The eggs in a basket doubled every minute. After an hour, the basket was full. When was the basket full to the half?
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The eggs in a basket doubled every minute. After an hour, the basket was full. When was the basket full to the half?
duh... that's a tough one :rolleyes: I won't answer, I'll leave it for others to think over for now! :D
Why does this sound like hw?
111011 minutes
Figure that out and u will get the answer.
Well, let me put it this way; our maths teacher gave us the question, it took about 2-3 minutes till I solved it:D
People started with variables for the time increasing, and almost everyone said the answer was 11110 minutes.
Too easy, but it did take me 59 minutes to figure it out.
Impossible. This thread was posted 05:35pm, you replied 06:27pm.
>> Impossible. This thread was posted 05:35pm, you replied 06:27pm.
hmm try looking at that post again. especialy the number. and compare it with golfinguy4 posts ;)
/*damn, should have wayted 5 min to post this*/
I think TGM is trying to hint at the answer. if it doubles, it is half full at 59 min, so when half the basket is full and then it is doubled you have two halves which eaquel a whole
I can't spell can I:)
Bingo!
olrite, you math masterminds, here is a similar problem for you:
A big water tank has a hole. Every hour, half of the water that is remaining leaks out. If the tank had 2000 L of water at 12:00 PM, what would be the exact time when all the water is gone?
answer that...
I have another one, but i will save it until someone answers the first one. The next one is a bit harder... bit of a challenge.
>A big water tank has a hole. Every hour, half of the water that is remaining leaks out. If the tank had 2000 L of water at 12:00 PM, what would be the exact time when all the water is gone?
I believe that would take an infinite amount of time (or until there was one water molecule left).
I second Sorensen's response
What grade are you in?Quote:
I have another one, but i will save it until someone answers the first one. The next one is a bit harder... bit of a challenge.
Anyways, here is my problem for you:)
Here is the definition you need
The function r(z) is equal to the sum of 1/(n^z), where n=1,2,3,...infinity. To those of you familiar with sigma notation (that funny Greek 'E'), you would have 'n=1' below the sigma, the infinity sign above the sigma, and 1/(n^z) to the right of the sigma. Here is a pathetic graphical representation.
infinity
---
\ 1/(n^z)=r(z)
/
---
n=1
So here is the problem:
When z is a complex number of the form z=a+it [i=sqrt(-1)], do all solutions to the equation r(z)=0 have a=0.5? (a=real part of z)
Constraint: candidate solutions must have a nonzero imaginary part ie, 't' != 0. Explain the rationale behind your answer.
Anybody have any ideas? ;)
Dude, what class is this?
wow!! i am just in grade 10. i dont get that, especially with that
"pathetic graphical representation" i have no idea.
anyway, that was an easy one.
this one is tricky:
on a very windy day, a snail is trying to climb a small hill.
this is the small hill:
/ \
/ \
a / \ b side a = 20 meters
/ \ side b = 20 meters
__/______c______\____ side c doesn't matter
the snail climbs 5m and then for sure a gust of wind comes and sweeps it 3m down. if it is 12:00 PM right now, what time will it be (to the closest hour) when the snail gets to the top or over it?
answer coming up after a few guesses.
d'oh!!!
my graphic design of the hilll did not work.
ok, here is another try (in words)
the side that the snail is climbing is 20 meters long (slope and all the other stuff included, so no calculations required)
the opposite side is also same length
the base length of the hill does not matter;
all you really need to know is that the snail needs to walk 20 meters to get to the top and with the wind... you know the question...
in graphic words:
a/\b
a is the distancesnail has to go (one side of the hill) and that is 20 meters
Isn't that a bit too large considering at t=60 min the egg basket is full.Quote:
People started with variables for the time increasing, and almost everyone said the answer was 11110 minutes.
Consider this, let N = number of eggs in the basket, t = the time in minutes.
Then we have
N = 2^t and since at t=60 the basket is full
N_full = 2^60 we want to find t when N = N_full/2 so we have
2^59 = 2^t. And so t = 59 minutes. Ok you probably could figure it out mentally now.
