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Category: | Math |

Submitted By: | javaguru |

Fun: | (2.42) |

Difficulty: | (3.05) |

In "4+4+4+4=1!" it was shown that just using addition, subtraction, multiplication, and division, you can form all the numbers from 0 to 9 using exactly four 4's.

For example: 7 = 4 + 4 - 4/4

Now if you add the operations square root, factorial, and exponent, what is the first number counting up from zero that can't be formed? How can each number be formed?

You may use parentheses, but you may not combine fours to make 44 or 444.

(An exponent is raising a number to the power of another number such as 4^4 = 256; a factorial is the product of all the integers from 1 to the integer such as 4! = 1x2x3x4 = 24.)

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Posted by fame | 01/17/09 |

1st comment! :) | |

Posted by Nerine | 01/18/09 |

(2nd comment) I'm not good at any sort of Maths, it's very confusing, but cool all the same. :-? :P | |

Posted by Zag24 | 01/19/09 |

I can make 39. (But only if you allow infinite progressions.) :D 4! + 4*4 - sqrt(sqrt(sqrt(sqrt(... sqrt(4) ...)))) After taking the square root an infinite number of times, you are left with the number 1. 24 + 16 - 1 = 39 | |

Posted by Elly_Bellsy | 01/23/09 |

thats really confusing!! :roll: | |

Posted by javaguru | 01/23/09 |

You can also make 39 if the multifactorial operators are allowed: 4!! * 4!! + (4!! / SQRT(4)) = 39 4!! is the double factorial of 4, which is 4 x 2 = 6. So 6 * 6 + (6 / 2) = 39 8) | |

Posted by javaguru | 01/23/09 |

Wow, really lame comment for my last one since 4 x 2 = 8. :oops: You can still make 39 (and probably most if not all of the rest of the numbers up to 100 or so) with multifactorials included. (4!)!!!!!!!!!!! / (4! - 4 - 4) = 39 24!!!!!!!!!!! / (24 - 4 - 4) = 39 624 / 16 = 39 A multifactorial is the product of the number and every nth number counting down from the number where n is the order of the multifactorial. The order is determined by the number of exclamation marks, so 24!!!!!!!!!!! (11 !s) is 24 x (24-11) x (24-2x11) which is 24 x 13 x 2 = 624. You can see how this allows you to create a wide variety of numbers with a single 4, and an even wider variety with four 4s. | |

Posted by stil | 01/26/09 |

Multifactorials are as different from factorials as horses are from donkeys; they represent a side excursion not invited by the question. | |

Posted by stil | 01/26/09 |

Which is to say if you start using/creating special "cases" of the allowed operations you could have an "infinite" answer. | |

Posted by javaguru | 01/26/09 |

Um, yeah, multifactorials ARE different, which is why I didn't allow them in the problem. I was pointing out that with them included the solution space expands to the point of trivializing the problem. My comment was not intended to suggest that there is a solution for 39. | |

Posted by bogdano88 | 01/09/13 |

You must correct the formulation of the problem, so you can write numbers using 4/4 only once. 4/4 is one unit, and you can add it with oneself to obtain any number. | |

Posted by Jimbo | 12/27/13 |

Well if you add exponent then you should be able to add the inverse operation to exponentiation and that is taking the logarithm. If that is allowed, there is a general solution to the four 4s problem. :D | |

Posted by Wimm | 03/14/15 |

Very well done!! I had then all except 1 (33), but it took me a lot of time. |

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