confirmed numbers = 1-51, 53-65, 70-72, 75, 79-85, 87-96, 99 & 100.
    Still missing (or only reported by one player) = 52, 66-69, 73-74, 76-78, 86, 97-98.
    % solved so far = 87%.

    I think I have solutions for 66, 69, and 73. But since no one has confirmed those, I will have to recheck my calculations after our company leaves…

    Reply

  10. I have found a way to keep the digits in order for the following numbers:
    16,20,26 and 32.
    And also 41 without the use of multi-digit numbers (with 5! as a starting point).
    66 is also on my list (it uses a power of 2 and no multi-digit numbers).

    Reply

  11. I’m doing this with my class as a welcome back/warm-up activity. So far, personally, I’ve got:
    1-51, 53-5, 59-61, 63-66, 70-72, 79-84, 87-94, 96, and 99-100. Only 19 left!

    Reply

  12. Ok, I’ve now caught up with the group, except I’m missing 56…but I can add 66 to the list, I used a similar trick to the one for 62.

    Reply
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  16. Corrie, that was the way I did 66. The easier method eluded me until you and Nth_X mentioned it. Funny how brains work!

    Nth-X, 56 was one of the last numbers I figured out, too. But as Qi Yanjun said, it is actually pretty easy, once you see it.

    Update: The numbers I am still missing are 52, 67-68, 74, 76-78, 86, and 97-98. Does anyone have any of those?

    Also, I am still using 2-digit expressions for 57, 84-85, 87-89, and 91-95. Has anyone found a trick for those without resorting to 2-digit numbers?

    Reply

  17. Oh, and if anyone else is trying to get the digits in order, I have made it up to 36 so far, plus an assortment of higher numbers. Corrie and Nth_X’s easy method for 66 helped me fill in a few of those.

    Reply

  18. Thanks Qi, I saw your comment and had the solution leap into my head practically fully formed 🙂

    Hmmm, I was already using the 2007 “trick” for 70 and 72…now I’m off to adapt it for 69 and 73.

    My students turn in their results tomorrow, so I’m looking forward to seeing how they did with the game.

    Reply

  19. ok, now using the 2007 trick and the unary negative mentioned under your rules clarification, I was able to get 69, 71, and 73 with the digits in order 🙂

    Reply

  20. A friend emailed me a few other “in order” solutions, using 2-digit numbers. (He even found a way to incorporate .9, which I knew ought to be useful for something, but I hadn’t yet figured out what.) With your entries, that brings us up to 58% and counting…

    Reply

  21. Can anyone hint me for 57. I don’t know why but i must have tried everything. I gathered from other comments that it requires 2 digit numbers but i’m still failing.

    Reply

  22. Nickserb: For 57, get 3 and 19, then multiply them together.

    Could someone give me a small hint for 52 without using two-digit numbers? Thanks!

    Anyways, the .2th root of a number is the same as taking the number to the fifth power. So you can get 32 using 2, 0, and 0 with that, which adds some more numbers in the “digits in order” section (41 and 96 and maybe some others).

    I can’t find anything else, has anyone tried dividing or multiplying by .02 or .09?

    Reply

  23. 52 is one of the numbers that no one has reported solved yet, with or without 2-digit numbers. Did you find a way to do it, Hannah?

    Hm, I never thought of using a .2th root. Can we do a decimal root like that? I’ve heard of fractional exponents, but not fractional roots.

    As for 2-digit decimals, I did use one in solving 59. But then I found another way to get it that used the normal numbers and operations.

    Reply

  24. Oh, sorry, I didn’t know no one got 52, I didn’t get it either 😦

    I don’t know, I think decimal roots make sense, they’re just fractional roots that happen to be less than one in this case.

    Third root of x = 3√x = x^(1/3)
    0.2nd root of x = 0.2√x = (1/5)√x = x^5 (since 5 is the inverse reciprocal of 1/5)

    I don’t know if it’s proper use, though 😦

    Reply

  25. While we are throwing out ideas here….

    [√(0!+0!)-.2bar] = 4/3

    I don’t think it gives us anything, but still, kind of neat.