Maybe use code tags
Code:/\
/ \
/ \
/ \
/ \
assuming that the snail climbs 5m per hour (since it was not mentioned--i didn't c it)
I would say it's 10 pm
Nick, dont' you recognise a bunch of ones and zeros? :DQuote:
Originally posted by Nick
Isn't that a bit too large considering at t=60 min the egg basket is full.
Consider this, let N = number of eggs in the basket, t = the time in minutes.
Then we have
N = 2^t and since at t=60 the basket is full
N_full = 2^60 we want to find t when N = N_full/2 so we have
2^59 = 2^t. And so t = 59 minutes. Ok you probably could figure it out mentally now.
Why so much of complications.. It takes 59 minutes.. Sine half + half is full... Half has to double and that takes one minute to double to become full... So 60 - 1 is 59.
that was not even a hard problem... someone else try a guess at my second question... one guess is not enough.i will reveal the ans after 1-2 more guesses.
1259 hrs
0.000000000001115465465246565432165498465 mins
look...2 more guesses...noe give me the damn answer!!!!!!:mad:
Now I do :)Quote:
Nick, dont' you recognise a bunch of ones and zeros?
yea ok, have it your way. te answer is soming up in my next post.
(I am practising methods to keep a moron in suspense. and it is working so far)
allright here is the answer:
the answer is 9 hours.Quote:
on a very windy day, a snail is trying to climb a small hill.
the side length of the hill is 20 meters long. the snail climbs 5m and then for sure a gust of wind comes and sweeps it 3m down. if it is 12:00 PM right now, what time will it be (to the closest hour) when the snail gets to the top or over it?
at the end of:
1st hour: climbs 5, drops 3. therefore climbed 2.
2nd hour: climbs 5, drops 3. therefore climbed 2+2=4.
3rd hour: climbs 5, drops 3. therefore climbed 4+2=6.
4th hour: climbs 5, drops 3. therefore climbed 6+2=8.
5th hour: climbs 5, drops 3. therefore climbed 8+2=10.
6th hour: climbs 5, drops 3. therefore climbed 10+2=12.
7th hour: climbs 5, drops 3. therefore climbed 12+2=14.
8th hour: climbs 5, drops 3. therefore climbed 14+2=16.
9th hour: climbs 5. therefore total climbed 16+5=21. Now the snail is over the top and therefore will not be swept back 3m.
see, that was not so easy.
Sorry, but you didn't give us enough info to solve the problem. You never mentioned how long it took the snail to climb the 5m. That is why I could not help you.
Jet_Master, what math class was this from?
Oh! Damn!!! I could have sworn that i said in one hour the snail climbs 5 m and drops 3m.... D'oh!! And no-one ever posted asking me that question.Quote:
You never mentioned how long it took the snail to climb the 5m. That is why I could not help you.
I wanted to give more time, but look back to page two, and you will see that commander was rushing too much. he had gone mad and was pestering me to give the answer.Quote:
Sorry, but you didn't give us enough info to solve the problem.
This was not from any math class- this one was just from me. by the way, i am in grade 10.Quote:
Jet_Master, what math class was this from?
go to this post --- here, I did mention that there was no time mentioned....read the posts more carefully next time.....Quote:
Oh! Damn!!! I could have sworn that i said in one hour the snail climbs 5 m and drops 3m.... D'oh!! And no-one ever posted asking me that question.
this was your previous post:Quote:
here, I did mention that there was no time mentioned....read the posts more carefully next time.....
when you said "assuming that the snail climbs 5m per hour..." i thought i already said it because you guessed it correctly. but i did not see that it was not mentioned; i also make mistakes you ass! i am human afterall. ........ happens. i apologized, what more do you want.Quote:
assuming that the snail climbs 5m per hour (since it was not mentioned--i didn't c it)
I would say it's 10 pm
oh, and commander, next time try providing a direct link, instead of just to the first page of the thread.
ok, try this math problem.
