*How many positive five-digit integers are there
consisting of the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, in which one digit appears
once and two digits appear twice? For
example, 41174 is one such number, while 75355 is not.*

Each number that satisfies these conditions will consist of three distinct digits, one which will appear once and two which will appear twice. The number of ways to select three digits out of nine is . In each set of three digits, any one digit (out of 3) can appear a single time. Thus, each case above will result in three possible combinations that can produce five digits satisfying the above conditions. So, we must multiply by 3 to obtain the number of possible ways to select five digits consisting of three distinct digits satisfying the above conditions.

We get .

Now, the number of ways we can permute 5 given distinct digits is 5!. However, in our case, two pairs of digits are identical. So, we can arrange 5 given digits that satisfy the given conditions , or 30 ways.

Since, for each combination there are 30 permutations, and
since there are 252 combinations, **there
are (252)(30),
or 7560 five digit numbers that satisfy the given conditions. **

*Determine, with proof, the positive integer whose
square is exactly equal to the number*

Let us take a general case .

So,

is the sum of the cubes of all odd numbers
from 1 to 2*n*. This can be written as the difference of the
sum of all cubes from 1 to 2*n* and the
sum of all even cubes from 1 to 2*n*. The sum of the cubes of all even numbers from
1 to 2*n* is .

We will use the fact that , which can be proven by induction. (See http://www.shu.edu/projects/reals/infinity/answers/induct.html and http://www.cse.ucsc.edu/classes/cmps102/Fall01/solutions2.pdf for proofs.)

So,

So, when *n* = 2001,

*S* = (4(2001)^{2}-1)^{2} = (**16016003**)^{2}

* *

*Factor the expression*

* *

If we expand an expression:

, we get

So, the factored form of

will have coefficients *w,
x, y, z, k, l, m, n* such that

From Eq. 1, we get

We can now write Eq.’s 2-7 in terms of *w, x, y, *and* z*:

, or , or

Solving these quadratic equations, we get

Without loss of generality, we can assume that *w* = 1.
(We can always divide the first factor by a constant and multiply the
second factor by the same constant to make *w*
= 1.

Now we have

*x* = 5/3 or 3/5

*y* = -1/2 or -2

*z* = -1/5 or -5

*x*/*y* = -5/6 or -6/5

*x*/*z* = -1/3 or -3

*y*/*z* = 2/5 or 5/2

So, if *x* = 5/3,
then *y* = -2 and *z* = -5

If *x* = 3/5, then *y* = -1/2 and *z* = -1/5

Thus, the factors are

(1*a* + (5/3)*b* 2*c* 5*d*)(30*a*
+ 18*b* 15*c* 6*d*),
or

(1*a* + (3/5)*b* (1/2)*c*
(1/5)*d*)(30*a* + 50*b* 60*c*
-150*d*)

Multiplying and dividing by suitable constants to get integer coefficients, we get

(3*a* + 5*b* 6*c* 15*d*)(10*a* + 6*b*
5*c* 2*d*),
or

(10*a* + 6*b* 5*c* 2*d*)(3*a* + 5*b*
6*c* 15*d*)

These are the same, so the factorization of

**is (3 a + 5b
6c 15d)(10a + 6b
5c 2d)**

We can verify that this factorization is correct by expanding it.

*Let X = (x _{1}, x_{2}, x_{3},
x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9})
be a 9-long vector of integers.
Determine X if the following seven vectors were all obtained from X by
deleting three of its components: *

*Y _{1} =
(0, 0, 0, 1, 0, 1), Y_{2} = (0, 0, 1, 1, 1, 0), Y_{3} = (0, 1,
0, 1, 0, 1), Y_{4} = (1, 0, 0, 0, 1, 1),*

*Y _{5} =
(1, 0, 1, 1, 1, 1), Y_{6} = (1, 1, 1, 1, 0, 1), Y_{7} = (1, 1,
0, 1, 1, 0)*

We know that X has at least five 1’s because Y_{5} and Y_{6}
have five 1’s. We also know that X has
at least four 0’s because Y_{1} has four 0’s. So, since X has only 9 terms, there must be
exactly five 1’s and four 0’s.

