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re: Someone explain this math brain teaser to me
Posted on 5/19/19 at 1:32 pm to fr33manator
Posted on 5/19/19 at 1:32 pm to fr33manator
I punch you in the face and take all of your doughnuts.
I'm homeless and hungry.
I'm homeless and hungry.
1
Posted on 5/19/19 at 1:34 pm to castorinho
43
Posted on 5/19/19 at 1:35 pm to ThatMakesSense
dude was confused. relax.
Posted on 5/19/19 at 1:37 pm to ThatMakesSense
Too far
This post was edited on 5/19/19 at 6:11 pm
Posted on 5/19/19 at 1:38 pm to castorinho
I finally figured it out.
Posted on 5/19/19 at 1:38 pm to castorinho
So you can buy a 6 box, 9 box or 20 box of donuts.
So the greatest number I can not purchase.
20 + 9 + 6 =35
If you buy another box of 9 that would be 44. 44-1 = 43.
That's the only way I can come up with a 43, but is the question poorly worded?
So the greatest number I can not purchase.
20 + 9 + 6 =35
If you buy another box of 9 that would be 44. 44-1 = 43.
That's the only way I can come up with a 43, but is the question poorly worded?
Posted on 5/19/19 at 1:40 pm to castorinho
McNugget Problem UGA.edu
Similar problem, but for McNuggets.
From site:
"For any desired number if it is divisible by 3 it can
easily be made with 6 and 9 packs, except if the
number is 3 itself. If you can't use all six packs then
use one 9 pack and you can do the rest with six packs.
If the number is not divisible by 3 then use one 20
pack. If the remaining number is divisible by 3 then
use the above method for the rest.
If the number still isn't divisible by 3 use a second 20
pack. The remainder must be divisible by 3, in which
case use the 6 and 9 packs as above.
The largest impossible number would be such that
you would have to subtract 20 twice to get a
remainder divisible by 3. However, you can't make 3
itself with 6 and 9 packs. So the largest impossible
number is 2*20+3=43."
Similar problem, but for McNuggets.
From site:
"For any desired number if it is divisible by 3 it can
easily be made with 6 and 9 packs, except if the
number is 3 itself. If you can't use all six packs then
use one 9 pack and you can do the rest with six packs.
If the number is not divisible by 3 then use one 20
pack. If the remaining number is divisible by 3 then
use the above method for the rest.
If the number still isn't divisible by 3 use a second 20
pack. The remainder must be divisible by 3, in which
case use the 6 and 9 packs as above.
The largest impossible number would be such that
you would have to subtract 20 twice to get a
remainder divisible by 3. However, you can't make 3
itself with 6 and 9 packs. So the largest impossible
number is 2*20+3=43."
Posted on 5/19/19 at 2:04 pm to castorinho
I can by 43...maybe not exactly 43...but 2 boxes of 20 and 1 box of 6
Throw away 3
Throw away 3
This post was edited on 5/19/19 at 2:08 pm
Posted on 5/19/19 at 2:09 pm to castorinho
Okay, think of it this way. Once you have 6 numbers in a row that you can add up using nothing but multiples of 6, 9, and 20, you can make any number higher than that by just adding boxes of 6 doughnuts as necessary to those 6 numbers. We know we can make at least 35 doughnuts using 1 box of each, so start there.
36 is 6 boxes of 6
Can't make 37
38 is 2 boxes of 9 and 1 of 20
39 is 3 boxes of 9 and 2 of 6
40 is 2 boxes of 20
41 is 1 box of 9, 2 of 6 and 1 of 20
42 is 4 boxes of 9, 1 of 6
can't make 43
44 is 4 of 6, 1 of 20
45 is 5 of 9
46 is 2 of 20, 1 of 6
47 is 3 of 9, 1 of 20
48 is 8 of 6
49 is 2 of 20, 1 of 9
That's 6 in a row and we can add multiples of 6 to all of those numbers to continue the pattern, and we're done.
It's 43.
ETA: Why the frick do these threads happen on Sundays so often? Save that shite for after lunch on Mondays, and you'll get a WAY more entertaining thread.
36 is 6 boxes of 6
Can't make 37
38 is 2 boxes of 9 and 1 of 20
39 is 3 boxes of 9 and 2 of 6
40 is 2 boxes of 20
41 is 1 box of 9, 2 of 6 and 1 of 20
42 is 4 boxes of 9, 1 of 6
can't make 43
44 is 4 of 6, 1 of 20
45 is 5 of 9
46 is 2 of 20, 1 of 6
47 is 3 of 9, 1 of 20
48 is 8 of 6
49 is 2 of 20, 1 of 9
That's 6 in a row and we can add multiples of 6 to all of those numbers to continue the pattern, and we're done.
It's 43.
ETA: Why the frick do these threads happen on Sundays so often? Save that shite for after lunch on Mondays, and you'll get a WAY more entertaining thread.
This post was edited on 5/19/19 at 2:17 pm
Posted on 5/19/19 at 2:12 pm to castorinho
Thank you for flying Delta.
Posted on 5/19/19 at 2:47 pm to fr33manator
Sometimes you go too far. This thread is an example of that.
Posted on 5/19/19 at 4:08 pm to castorinho
quote:
Suppose you can purchase doughnuts in boxes of 6, 9 and 20.what is the greatest number of donuts you cannot purchase.
So the real question is, what is the largest number not divisible by 6, 9, or 20. There are an infinite set of numbers that meet this criterion.
Posted on 5/19/19 at 4:12 pm to Diver Diva
quote:
So the real question is, what is the largest number not divisible by 6, 9, or 20. There are an infinite set of numbers that meet this criterion.
It's 43, woman.
Posted on 5/19/19 at 4:16 pm to castorinho
Easy, how many donuts to they have? order one more than that.
Posted on 5/19/19 at 4:17 pm to ThatMakesSense
quote:
It's 43, woman.
So why is 101 incorrect? Or 121? Or 141?
Posted on 5/19/19 at 4:22 pm to castorinho
how much money do I have
what kind of donuts
what if I want Kolaches
what kind of donuts
what if I want Kolaches
Posted on 5/19/19 at 4:26 pm to castorinho
quote:The largest box is 20, so you cannot purchase 21 donuts in 1 box. Next.
Suppose you can purchase doughnuts in boxes of 6, 9 and 20
Posted on 5/19/19 at 4:31 pm to Diver Diva
quote:
So why is 101 incorrect? Or 121? Or 141
101 = 7x9 , 3x6, 1x20?
ETA : Silly, women.
This post was edited on 5/19/19 at 4:33 pm
Posted on 5/19/19 at 4:34 pm to dawgsjw
The largest box is 20, so you cannot purchase 21 donuts in 1 box. Next.
What about 22?
What about 22?
Posted on 5/19/19 at 4:36 pm to Diver Diva
quote:
So why is 101 incorrect? Or 121? Or 141?
Because you can buy 4,5 or 6 boxes of 20 for each of these numbers respectively, then add 2 boxes of 6 and 1 box of 9.
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