*Note: If you get a terrible headache doing this, dont sue me. I will not force you to do this.*
You are driving a bus from point A to point K. When you start, you have 15 people aboard. At point B, 3 get off, 6 get on. At point C, 3 get off, 5 get on. At point D, 5 get off,1 gets off. At point E, 8 get off, 4 get on. At point F, 5 get on. At point G, 3 get off, 4 get on. At point H,3 get off, 9 get on. At point I, 2 get off. At point J, 3 get off. When you get to point K, what is the name of the driver?
that post was supposed to take you to the post I posted.....
shut up please. now leave that topic.
by the way, iforgot to post something....
it's not my fault that ur comp can't recognize anchors or you tried to scroll down b4 the browser had time to take yo to that anchor or you stopped the transfer or something else
here is one that i found quite trickey..
I hear that Bill Gates actually used this at a conference of about 200 programmers who were applying to Micro$oft.. only 3 got in or so:D
There are two rooms completly isolated from each other, there is no way you can see into the other room at all.... one room has 3 light switches and the other room has a light in it... it is your job to find out which of the 3 switches turns on the light in that room.....
you can only walk down to the other room (with the light) ONCE!! however, you can flip the switches as many times as you like for as long as you like.
anyone? if u know this dont post the answer just yet for those who do not know it!!!
-good luck!
Walk to the other Room, turn on Switches 1 and 2, leave them on for a few minutes.
Turn off switch 1 off, and switch 3 on.
Walk back into the other room, reach up, touch the light. If it's off, switch 1 turns it on. If it's on and warm, switch 2 turns it on. If it's on and cold, switch 3 is the one.
Here's one for ya:
A guy walks into a 7-11 store and selects four items to buy. The clerk at the counter informs the gentleman that the total cost of the four items is $7.11. He was completely surprised that the cost was the same as the name of the store. The clerk informed the man that he simply multiplied the cost of each item and arrived at the total. The customer calmly informed the clerk that the items should be added and not multiplied. The clerk then added the items together and informed the customer that the total was still exactly $7.11.
What are the exact costs of each item?
for this question, does each item cost have to be in two decimal points (example: 1.45) or can it go beyond two (example: 1.4532623345)? I need to know that. because if it has to be two decimal points sharp, the closest total i could get was
sum: 7.11
product: 7.115724
but you said it has to be exact.
please reply asap.Quote:
A guy walks into a 7-11 store and selects four items to buy. The clerk at the counter informs the gentleman that the total cost of the four items is $7.11. He was completely surprised that the cost was the same as the name of the store. The clerk informed the man that he simply multiplied the cost of each item and arrived at the total. The customer calmly informed the clerk that the items should be added and not multiplied. The clerk then added the items together and informed the customer that the total was still exactly $7.11.
What are the exact costs of each item?
Yes, it has to be exact. (2 Decimal places) Hint: you can use programming for this :)
In the post below, I will be going off of the assumption that every item being sold can be expressed as some integer in terms of cents, ie you CANNOT have a fraction of a penny for a price of an item. If that is incorrect to assume so, please do tell me and ignore the post below.
I know that it might be possible to have fractional prices (in cents), such asIn any case, taht is what I assume, if I'm wrong, I'm wrong, won't be the first time.Code:3 for $0.25
[EDIT]Highlited in red is incorrect!
I pick A, B, C, and D to be prices for the items being sold (in cents). The product will look like this:
A * B * C * D = 711
(A * B * C) * D = 711
>> 711 % (A * B * C) == 0
What I mean by above is that EACH of the items bust be an INTEGER FACTOR of 711, in order to divide 711 cents by either of the prices and get an integer value in cents again. This is based on the assumption that the prices are integer values in cents, can't have a dicimal of a cent in price.
Now, the integer factors of 711 are: 1, 3, 9, 79 and 237.
