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Forums - General - The riddle and english language thread

zexen_lowe said:
twesterm said:
How about a fun one I found on Professor Layton:

You have five identical weights and a scale. Four of the weights weigh exactly the same while one weighs slightly more than all the others. How do you find which weight is the heavier one and only use the scale a max of two times?

Haven't played Layton, but here goes. Pick two and weigh one against one, if one's heavier, you have it, if not of the three remaining, pick two and weigh one against one, if one's heavier, that's the one, if not, then the one you didn't put in the scale is the one

 

 

lol, that's actually simpler than the way I worked it out >_>

You have five weights, A B C D and E.  Weigh AB vs DC

If AB and DC weigh the same then E is the heavier.  If AB weighs more than A or B is heavier (and same with CD but assuming AB weighs more for this).

Weigh A vs E.  If they weigh the same B is heavier.  Otherwise, whichever one of A or E weighs more, that's the heavier one.



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twesterm said:
zexen_lowe said:
twesterm said:
How about a fun one I found on Professor Layton:

You have five identical weights and a scale. Four of the weights weigh exactly the same while one weighs slightly more than all the others. How do you find which weight is the heavier one and only use the scale a max of two times?

Haven't played Layton, but here goes. Pick two and weigh one against one, if one's heavier, you have it, if not of the three remaining, pick two and weigh one against one, if one's heavier, that's the one, if not, then the one you didn't put in the scale is the one

 

 

lol, that's actually simpler than the way I worked it out >_>

You have five weights, A B C D and E.  Weigh AB vs DC

If AB and DC weigh the same then E is the heavier.  If AB weighs more than A or B is heavier (and same with CD but assuming AB weighs more for this).

Weigh A vs E.  If they weigh the same B is heavier.  Otherwise, whichever one of A or E weighs more, that's the heavier one.

Yeah, I knew a similar riddle with 9 weights where you have to find the one that's heavier in two attempts

Most people would try 4vs4 then 2vs2, but it doesn't give the answer, the right one is doing 3vs3 and then 1vs1, you can figure exactly how

 




zexen_lowe said:
twesterm said:
zexen_lowe said:
twesterm said:
How about a fun one I found on Professor Layton:

You have five identical weights and a scale. Four of the weights weigh exactly the same while one weighs slightly more than all the others. How do you find which weight is the heavier one and only use the scale a max of two times?

Haven't played Layton, but here goes. Pick two and weigh one against one, if one's heavier, you have it, if not of the three remaining, pick two and weigh one against one, if one's heavier, that's the one, if not, then the one you didn't put in the scale is the one

 

 

lol, that's actually simpler than the way I worked it out >_>

You have five weights, A B C D and E.  Weigh AB vs DC

If AB and DC weigh the same then E is the heavier.  If AB weighs more than A or B is heavier (and same with CD but assuming AB weighs more for this).

Weigh A vs E.  If they weigh the same B is heavier.  Otherwise, whichever one of A or E weighs more, that's the heavier one.

Yeah, I knew a similar riddle with 9 weights where you have to find the one that's heavier in two attempts

Most people would try 4vs4 then 2vs2, but it doesn't give the answer, the right one is doing 3vs3 and then 1vs1, you can figure exactly how

 

Oh!  That was it, it was supposed to be 9.  I'll edit my post >_<

 



How about this?
The first man quoting the second man quoting the third man quoting the fourth man quoting the fifth man said "He said something and "And '"'And "'And 'And that's all'"''"'".
:p
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