In Y_{1}, 1 is the rightmost term. There are four 0’s to the left of it. Since there are only four 0’s in X, no 0 can
be right of the 1 in X. Thus, the last
term of X is 1. **x _{9} = 1**

In Y_{7}, 0 is the rightmost term. There are four 1’s to the left of it. However, we know that the last term of X is
1, so the 0 must be followed by a 1.
Since there are five 1’s in X, and four are to the left of the 0, there
is only one 1 in front of the 0. Thus,
the second-last term of X must be a 0. **x _{8}
= 0**

Now, Y_{1} contains four 0’s.
The rightmost 0 must be the second last term, because there can be no 0
to the right of it in X, and we know that there
must be a 0 before the rightmost
1. There is a 1 to the left of this 0. Since there can be no more 0’s between this 0
and the 1 to the left of it, the third last term of X must be a 1. **x _{7} = 1**

In Y_{4}, 1 is the leftmost term.
There are three 0’s to the right of it.
To the right of these three 0’s are two consecutive 1’s. Since X must end in (…1, 0, 1), the fourth 0
must either be between the two consecutive 1’s, or after the leftmost 1. In either case, it cannot be to the left of
the leftmost 1. Since there are only
four 0’s in X, no 0 can be left of the leftmost 1. Thus, the leftmost term in X is 1. **x _{1} = 1**

In Y_{2}, both the leftmost and the rightmost terms are 0’s. The terms between these 0’s include three
1’s. There must be a 1 left of the
leftmost 0 and a 1 right of the rightmost 0 in X. Since all our 1’s are now “used up”, and the
term after the leftmost 0 in Y_{2} is also a 0, the leftmost 1 in X
must be followed by two consecutive 0’s.
Thus the second and third terms of X are 0’s. **x _{2} = 0, x_{3} = 0**

In Y_{7}, there are four 1’s.
Since the rightmost term is a 0, there must be a 1 to the right of it in
X. Also, there must be two consecutive
0’s after the first 1 and before the second 1.
Thus, the fourth, fifth, and sixth terms of X are 1, 0, and 1. **x _{4} = 1, x_{5} = 0, and
x_{6} = 1**

We now have all nine components of X:

** **

*X = *(1, 0, 0, 1, 0, 1, 1, 0, 1)

(It is easy to see that Y_{1}, Y_{2}, … Y_{7} can
all be obtained by removing three components from X..)

*Let R
and S be points on the sides and ,
respectively, of , and let P be the intersection of and . Determine the area of ** if the areas of ,
**, and
** are 5, 6, and 7, respectively.*

We place the figure on a coordinate plane, as shown

in the figure to the right. lies on the *x* axis, with *P* as

the origin. Let point *A*
lie on (*p*, *q*).

Since the area of is 5, and

the height from to *A*
is *q*, the

length of must equal .

Similarly, since the area of

is 6, the length of must equal .

Since the area of is 7, and is , the height

of from to *R*
must be .

Since the triangle formed by ,
the altitude from *A* to ,
and the length from the altitude to *P*
is similar to the triangle formed by ,
the altitude from *R* to ,
and the length from the altitude to *P*,
the ratio of the lengths of the altitudes from *R* to and from *A*
to must be equal to the ratio of the lengths from
the altitudes to *P*. The length of the altitude from *R* to is ,
and the length of the altitude from *A*
to is *q*. Therefore, since the distance from the
altitude from *A* to to point *P*
is *p*, the distance from the altitude
form *R* to to point *P*
must be .

Thus the coordinates of *A, B, S, *and *R* are *A*(*p*, *q*),
*B*( *,0),
**S *(
,
0), and *R*( ,
), with the origin at *P*.

The equation of the line is

, or

, or

, or

The equation of the line is

, or

, or

, or

Lines and intersect at point *C*.

So, for *C*(*x*, *y*),

and , so

, or

, or

, or

, or

, or

So, point *C* lies 77*q* units below
the *x*-axis.

The area of is . We know that *BS * is ( ), and *h*
is (77

So, point *C* lies 77*q* units below
the *x*-axis.

The area of is . We know that *BS * is ( ), and *h*
is (77*q*).

So,

(It is independent of *p*, *q*)

** **

**The
area of is 858.**

** **