This is the same as:
$0.01, $0.03, $0.09, $0.79 or $2.37
Let's theorize that all items are priced below $2.37, that is either $0.01, $0.03, $0.09 or $0.79. This would mean that you must be able to get a sum of $7.11 out of 4 items that are $0.79 or below, which is impossible. If only 1 item is $2.37, then the other 3 must add to 7.11 = 2.37 = 4.74 >> impossible again with 3 items below $1. So, at least 2 items must be priced at $2.37 each, leaving other 2 to be: 7.11 - (2 * 2.37) = 2.37
As you can see, it's impossible to have 2 items priced $0.01, $0.03, $0.09, $0.79 or $2.37 to add to $2.37 without them being $0 and $2.37. Now, this again is impossible, because the product must be a non-zero value.
I see no solution for this probem.
YES!!! Ha Ha Ha!!!
This IS getting INTERESTING!!! Ha Ha Ha !!!
I LOVE this thread!!!
dbaryl, thanks a lot for the info you have provided. I WILL use all that to find any possible solution if possible. if not possible, i will not possibly find an answer that is possible and is correct. but if there is a possible answer, i possibly can find it! just wait for the possible; or for some, my accptance of the impossible.
(either way, i am sure that i have confused you by now)
There is an absolute answer... I once had a program written in C which calculated the answer(s) but I lost it when my HD crashed.
I know the values, but the program would be damn cool to have now :(
the closest i could get was:
the items cost $1.00, $2.91, $1.94, $1.26
the sum is $7.11
but the product is $7.113204 - when you round this up to two decimal points, i get: $7.11!!
but, i didnot use programming. just my mind and a calculator.
Well, sadly, that's wrong. It's not an exact value..
I may be able to recreate the program which calculated the answer, but I doubt it :/
>>Well, sadly, that's wrong. It's not an exact value..
i thought so. my program cannot calculate anything. this is how my program is:
i declared 4 variables - a,b,c,d
a+b+c+d=7.11
a*b*c*d=7.11
isn't that right. correct me if i am wrong.
also,
a=7.11-(b+c+d)
b=7.11-(a+c+d)
c=7.11-(a+b+d)
d=7.11-(a+b+c)
when i do that, it calculates the first statement first. so, it gives my a value=7.11 and all others 0. so it satisfies the first part of my function. but i dont know why it neglects the part (a*b*c*d=7.11)
i hope you understand what i am saying...
dbaryl is right
but here's a piece of code that'll prove it:
Code:#include <stdio.h>
int main()
{
int i,j,k;
int a,b,c,d;
for(i = 1, a = 1; i <= 711; i++, a++)
{
for(j = 1, b = 1; j <= 711; j++, b++)
{
for(k = 1, c = 1; k <= 711; k++, c++)
{
d = 711 - a - b - c;
if(a * b * c * d == 711)
{
printf("a = %d\nb = %d\nc = %d\nd = %d\n", a, b, c, d);
return 0;
}
}
}
}
printf("No such answer\n");
return 0;
}
oops my bad
here's the revised code:
Code:#include <stdio.h>
int main()
{
double a,b,c,d;
for(a = 0; a <= 7.11; a += 0.01)
{
for(b = 0; b <= 7.11; b += 0.01)
{
for(c = 0; c <= 7.11; c += 0.01)
{
d = 7.11 - a - b - c;
if(a * b * c * d == 7.11)
{
printf("a = %d\nb = %d\nc = %d\nd = %d\n", a, b, c, d);
return 0;
}
}
}
}
printf("No such answer\n");
return 0;
}
No, you didn't. :)Quote:
Originally posted by Jet_Master
dbaryl, thanks a lot for the info you have provided. I WILL use all that to find any possible solution if possible. if not possible, i will not possibly find an answer that is possible and is correct. but if there is a possible answer, i possibly can find it! just wait for the possible; or for some, my accptance of the impossible.
(either way, i am sure that i have confused you by now)
Oh, a couple more things... those of you who say that there is a possible answer, the only way I see it is if the prices are some ugly "non-nice" decimal. If you check the first post on this page, that is where I show my thoughts on this... if you see anthing wrong, tell me, otherwise, my answer stands: no solution.
Unregistered, how about this one?
:DCode:#include <stdio.h>
int main()
{
printf("No such answer!\n");
return 0;
}
Unregistered is on the right track... that is somewhat what the program looked like, but the logic is still off.
