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Thread: ?5?7?7?4? = 34

  1. #1

    Default ?5?7?7?4? = 34

    So we just had the Jewish New Year 5774, and one of my local newsletters has challenged its readers to use the digits in the new year in various combinations to get every integer from 1 onward. For instance:

    5*7/7-4 = 1
    5+7/7-4 = 2
    5*7/7-√4 = 3
    (-5+7)*((7-4)!)= 12
    √((5+7+7))^(√4)) = 19
    √((5*√(7*7))^√4) = 35
    5!-7*7-4! = 47
    et cetera.

    I've so far managed to get every number from 1 to 55 and climbing. But I can't get 34 (and there's no guarantee it's possible).

    Anyone want to help me get 34?
    Last edited by Yerushalmi; 09-09-2013 at 10:57 AM.
    <img src="https://dl.dropboxusercontent.com/u/3451091/dot.gif" :bou::bou::bou::bou::bou::bou:="var e=$(this);var se=$(e).closest('li').prev().find('.postcontent').parent();var te=$(e).parent();var tc=se.html();var th=120;var sh=parseInt(se.height());var r=th/sh;te.html(tc);te.css({'transform-origin':'0% 0%','-webkit-transform-origin':'0% 0%','transform':'scaleY('+r+')','-webkit-transform':'scaleY('+r+')','height':th+'px'});" />

  2. #2
    'Just Friends' Formalhaut's Avatar
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  3. #3
    pirate heartbreaker The Man's Avatar
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    :monster:

    Can you only use each number once? If not then it should be easy. If you can only use each number once, then yeah, it might be impossible.
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  4. #4

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    Not only can you use them once each (because otherwise it's idiotic ), but they need to be used in order as well. Otherwise I could've just done 5*4+7+7.

    34 is trivial to obtain if you can figure out how to do any of the following:
    -Eliminate the 5, the way I eliminated the 4 in obtaining 19, and eliminated a 7 and the 4 in obtaining 35 (see the original post)
    -Turn the two 7s into a 2 or 5
    -Turn the 5 and first 7 into a 3 or a 6
    -Turn the second 7 and the 4 into a 0 or 1
    Last edited by Yerushalmi; 09-09-2013 at 12:01 PM.
    <img src="https://dl.dropboxusercontent.com/u/3451091/dot.gif" :bou::bou::bou::bou::bou::bou:="var e=$(this);var se=$(e).closest('li').prev().find('.postcontent').parent();var te=$(e).parent();var tc=se.html();var th=120;var sh=parseInt(se.height());var r=th/sh;te.html(tc);te.css({'transform-origin':'0% 0%','-webkit-transform-origin':'0% 0%','transform':'scaleY('+r+')','-webkit-transform':'scaleY('+r+')','height':th+'px'});" />

  5. #5
    pirate heartbreaker The Man's Avatar
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    :monster:

    smurf. Yeah, it's much too early in the morning for me to be able to figure any of that out.

    I did think it would be rather silly if you could use each number more than once, but having to use each number in order makes it tricky.
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  6. #6
    ¯\_(ツ)_/¯ Tigmafuzz's Avatar
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    Quote Originally Posted by Yerushalmi View Post
    use the digits in the new year in various combinations
    Judging from your examples, I'm assuming the digits have to stay in order. An easy solution would be 5*7-((7+4)^0) but that's assuming you can add a zero at the end as an exponent. If that's not an option, then the (SPOILER)just as easy solutions would be either:

    ⌈5√7 + 7 * 4⌉ which is a ceiling function of the 5th root of 7 (≈1.476) plus 7 (which gives us 8.476) multiplied by 4 (which gives us 33.9, and the ceiling function automatically rounds that up to the nearest integer, which is 34.)

    or, if you're putting it in a calculator:

    ⌈5 yroot 7 + 7 * 4⌉ which gives the same result, because the ceiling function is cool like that.
    EDIT: if you're putting the previous function in the calculator, you would actually press 7 yroot 5. This function is a different equation entirely that just so happens to also end up with a number between 33 and 34.

    Another weird solution, using the floor function instead, would be:

    ⌊5! / 7 - 7⌋ + 4! which is a floor function of the factorial of 5 (120) divided by 7 (which gives us ≈17.143) minus 7 (leaving us with ≈10.143, which the floor function rounds down to 10) plus the factorial of 4 (which is 24, giving us a total of 34.)

