**Find, with proof, the smallest positive integer n for
which the sum of the digits of 29n is as small as possible.**

Let *m*
= 29*n*

Since 29 is prime
to 10, the units digit of *m*
must be greater than 0. Also, the
leftmost digit of *m* by definition must be greater than 0.

So, the sum of the digits of *m* is greater than or
equal to 2. In fact, the sum of the
digits of *m* is 2 if and only if *m* is of the form (10* ^{k}*
+ 1).

We see that such a case does exist when *k* = 14. That is, 10^{14} + 1 is divisible by
29.

(See note below we can use Fermat’s Little Theorem to help
find this value.)

**Thus, when n = (10^{14} + 1) / 29 =
3448275862069, we get m = 10^{14} + 1, and the sum of the digits
is 2.**

We have checked that for all values of *k* < 14, 10* ^{k}*
+ 1 is not divisble by 29. Thus,

Note:

Since 29 is prime, from Fermat’s Little Theorem, we know that:

10^{28} ≡ 1 (mod
29).

Let *h* be the smallest positive integer where

10* ^{h}* ≡ -1 (mod
29)

Then, 10^{2h} = 1 (mod 29)

So, 28 must be divisible by 2*h*, or 14 must be
divisble by *h*.

We check the possible values of *h* 1, 2, 7, and 14,
and find that *h* = 14 is the smallest number for which

10* ^{h}* ≡ -1 (mod
29)

**For four integer values of n greater than six, there
exist right triangles whose side lengths are integers equivalent to 4, 5, and 6
modulo n, in some order. Find those
values. Prove that at most four such
values exist. Also, for at least one of
those values of n, provide an example of such a triangle.**

Let the triangle side lengths be:

*an* + 4

*bn* + 5

*cn* + 6

where *a*, *b*, and *c* are all nonnegative
integers.

We have three cases, depending on which of these sides is the hypotenuse:

1: (*an* + 4)^{2} + (*bn*
+ 5)^{2} = (*cn* + 6)^{2}

2: (*an* + 4)^{2} + (*cn*
+ 6)^{2} = (*bn* + 5)^{2}

3: (*bn* + 5)^{2} + (*cn*
+ 6)^{2} = (*an* + 4)^{2}

Modulo *n*, these reduce to:

*n*)

*n*)

*n*)

or

1: 5 ≡ 0 (mod *n*)

*n*)

*n*)

** **

Since *n* > 6, the possible values of *n* are
(from 2) 9 and 27 or (from 3) 9, 15, and 45.

** **

**Thus, the possible values of n are 9, 15, 27, and
45.**

(An example of such a right triangle, where *n* = 9, is
the Pythagorean triplet {24, 32, 40}

(or 8 × {3, 4, 5}). In this case, 24 =
(2(9) + 6), 32 = (3(9) + 5), and 40 = (4(9) + 4).)

**Find a nonzero polynomial f(w, x, y, z) in the four
indeterminates w, x, y, and z of minimum degree such that switching any two
indeterminates in the polynomial gives the same polynomial except that its sign
is reversed. For example, f(z, x, y, w)
= -f(w, x, y, z). Prove that the degree
of the polynomial is as small as possible.**

In the example above, indeterminates *w* and *z*
are switched, and we have

*f*(*z*, *x*, *y*, *w*)
= - *f*(*w*, *x*, *y*, *z*).

So, for ,

when *w* = *z*, ,
or .

Thus, (*w* *z*) must be a factor of .

Similarly, when *w* = *x*, when *w* = *y*,
when *x* = *y*, when *x* = *z*, and when *y* = *z*,
= 0.

Thus,

, , , , , and must all be factors of .

So, a polynomial with this property with minimum degree has factors

, , , , , and , and can be expressed as:

.

**For each nonnegative integer n define the function f _{n}(x)
by**

**f _{n}(x)
= sin^{n} (x) + sin^{n} (x + 2π/3) + sin^{n} (x + 4π/3)**

**for all real numbers x, where the sine function uses
radians. The functions f _{n}(x)
can be also expressed as polynomials in sin (3x) with rational
coefficients. Find an expression for f_{7}(x)
as a polynomial in sin (3x) with rational coefficients, and prove that it holds
for all real numbers x.**

*f*_{7}(*x*) is defined as:

Expanding this, we have

or,

We know that

Since , and ,

**Triangle
ABC is an obtuse isosceles triangles with the property that three squares of
equal size can be inscribed in it as shown on the right. The ration AC / AB is an irrational number
that is the root of a cubic polynomial.
Determine that polynomial.**

Let *m**BAC*
= α (0° <
α <
45°), and let the side length of one of the inscribed squares

equal 1.

Then, *AC* = 1
+ 2 cot α

and *AB* = 1 + cot α

*AC* / *AB* = *x* = 2 cos α

, or

So,

Assuming that *x* ≠ 2 (if *x* = 2, α = 0), we
have:

**Thus, x = AC / AB is a root of 2x^{3}
2x^{2} 3x + 2 = 0.**