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Noisician
Harsh electronic purity

Registered: Aug 2001
Location:
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| quote: | Originally posted by A_M_8
I got one that took me a while to solve.......
You have 9 machines (numbered one through nine) that produce screws. However, one of them is producing screws that are .01 grams overweight (a difference that you couldnt possibly recognize without an accurate scale)...... Now, you are given a very accurate scale, however, you can only use the scale once (only one load can be placed on the scale and it can't be touched/changed)....... How do you find out which machine is the one producing the overweight screws? |
that's a good one. here's what i think:
u need to take 1 screw produced by the 1st machine, 2 screws produced by the 2nd machine, ... , 9 screws produced by the 9th machine and put all of them on the scale. now, if all the machines produced the same screws, the ideal total weight would be (45x) grams, where x is the weight of one screw. by comparing the actual weight to the ideal one, u can figure out which machine produces defective screws because, in this case, the actual weight will vary, depending on which machine contributed to the overweight.
if the actual weight is (45x+0.01) grams - it's machine 1
if the actual weight is (45x+0.02) grams - it's machine 2
if the actual weight is (45x+0.03) grams - it's machine 3
...............................................
if the actual weight is (45x+0.09) grams - it's machine 9
make sense?
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Last edited by Noisician on Dec-15-2003 at 00:16
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Dec-15-2003 00:11
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moncster
Guest
Registered: Not Yet
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Family guy:
Ok, here's another riddle. A woman has two children. A homicidal murderer tells her she can only keep one. Which one does she let him kill?
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Dec-15-2003 00:15
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Durafei
the crazy russian

Registered: Oct 2000
Location: San Francisco, California
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| quote: | Originally posted by Noisician
just remembered one decent math problem/riddle i had to solve recently.
there are a certain number of books lying on the floor that u need to pack in boxes via tying them up into packages. u first try to make some packages with only 4 books in each, but then u notice that, at the end, 1 book is left unpacked. u then try to make a number of packages with 5 books in each, which also leaves only 1 book left. so u make packages with 6 books in each, which, once again, yields one single book remaining on the floor. finally, tying up 7 books together leaves u no additional books whatsoever. my question is, *at least* how many books had to be lying on the floor?
it's pretty easy, though |
4a + 1 = n
5b + 1 = n
6c + 1 = n
7q = n
n is divisible by 7.
n - 1 divisible by 4,5 and 6. lcm(4,5,6) = 60. thus n-1 is divisible
by 60.
so need only consider those n, such that n-1 divisible by 60, and n itself is divisible by 7.
n - 1 = 60a
n = 7b.
hence: 60a + 1 = 7b. Now we can solve this using Euclid's GCD algorithm.
done.
___________________
My Blog: durafei.blogspot.com - Last Update March 23, 2006
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Dec-15-2003 00:33
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A_M_8
Supreme tranceaddict

Registered: May 2003
Location: San Salvador/ Miami/ Boston
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| quote: | Originally posted by Noisician
that's a good one. here's what i think:
u need to take 1 screw produced by the 1st machine, 2 screws produced by the 2nd machine, ... , 9 screws produced by the 9th machine and put all of them on the scale. now, if all the machines produced the same screws, the ideal total weight would be (45x) grams, where x is the weight of one screw. by comparing the actual weight to the ideal one, u can figure out which machine produces defective screws because, in this case, the actual weight will vary, depending on which machine contributed to the overweight.
if the actual weight is (45x+0.01) grams - it's machine 1
if the actual weight is (45x+0.02) grams - it's machine 2
if the actual weight is (45x+0.03) grams - it's machine 3
...............................................
if the actual weight is (45x+0.09) grams - it's machine 9
make sense? |
did you know this one already or did u figure it out right now? bec. thats the fastest time someone has anwered that riddle for me
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Dec-15-2003 01:52
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Noisician
Harsh electronic purity

Registered: Aug 2001
Location:
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| quote: | Originally posted by A_M_8
did you know this one already or did u figure it out right now? bec. thats the fastest time someone has anwered that riddle for me |
nah. about a year ago, there was a semi-similar problem in one of my math classes. it gave me a general idea how to do yours.
anyway, here's another riddle that i think is pretty good (and hard)
it's a riddle about two mathematicians, mister a and mister b.
a certain person picked two natural numbers, each of which was greater than 1 and smaller than 100. to mister a, that person gave only the sum of those numbers. to mister b, the person gave only their product. they then were asked to find the numbers, but they were not allowed to tell each other what they initially knew about those numbers.
when they met, they had this short dialogue:
i'm afraid i can't deduce the right answer. there is not enough information given.
yes, i know that u can't possibly know the answer. i can even prove that.
in that case, i know what the numbers are, as u just unintentionally gave me a great hint.
but now i, too, know the answer, as what u just said to me was enough for me to pick the right numbers.
believe it or not, that dialogue should be enough to enable u to figure out the initial numbers, without even knowing their sum or product. so what were they?
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Dec-15-2003 20:46
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