Well, there is a solution, I have it right here... so I guess you're wrong :/Quote:
If you check the first post on this page, that is where I show my thoughts on this... if you see anthing wrong, tell me, otherwise, my answer stands: no solution.
Is my assumption incorrect? Or are the 3 values something other tahn int values of cents? Tell me that much.... pliz.
I worte a progra the second i saw thins q, to figure out the ans I'm attaching it, I ran the damn thing for 8 hours, still no hit!!!!!!!!!!!!
And here's the codeSo what if it uses random numbers!!!!!!!!!!!!!!! it helps me not to fry my brain!!PHP Code:#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
#include <string.h>
#include <math.h>
void main(){
float a,b,c,d;
for(;;){
a= (rand() / 65654.356); // I got these Numbers
c= (rand() / 3215.265); //by the rand() function
b= (rand() / 3541.065); // Therefore, they are
d= (rand() / 12456.265); //random numbers!
printf("\n%.2f + %.2f + %.2f + %.2f == %.2f", a,b,c,d, a+b+c+d);
printf("\n %.2f * %.2f * %.2f * %.2f == %.2f", a,b,c,d, a*b*c*d);
if( ((a+b+c+d)==7.11) && ((a*b*c*d)==7.11) ){
system("pause");}
}
}
No, all four values are expressed to the cent.
Is one of the numbers over $7.11? And for those of you who think this is impossible, it isn't. You can have a negative number to bring down the number above $ 7.11-
ex: $10+($-5)=$5
However this creates a problem- when you multiply, no matter what you will come out a negative number ( a negative number times a positve comes out negative, negative * positive * positive * positve is still negative ) To solve the problem we just add another number.
Dope, that was me, forgot to log in.
A couple of things:
First, I want to make sure that this is not one of those wording problems, where there is some hidden thing like: He multiplied the name of the story by ... Is it right to assume that if the 4 items are priced a, b, c, and d then a + b + c + d = 7.11 and a * b * c * d = 7.11?
Now, if that is the correct case, I seriously don't see any other solution to this... except for one: negative values. I'm not sure if it's possible to achieve the result that way, but you can try by having 2 of the numbers be negative [negative goes away that way].
Can anyone who knows the answer at least hint as to what I'm doing wrong in my guess at the top of page 4?
There are no negitive values. No value is over $7.11
$7.11 = a + b + c + d
$7.11 = a * b * c * d
there is no solution, apart from the clerk is on crack.
as dbarry said
to go futher
sqrt(711) < 27
if more than one one object is priced at greater than 0.27 then the product will be greater than 711 ie 0.27 * 0.27 * 0.01 * 0.01 = 7.29
so no two values can be greater than 0.27
The largest factor of 7.11 is 2.37 so this is the largest value possible. (0.01*0.01*0.03*2.37=7.11)
but if 2.37 is one value
7.11 - 2.37 = a+b+c
4.74 = a+b+c
as we have proved only one value can be above 0.27 there is no possible integer solution.
damnit, i'm gone for one day and i thought you people wouldn't know my problem.. heh, figures the first person to read it would get it :D
I worked my ASS off.
Yesterday, i worked for 1.5 hours on paper and these are the results i got.
bcd=2.
so a=(7.11/2) but that gives 355.5 cents. so i guess that is not possible.
also i managed to eliminate a. but the equation i got further afterthat resembles this: (everything is in cents)
3555+2(b+c+d)=bcd=2
.: 3555+2(b+c+d)=2
.: 2(b+c+d)=(-3553)
.: b+c+d=(-1776.5)
.: a=711+1776.5=355.5
.: 1065.5=355.5 ?!?!
quite illogical isn't it???
does that prove my theory that 0=infinity??
c'mon, accept it. my theory is CORRECT!!!
YES!!!
0=infinty
Ha ha ha!!!
I simply LOVE this post. its the greatest. MATH rules!!!
Apparently everyone thinks its impossible... Well, it's not, and I shall show you.
