anonymous
  • anonymous
You have 8 objects of similar size, but one object weighs a little bit more than the others. Given a balance scale, you must find out which object it is, but you may only use the scale twice. What do you do?
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

2bornot2b
  • 2bornot2b
If you allow me thrice, then I have the answer
anonymous
  • anonymous
Great; they say it can actually be done with only 2 "weighing" steps but still, I wanna see your answer!
2bornot2b
  • 2bornot2b
Just divide the 8 things into two groups, and place it in each of the pans of your balance.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

2bornot2b
  • 2bornot2b
*of course two equal groups of 4
2bornot2b
  • 2bornot2b
So now you know when 4 contains the heavier one
2bornot2b
  • 2bornot2b
Divide that group into 2 groups each of 2,
2bornot2b
  • 2bornot2b
Now you know which two contains the heavier one
anonymous
  • anonymous
how about 3 and 3 for the first weighing?
2bornot2b
  • 2bornot2b
Now again place them, and you know the culprit
anonymous
  • anonymous
@satellite Great; what's the rest
anonymous
  • anonymous
then if they are equal it is one of the other two, one weighing does it
anonymous
  • anonymous
and if it is one of the groups of 3, then weight two out of the three: if they are equal then it is the third
2bornot2b
  • 2bornot2b
satellite73 solved it
anonymous
  • anonymous
i think that does it
anonymous
  • anonymous
Very cool!
anonymous
  • anonymous
here is a question, that i can do but always have to start fresh because i always forget. how do i multiply two 2 digit numbers with only 3 multiplications? example 24 x 53
anonymous
  • anonymous
i am guessing adg knows because it is related to computer algorithms.
2bornot2b
  • 2bornot2b
Now solve this: There are 10 cigarette packets, each containing 10 cigarettes. One of the packets contain cigarettes which are a bit heavier than the cigarettes in any of the other packets. Although each cigarette in anyone packet weighs the same. Now with just one weighing you need to find out the culprit box.
anonymous
  • anonymous
i have seen this one before so i keep quiet
2bornot2b
  • 2bornot2b
great satellite73!
anonymous
  • anonymous
just select cigarettes from each box and compare!
2bornot2b
  • 2bornot2b
You are allowed to weight only once
2bornot2b
  • 2bornot2b
However, I must mention, you are allowed to use standardized weights
anonymous
  • anonymous
right empty one box, leaving 1 cigarette in, and stuff that box with cigarettes from other boxes. and then..... I'm stuck :(
2bornot2b
  • 2bornot2b
Oh! Just don't try it that way. Keep it with you for a few days, then come back. You will surely have solved it by then
anonymous
  • anonymous
lol
anonymous
  • anonymous
alright... just look at the boxes and compare the weights...
anonymous
  • anonymous
ok i remember now. you can multiply 53 times 24 with 3 multiplications instead of 4

Looking for something else?

Not the answer you are looking for? Search for more explanations.