    Jonathan

    Reply

  26. That is neat, Jonathan. The rules don’t say we can use the bar, but it’s fun to play around with.

    Nth_X, can you confirm that you did get 71 with the digits in order? I haven’t figured that one out, yet. But I did get 72, and I found a way to do 60 and 80 in order without using 2-digit numbers.

    Reply

  27. 71 is the second root (that uses the two) of [a certain number, which is conveniently just 1+ a certain factorial—ed.].

    [from Denise: Sorry for the edit, Hannah, but you can’t give answers here, only semi-cryptic hints. But thanks for the hint! I should have figured it out myself, since I use that root in other calculations. BTW, you don’t have to use the 2 with the root symbol; it can also go in as part of the “1”.]

    Reply

  28. I wrote an algorithm to solve this problem: I generate recursively valid expressions that I evaluate.
    After trying all the combination I also get the same 90%.
    If someone is interested I can give my code or post all the results.

    Laurent

    Reply

  33. Jennifer: 32 can be reached in at least two ways — one using addition, and one using a power of 2.

    Laurent: Please do not post your results, at least until after the Math Forum starts posting student entries in February. Some teachers use this as a homework problem, and we don’t want to make life too easy for the kiddies!

    I just dropped by to add 46 to the “in order” list, and I see I’m running behind. A few on my “in order” list have to use 2-digit numbers: 38 (thanks to Mr. Zahman!), 44, and 100. Has anyone found a way to make those, keeping the digits in order, without using double digits?

    Reply

  34. To summarize, I have the same 90% of the digits and 79% without the multi-digit.

    For the ordered number, I have a few digits missing that you have on your list. But I have 81 (I am sure).

    I don’t have 38, 44 and 100 in order and with one digit either.

    Reply

  44. There are two ways to get 66: directly, as 11×6; or indirectly, as 64+2. The latter is easier, requiring only a power of 2.

    To get 40 with the digits you mentioned, you will need the equivalent of 8×5. You can do that by using a decimal point with one of the digits.

    Reply

  45. I got 1-33, 35-6, 39-41, 43-51, 53-55, 58-61, 63-65, 79-83, 88-94, and 100

    This is my first year doing this puzzle and I was wondering if you guys could give some hints on the puzzle. I don’t know any of the tricks from the previous years either.

    Reply

  50. That’s my fault. I’m too lazy to put in the link manually, so I posted the comment and then used WordPress.com’s editing page to add the link. But my browser is glacially slow tonight, so it took several minutes instead of just a few seconds.

    Reply

  52. It would depend: Is the 8th grader obsessed with math puzzles, or is he just doing this for a class assignment? And does he have experience with math contest puzzles? I would not expect a “normal” 8th-grader to get any of the more complicated answers on his own — for instance, the numbers that depend on first getting 71. On the other hand, none of the solutions are too difficult for a determined 8th grader who reads the hints and enjoys playing with numbers.

    I am giving this puzzle to my math classes Friday, and we’ll see how they do. Any other teachers want to chime in with your experience?

    Reply

  53. Two questions:
    – Can you use a digit more than once ? say use 2 twice ?
    – Is division integer division (so 9/2=4 ?)

    Reply

  54. You may use a digit as many times as it is present in the year. So you may only use 2 once, but you have two 0’s to work with.
    Division uses the standard definition, the inverse of multiplication: 9/2=4.5 — no truncating.

    Reply

  58. I need 69, 73, and 75…
    I don’t get how you can tweak the below method to get 69 and 73…

    To rgdh: Here’s a hint for 71: Square root of (factorial + 1).

    Reply

  59. I’m using the game with 7th graders. They have come up with expressions for almost 80% of the numbers, arranging the 2, 0, 0, and 9 in any order. I’m impressed with their interest and perseverance.

    Reply

  60. K: The hint here is as far as I am willing to go this early in the year. You have to play with the numbers for awhile.

    rgdh and Krvan: For 69 and 73, first, you need to figure out how to get 71 in general. The hint should help you with that. Then the challenge is to figure out how to get the same thing without using the 2, so you can have it left over to add or subtract.

    75 is a different animal, and as far as I know can only be calculated as 3×25. I found two (nearly identical) ways to do that, both of which involved a power of a decimal.

