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Hence for we get paths, for we get
paths, and for
paths. This gives us
hence the total number of paths is
we get
paths through the second quadrant,
.
Combinatorial Solution 2
Each path that goes through the second quadrant must pass through exactly one of the points
,
, and
.
There is exactly path of the first kind, paths of the second kind, and
paths of the third type. The conclusion remains the same.
Problem 19
Each of 2010 boxes in a line contains a single red marble, and for box in the
, the
position also contains white marbles. Isabella begins at the first box
be the probability that Isabella
and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let
stops after drawing exactly marbles. What is the smallest value of for which
?
Solution
The probability of drawing a white marble from box is drawing a red marble from box is
.
. The probability of
The probability of drawing a red marble at box is therefore
It is then easy to see that the lowest integer value of that satisfies the inequality is
.
Problem 20
Arithmetic sequences value of ?
and and
have integer terms with
for some . What is the largest possible
Solution Solution 1
Since
and
have integer terms with
, we can write the terms of
each sequence as
where and ( Since
) are the common differences of each, respectively.
it is easy to see that
.
Hence, we have to find the largest such that
The prime factorization of have a product of
and soon find that the largest largest value is
.
value is for the pair
, and so the
is
. We list out all the possible pairs that
and
are both integers.
Solution 2
As above, let
and
for some
.
Now we get
. Therefore
divides
, hence
. And
as the second term is greater than the first one, we only have to consider the options For other
.
we easily see that for it is way too large.
the right side is less than
and for any
For that
we are looking for such that . Note
must be divisible by . We can start looking for the solution by trying the
, and we easily discover that for
.
anymore.) we get
possible values for
, which has a suitable solution
Hence
is the largest possible . (There is no need to check
Problem 21
The graph of
largest of these values?
lies above the line
except at three values of , where the graph and the line intersect. What is the
Solution
The values in which are the same as the zeros of
intersect at
.
Since there are zeros and the function is never negative, all zeros must be double roots because the function's degree is .
Suppose we let , , and be the roots of this function, and let
be the cubic polynomial with roots , , and .
In order to find out the terms of
.
we must first expand
[Quick note: Since we don't know , , and , we really don't even need the last 3 terms of the expansion.]
All that's left is to find the largest root of .
Problem 22
What is the minimum value of
?
Solution Solution 1
If we graph each term separately, we will notice that all of the zeros occur at where
is any integer from to
, inclusive.
,
The minimum value occurs where the sum of the slopes is at a minimum, since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some
The sum of the slope at
is
.
Now we want to minimize means the slope is where
.
. The zeros occur at
and
, which
We can now verify that both and yield .