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Puzzles: 4, 3, 2, 1 and maybe (())

June 7, 2007 am30 2:58 am

The Four fours puzzle is famous. I learned the annual variation (edition 2007) from Denise, and that was fun.

Every once in a while I try another variation, and here’s todays. Use 4, 3, 2, 1 in exactly that order, and combine them with +, -, * and /, not necessarily in that order. (operations may be repeated, parentheses may be inserted). How many distinct numbers can you create?

Before you start, quiz yourself:

  • How many do you expect to get?
  • What percentage do you expect to be negative?
  • What percentage do you expect not to be integers?
  • And of the integers, what percent do you expect to be even?

Use the comments section for your guesses, but click here for discussing the actual answers.

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20 Comments leave one →
  1. June 8, 2007 pm30 5:13 pm 5:13 pm

    Wild guesses:
    * I expect to make a surprising number of numbers, but not as many as in the 2007 game, since you didn’t allow powers or factorials and since the order of the digits has to stay the same.
    * Assuming that I can put a negative sign in front of the first number as well as in between numbers, I would expect to be able to make about the same number of positives and negatives.
    * I expect more than half of the numbers to be even, because when I get past all that I can do with adding and subtracting, multiplication by the 2 or 4 will give evens.
    * And I didn’t even think about the fact that I could make fractions. I guess I still had the 2007 game stuck in my mind. Hmm…
    Okay, time to get to work and see how it goes!

  2. June 9, 2007 am30 7:33 am 7:33 am

    I will guess (I have not tried this yet)
    About 200 distinct numbers.
    40% negative (I assume that I can’t put a minus sign in front of the first, accounting for the difference between Denise and myself)
    10% integers
    70% of the integers will be even (same reasoning, but I think the effect will be greater)

    Now I need to go try this, too.

  3. mathercize permalink
    June 15, 2007 am30 5:42 am 5:42 am

    New reader.. like the blog.

    Even if you don’t allow “-4”, using parantheses and a negative, you can still get other numbers to be negative and multiplied by using parantheses (for instance 4(-3)+2 + 1 is, of course, different from 4-3+2+1. So, I’m thinking for my calculation, I’ll exclude using the “-” as a negative as stated above and not use () for multiplication (unless there is a multiplication symbol placed between two paranetheses).

    If I’m thinking of it right (while my calc students are taking an exam), There are 3 spots for an operation 4 ? 3 ? 2 ? 1, since we’re allowing duplicates of operations, so we have 4^3 =64 possible arrangements of just operations.

    I ‘m a little less sure on the parantheses, but here’s my preliminary look, ignore operations for a moment. Also ignore the trivial cases of parantheses.
    : 0 ()’s renders just 1 arrangement
    : 1 set of () renders 5 arrangements – (432)1, 4(321), (43)21, 4(32)1, and 43(21)
    : 2 sets of () also renders 5 arrangements – (43)(21), ((43)2)1, (4(32))1, 4((32)1), and 4(3(21)).
    :3 sets of () contain no new non-trivial cases.
    That leaves a total of 11 sets of parantheses arrangements.

    So, combining parantheses and operations, we have 64*11 = 704 possible arrangements. So, at most there will be 704 different solutions. Now, to check for (multiple, I’m sure) duplicates, I think I could set-up a spreadsheet for the parantheses and operations, then arrange all of the solutions in a frequency table (which could be problematic with fractional answers) or just remove the duplicates somehow…

  4. June 15, 2007 am30 7:29 am 7:29 am

    Why are you distinguishing (432)1 from ((43)2)1 ?

  5. mathercize permalink
    June 15, 2007 am30 8:10 am 8:10 am

    >> Why are you distinguishing (432)1 from ((43)2)1 ?

    I forgot to mention my original assumption, that being order of operations. If there is addition/subtraction between 4 and 3 and either multiplication or division between 3 and 2, this would matter.

    I was looking for a maximum number of different solutions and with my assumptions, I think 704 is the max.

    I was really curious how you came up with your % guesses (integers, negatives, etc). Was this an educated guess from puzzle experience or just some number sense…?

  6. June 15, 2007 am30 9:16 am 9:16 am

    My guesses are just that, guesses.

    But back to those parentheses. Assume for a moment that our operations are subtraction, division, and addition, in that order. We get:

    ((4-3)/2)+1
    (4-3)/(2+1)
    (4-(3/2))+1
    4-((3/2)+1)
    4-(3/(2+1))

    skip this which repeats the 2nd:
    (4-3)/(2+1)

    Principle: I used parens to establish order of operations.
    -/+
    -+/
    /-+
    /+-
    +/-
    (+-/ is really a repeat)

    So that’s 5 parenthesizations for each group of operations (max), or 5*64 = 320 maximum possible values

  7. December 9, 2007 pm31 9:51 pm 9:51 pm

    ok does anyone know where i can get some anwsers. Im 13 and doing this as a project for school oonn maths. I have got up to 356 but am now quite stuck if anyone can help post a comment back on this blog.

  8. Hatty permalink
    January 2, 2008 pm31 7:04 pm 7:04 pm

    Help me!!!!!!
    Numbers 73 to 87 are really puzzling me!!!! can anyone help?????

  9. morgan permalink
    February 11, 2008 am29 5:49 am 5:49 am

    Can anyone help me? I am working on this project but can not figure it out. I have 1-10, and other random numbers. I have looked at other sites but can not figure out what they are saying. If you can help, please e-mail me at masadler90@yahoo.com

  10. mikaela permalink
    April 4, 2008 am30 4:29 am 4:29 am

    answers please!!!!!!!!! my daughter really needs fifth grade answers

  11. September 16, 2008 am30 8:27 am 8:27 am

    go to, http://www.wheels.org/math/44s.html and it will give you the answers

  12. January 2, 2009 pm31 8:44 pm 8:44 pm

    Here’s another variation:
    Use three 9’s to make the integers from 1 through 12.
    They can then be used as numbers on a clock. See:
    http://www.mrlsmath.com/uncategorized/triple-nine-wall-clock-student-math-challenge/
    for a picture of this Triple Nine Wall Clock from cafepress.
    Cordially,
    Bill Lombard (Mr. L)

  13. Anonymous permalink
    January 7, 2009 am31 5:02 am 5:02 am

    Hi I need 43,but without using 44, otherwise the answer would be obvious

  14. Anonymous permalink
    January 7, 2009 am31 5:05 am 5:05 am

    kj, whats 43, but not using 44???? I am not allowed to use 44

  15. Anonymous permalink
    January 7, 2009 am31 5:06 am 5:06 am

    Ahhh i need this fast!

  16. Anonymous permalink
    September 2, 2009 pm30 8:06 pm 8:06 pm

    Easy i know how to get 324 numbers

Trackbacks

  1. Solutions: 4, 3, 2, 1, and maybe (()) « JD2718
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