    Math is fun.

    EDIT2: 5+7+7+4x where x=3.75
    lol cheating
    Last edited by Tigmafuzz; 09-09-2013 at 12:37 PM. Reason: I've been awake 28 hours, gimme a break

  7. #7

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    Haha! No, I can't add 0s. If I could, then I'd simply do 5+7+7+4+0!+0!+0!+0!+0!.... etc.

    I like your other solutions! I hadn't known floor and ceiling functions had symbols. I'm a little afraid to use them - I think I'm pushing it just having used factorials - but I'll list it for now, and replace it if I can find anything 'simpler'.

    In the meantime, I'm only missing two of the numbers between 55-69.
    <img src="https://dl.dropboxusercontent.com/u/3451091/dot.gif" :bou::bou::bou::bou::bou::bou:="var e=$(this);var se=$(e).closest('li').prev().find('.postcontent').parent();var te=$(e).parent();var tc=se.html();var th=120;var sh=parseInt(se.height());var r=th/sh;te.html(tc);te.css({'transform-origin':'0% 0%','-webkit-transform-origin':'0% 0%','transform':'scaleY('+r+')','-webkit-transform':'scaleY('+r+')','height':th+'px'});" />

  8. #8
    ¯\_(ツ)_/¯ Tigmafuzz's Avatar
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    What are they? Lemme at 'em.

  9. #9

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    Nah, I want to give myself a fair crack at them first. I only posted 34 up here because it's been almost a week and I'm desperate

    EDIT: Grumble. I thought I had it. I need to turn the last 7 and 4 into a 1 or 0 so I can subtract 1 or 0! from 5*7. I've managed to eliminate the final 7 and 4 by doing 7!/(7-4)! = 7, but that doesn't turn them into a 0, it just erases them. Damn!
    <img src="https://dl.dropboxusercontent.com/u/3451091/dot.gif" :bou::bou::bou::bou::bou::bou:="var e=$(this);var se=$(e).closest('li').prev().find('.postcontent').parent();var te=$(e).parent();var tc=se.html();var th=120;var sh=parseInt(se.height());var r=th/sh;te.html(tc);te.css({'transform-origin':'0% 0%','-webkit-transform-origin':'0% 0%','transform':'scaleY('+r+')','-webkit-transform':'scaleY('+r+')','height':th+'px'});" />

  10. #10
    Mold Anus Old Manus's Avatar
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    It's times like these that confirm to me that I was never cut out for grad school anyway.


    there was a picture here

  11. #11
    ¯\_(ツ)_/¯ Tigmafuzz's Avatar
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    >mfw Americans call onesies-twosies-redsies-bluesies "Mathematics"

  12. #12
    Prinny God Recognized Member Endless's Avatar
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    If I'm allowed letters :
    a7f55651a7320114cb673a3f2672c25f.png

    Edit : another way, modulus ftw.
    5*7 - (7! mod 4!)! = 34
    Last edited by Endless; 09-09-2013 at 11:18 PM.

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  13. #13

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    Quote Originally Posted by Endless View Post
    If I'm allowed letters :
    a7f55651a7320114cb673a3f2672c25f.png

    Edit : another way, modulus ftw.
    5*7 - (7! mod 4!)! = 34
    Oooh! Except instead of letters, using % for modulus would work:
    5*7-(7!%4!)!

    Awesome!
    <img src="https://dl.dropboxusercontent.com/u/3451091/dot.gif" :bou::bou::bou::bou::bou::bou:="var e=$(this);var se=$(e).closest('li').prev().find('.postcontent').parent();var te=$(e).parent();var tc=se.html();var th=120;var sh=parseInt(se.height());var r=th/sh;te.html(tc);te.css({'transform-origin':'0% 0%','-webkit-transform-origin':'0% 0%','transform':'scaleY('+r+')','-webkit-transform':'scaleY('+r+')','height':th+'px'});" />

  14. #14
    Recognized Member Shorty's Avatar
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    I have nothing of value to contribute to this thread, but it is making my head spin.

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    Actual cannibal Pheesh's Avatar
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