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If you allow me thrice, then I have the answer
Great; they say it can actually be done with only 2 "weighing" steps but still, I wanna see your answer!
Just divide the 8 things into two groups, and place it in each of the pans of your balance.
*of course two equal groups of 4
So now you know when 4 contains the heavier one
Divide that group into 2 groups each of 2,
Now you know which two contains the heavier one
how about 3 and 3 for the first weighing?
Now again place them, and you know the culprit
then if they are equal it is one of the other two, one weighing does it
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
satellite73 solved it
i think that does it
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
i am guessing adg knows because it is related to computer algorithms.
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.
i have seen this one before so i keep quiet
just select cigarettes from each box and compare!
You are allowed to weight only once
However, I must mention, you are allowed to use standardized weights
right empty one box, leaving 1 cigarette in, and stuff that box with cigarettes from other boxes. and then..... I'm stuck :(
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
alright... just look at the boxes and compare the weights...
ok i remember now. you can multiply 53 times 24 with 3 multiplications instead of 4