    Reply

  62. My 5 classes have been working on this since Jan. 5, our first day back from holiday break. We ended our competition (class against class) today, although we do not have expressions for all of the numbers. The top class has 83 expressions! Some classes found expressions for numbers that other classes had not found. Overall, there are 13 numbers that no one has found: 52, 62, 66-69, 71, 73, 74, 77, 78, 97, and 98.

    Reply

  65. K: To get 39, I know one way is to get 40, then -1. Well, thats how I did it. To get 66, you have to get 64, and then add 2.
    HINT: For 66, it does not matter whether you use 2 or 0!+0!.

    Reply

  67. IW: Your classes are doing fantastic! I would also like a hint to help with 76 and 86. They are still missing on my list.

    Krvan: By “power of a decimal,” I meant that the decimal is the base.

    Reply

  68. I just got 75 in a whole different way…
    I used 3, then divided that by a decimal to a power. Which makes an even smaller decimal.

    Reply

  69. I finally figured out a way to get 81 with the digits in order and using only single digits. It was actually so obvious once I did it (hint: 81 = 3^4), but it took me a long time to find.

    Reply

  70. When you say you use the digits in order, does it mean you use 2, then 0, then 0, then 9, or can they be in 9, 0, 0, 2 order?

    Reply

  72. Hint for 86: (the 2,0,0,9 won’t be used in order) 81+5, for the 5 use a division problem involving a decimal

    Hint for 76: Read the hint for 86 and make a small change

    Reply

  74. Hannah: Wow! I’ve used powers for a few other numbers, but I didn’t even think of using a power of 3 to put 81 in order. I think my brain is turning to mush.

    Krvan: “In order” means 2, 0, 0, 9, just as they appear in the year.

    Sam: I used 90+5=95, but getting the five is a bit tricky. Not as difficult as getting 75, though. And no, you cannot use permutations, combinations, trig, or any other functions that aren’t explicitly allowed in the rules. [But you can manually create the equivalent of a permutation to get 56.]

    IW: Hmmm. I’m going to have to mull over your hints. At first glance, I can’t see any way to calculate 81+5 without at least one extra digit.

    Reply

  75. I think I understand where IW is going with 86. You want to do ^(.5^-1) = ^2, but I don’t know if there is a correct way to write that with ‘sqrt’.

    Reply

  76. I think you may have mistyped something, Corrie. Did you mean that “+.2^(-1)” gives you “+5”? But that uses up two of the digits, and I cannot see how to make 81 with the remaining 0 and 9.

    Reply

  77. Oops! I have to retract my hint. We did not find expressions for 86 and 76 (well we thought we did, but we used too many 2’s).

    Reply

  78. No, what I meant was the following. You want to get 81 with only the 9 and one 0 (so that you can use the 2 and the other 0 to get 5). And I would like to do 9^(.5^-1). The 1 would be written as 0! and the ^.5 as sqrt. But again, I don’t think there is a correct notation with sqrt instead of ^.5.

    Reply

  79. I see. I haven’t ever heard of a way to do something like that — but there’s plenty I don’t know! If anyone else knows a way to make it work, feel free to chime in…

    Reply

  83. Sam: You can use the square root symbol (the radical sign) without using up any of your digits, but you cannot use “^1/2” unless you create it using a 0! and the 2.

    Also, the numbers you listed are the ones we are all missing. They may be impossible.

    Reply
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  85. Puzzled a bit for myself until more and more gaps appeared in my list, then wrote a program, which was quite a puzzle in itself. 🙂
    Solved 90 / 100, and got the same list of unsolved numbers.

    Also, 95 is possible without using multi-digit numbers.

    Reply

  90. Snow day last week, but my classes played the 2009 Game this Friday. The middle school class (mostly 6-7th graders) got 46 numbers in about 45 minutes. The high school class (mostly 9-10th graders) worked more efficiently and came up with 74 answers before they ran out of time.

    My biggest surprise was that the middle school class made no use of any expression for 6, even though I gave them (2+0!)! in our introductory discussion. The high school boys, likewise, did not take advantage of 4! until more than halfway through the class, when I specifically suggested they give it a try.

    The middle school kids beat the high schoolers on one number: They found an expression for 41. The older kids couldn’t get it even after I told them the other class had and hinted that it involved division.

