Hi, there's just one thing which I'm not sure with this problem :
There are 24 pounds of nails in a sack. Can you measure out 9 pounds
of nails using only a balance with two pans?
I answered yes and here's the reason. Consider a number X of nails which weights 24 pounds. This number X has atleast a factor of 2^3 (It can be higher.) in its factors. Why ? Because we need to divide evenly the number of nails so we can have equal weight on each side of balance.
x/2=12 pounds x/4=6 pounds x/8=3 pounds
So we have the following groups (after successive divisons.):
Clearly, x/4+x/8 nails would give 9 pounds
Now, the only thing bugging me is that my solution is founded on the assumption that all nails must have equal weights. The question isn't clear if the nails do have all the same weight or not...
What do you think ???
All the proportions which we have deduced from a:b::c:d may be represented generally as follows :
ma+nb:pa+qb::mc+nd: pc + qd
For the product of the extreme terms is mpac+npbc+mqad +nqbd; which, since ad = bc, becomes mpac+npbc+mqad +nqbd; also the product of the mean terms is mpac+mqbc+npad+nqbd; or, since ad = bc, it is mpac+npbc+mqad +nqbd: so that the two products are equal.
Here are some of the proportions deduced from a:b::c:d :
and some others (I'm not putting all of them.)
I don't understand the part : "All the proportions which we have deduced from a:b::c:d may be represented generally as follows :
ma+nb:pa+qb::mc+nd: pc + qd"
What does it mean ??? I'm not following for a certain reason....
Hi, basic question here:
Now, I know that this reduces to :
which gives infinity over -5. Now, how do I evaluate infinity over a constant ??? Is there a fact or definition that tells us how to deal with this ?
Well, I think I could show you an example of what I was talking about. I read it somewhere :
We have :
Now we know that the first term of the root will be "a"
"a" squared cancels the other a^2
Now, we know that 2ab+b^2=(2a+b)b and so if we can figure out a divisor, we will also know the root. We also know the first term of the divisor because it is always the double of the firm term of the root. (Look at the square root algorithm :http://www.basic-mathematics.com/square-root-algorithm.html)
2ab+b^2(a+b (we find, by luck I guess, that we have "b" for the second term.)
Anyway, i was searching for something like what I described.