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anonymous

  • 5 years ago

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?

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  1. 2bornot2b
    • 5 years ago
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    If you allow me thrice, then I have the answer

  2. anonymous
    • 5 years ago
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    Great; they say it can actually be done with only 2 "weighing" steps but still, I wanna see your answer!

  3. 2bornot2b
    • 5 years ago
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    Just divide the 8 things into two groups, and place it in each of the pans of your balance.

  4. 2bornot2b
    • 5 years ago
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    *of course two equal groups of 4

  5. 2bornot2b
    • 5 years ago
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    So now you know when 4 contains the heavier one

  6. 2bornot2b
    • 5 years ago
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    Divide that group into 2 groups each of 2,

  7. 2bornot2b
    • 5 years ago
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    Now you know which two contains the heavier one

  8. anonymous
    • 5 years ago
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    how about 3 and 3 for the first weighing?

  9. 2bornot2b
    • 5 years ago
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    Now again place them, and you know the culprit

  10. anonymous
    • 5 years ago
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    @satellite Great; what's the rest

  11. anonymous
    • 5 years ago
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    then if they are equal it is one of the other two, one weighing does it

  12. anonymous
    • 5 years ago
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    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

  13. 2bornot2b
    • 5 years ago
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    satellite73 solved it

  14. anonymous
    • 5 years ago
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    i think that does it

  15. anonymous
    • 5 years ago
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    Very cool!

  16. anonymous
    • 5 years ago
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    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

  17. anonymous
    • 5 years ago
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    i am guessing adg knows because it is related to computer algorithms.

  18. 2bornot2b
    • 5 years ago
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    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.

  19. anonymous
    • 5 years ago
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    i have seen this one before so i keep quiet

  20. 2bornot2b
    • 5 years ago
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    great satellite73!

  21. anonymous
    • 5 years ago
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    just select cigarettes from each box and compare!

  22. 2bornot2b
    • 5 years ago
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    You are allowed to weight only once

  23. 2bornot2b
    • 5 years ago
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    However, I must mention, you are allowed to use standardized weights

  24. anonymous
    • 5 years ago
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    right empty one box, leaving 1 cigarette in, and stuff that box with cigarettes from other boxes. and then..... I'm stuck :(

  25. 2bornot2b
    • 5 years ago
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    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

  26. anonymous
    • 5 years ago
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    lol

  27. anonymous
    • 5 years ago
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    alright... just look at the boxes and compare the weights...

  28. anonymous
    • 5 years ago
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    ok i remember now. you can multiply 53 times 24 with 3 multiplications instead of 4

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