    I don’t know whether any of the kids will continue this puzzle on their own time. If they do, I’ll post an update to their totals.

    Reply
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  92. Your classes found a lot of expressions in 45 minutes. Each of my classes worked for about 45 minutes the first day and then we added to each class list over the next several days. In one class, one 7th grade boy was especially good at finding the more complex expressions–using an expression for an exponent, using 4!, using .2 as a divisor. But I was also pleased with the effort put forth by all students. Many times a student found an expression for an available number and they were so pleased with themselves. The 2009 Game really boosted many students’ attitudes about their own math abilities. …and we are still looking for a few more numbers.

    Reply

  93. I’ve been working on the extended edition, trying to write 17 (for example) with the digits in order. Is there a way to do it without using a multi-digit number (20)?

    Reply

  94. For 17, I was able to write the digits in order by using an expression for 11+6. But it’s pretty complicated: one decimal, four factorials, two square roots, and a unary negative sign.

    Reply

  98. Nathan: Yes, negative numbers are allowed. And yes, this puzzle can be quite infuriating. But just think of how bad it will be next year…

    anies: Those are some tough numbers. Start with the hint here.

    Reply

  99. i am having real trouble with 39 41 43 51 52 56 62 66 67 73 74 75 76 77 78 85 86 95 97 98 I HAVEsolved the rest but have spent ages on these but cant seem to solve them can any one help

    Reply

  101. For several of those numbers, playing around with decimals like .2 or .02 will help. Also, check the list above — many of those numbers are on our “probably impossible” list.

    Can you tell us how you got 68?

    Reply

  105. If you use a decimal .2, that is the same as 2/10 or 1/5. If you want to use a 5, you need the reciprocal of that fraction, so how can you use the 0 to make a reciprocal? I can think of at least two ways to do it…

    Reply

  110. which of the following that are possible can you get with out using decimals 41, 43 ,51, 52, 56, 62 67 ,68, 74, 75 ,76 ,77, 78 ,85 ,86, 95, 97 ,98

    Reply

  111. To find 64, first determine what power of 2 it is, then figure out a way to get that exponent using the 9.

    Please read through the article (which includes a link to hints from last year) and the comments above for hints. Many of the numbers on your list do NOT have answers.

    Reply

  116. You do not need an 8 to get either 64 or 66. You can get 64 with an 8 (made from the 9 and one 0), if you want, but you can also get it as a power of 2. No more hints on this one — figure it out!

    Reply

  118. If you are trying to get 66 by adding 2 to something, then you need to first get 64, don’t you?

    By the way, 66 = 11 x 6, too, if you’d rather do it the hard way.

    Reply

    1. This is not a help forum. If you want instant answers to your questions, you need to be asking them at a place designed for that sort of interaction. Several online help forums are listed on the Resources page.

      And when I say, “Think about it for awhile,” I mean that you need to get out some scratch paper and a pencil. Write down lots of different ways to combine the numbers. Try factorials and square roots and multiplying and dividing and everything else you can think of. How many different numbers can you make using the 9 only? Little numbers AND big numbers — you don’t know what will come in handy! How many different numbers can you make using the 9 and a 0? 2 and 0? etc. Then combine these options in different ways.

      Don’t just ask for us to give you the answer. It’s much more fun to find it for yourself!

      Reply

  125. For 41, I used an expression to make 40 that included dividing by a decimal. Then add 1. Remember that dividing by a decimal has the effect of multiplying by a larger number. Remember that you can use the sqrt function and factorials. You will find 0! to be very helpful. 0! is defined as 1.

    Reply

    1. It is a way of showing, when you don’t have formatting, that whatever comes next is an exponent. If you are working in a word processor, you could just use superscript, but in email or on forums the ^ symbol works.

      Reply

  129. I have to confess ignorance here, James. I tried Wolfram|Alpha to look up what the H function might be, but WA didn’t recognize it either. At any rate, this isn’t among the elementary operations that were allowed in the rules, so I don’t think we can count it for this game.

    Reply

  130. how do you get to the number 53 using 2,0,0,9???????????????????????????????????????????????????
    Helllllllppppppp!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    Reply
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