$3.16 $1.20 $1.25 $1.50 seem to work out to $7.11
http://mathforum.org/library/drmath/view/55897.html
http://www.mozart-oz.org/documentation/fdt/node21.html#section.elimination.grocery
There are many ways to do it... I used the now non-existant program.
O2, if you are talking about thee 3-lightbulb problem, that one is a bit overused and I think evewryone knows it by now :)Quote:
Originally posted by OxYgEn-22
damnit, i'm gone for one day and i thought you people wouldn't know my problem.. heh, figures the first person to read it would get it :D
>>0.27 * 0.27 * 0.01 * 0.01 = 7.29
I'm not sure that this is correct :D
Now, for Jet_Master: where did you come up with the fact that one of the values is 2? Is that a random guess?
One thing I'm sticking to is that the values have to be either $0.01, $0.03, $0.09, $0.79 or $2.37... those are the only integer factors of 711. I simply don't see any other way.
Now, about my reasoning a couple of pages back, I made a mistake in one of the assumptions, but this above does stay true, as far as I know.This is what I said, and I thinkit's correct. See if anything is wrong with this reasoning.Quote:
Now, the integer factors of 711 are: 1, 3, 9, 79 and 237.
This is the same as:
$0.01, $0.03, $0.09, $0.79 or $2.37
Let's theorize that all items are priced below $2.37, that is either $0.01, $0.03, $0.09 or $0.79. This would mean that you must be able to get a sum of $7.11 out of 4 items that are $0.79 or below, which is impossible. If only 1 item is $2.37, then the other 3 must add to 7.11 = 2.37 = 4.74 >> impossible again with 3 items below $1. So, at least 2 items must be priced at $2.37 each, leaving other 2 to be: 7.11 - (2 * 2.37) = 2.37
As you can see, it's impossible to have 2 items priced $0.01, $0.03, $0.09, $0.79 or $2.37 to add to $2.37 without them being $0 and $2.37. Now, this again is impossible, because the product must be a non-zero value.
I see no solution for this probem.
Anyone know why the code before with the floats didn't work? I converted the code to cents so all 4 items were multiplied by 100. So when multiplying the four items I just compared it with
$7.11 *100 ^ 4
Here's the code and it surprisingly works:
Code:#include <stdio.h>
int main()
{
int a,b,c,d;
for(a = 0; a <= 711; a++)
{
for(b = 0; b <= 711; b++)
{
if(a + b > 711)
break;
for(c = 0; c <= 711; c++)
{
if(a + b + c > 711)
break;
d = 711 - a - b - c;
if(a * b * c * d == 711000000)
{
printf("a = %d\nb = %d\nc = %d\nd = %d\n", a, b, c, d);
return 0;
}
}
}
}
printf("No such answer\n");
return 0;
}
Here's a possibility, there is an answer:
all numbers are null set. In math, null set doesn't equal zero, it means no real numbers. Therefore, I say all four values are null set
Dual-catfish already showed us the answer, and will be revealed only to smart people... :D
lol, I didn't notice the links were in white until you said that, took me about 3 seconds after I read that to figure it out
Thanks :)Quote:
Originally posted by Dual-Catfish
Apparently everyone thinks its impossible... Well, it's not, and I shall show you.
$3.16 $1.20 $1.25 $1.50 seem to work out to $7.11
http://mathforum.org/library/drmath/view/55897.html
http://www.mozart-oz.org/documentation/fdt/node21.html#section.elimination.grocery
There are many ways to do it... I used the now non-existant program.
Lets not forget that these are prices. Prices of actual items, like bananas, apples, pears or slaves.Quote:
all numbers are null set. In math, null set doesn't equal zero, it means no real numbers. Therefore, I say all four values are null set
I don't know about you, but i've never saw something priced null.
That program is similar to the other which I used... however different. Anyone else have any brainteasers?
You have a bag that contains two balls, each of which is either red or blue with equal probability. You add two balls, one red and one blue, and then draw out a ball and notice it is blue. Then you draw another ball. What is the probability that you will be fired for wasting your time on stupid puzzles like this when you should be working?
Zero, as I do this during school ;)