USAMTS Problem 1/3/14

The integer n, between 10000 and 99999, is abcde when written in decimal notation.  The digit a is the remainder when n is divided by 2, the digit b is the remainder when n is divided by 3, the digit c is the remainder when n is divided by 4, the digit d is the remainder when n is divided by 5, and the digit e is the remainder when n is divided by 6.  Find n.

The digit a is the remainder when n is divided by 2; a can be either 0 or 1.  Because
n > 10000, a must be 1.

The digit b is the remainder when n is divided by 3; b can be 0, 1, or 2.  Furthermore, because the sum of the digits of a number modulo 3 is the same as that number modulo 3,
a + b + c + d + eb  (mod 3), or

(a + c + d + e) is divisible by 3

The digit c is the remainder when n is divided by 4, so it must be congruent to a modulo 2; that is, c can be either 1 or 3.  Furthermore, because 100 is divisible by for, we can take

c as the remainder when the two digit number de is divided by 4.

The digit d is the remainder when n is divided by 5; d can be 0, 1, 2, 3, or 4.  Also, d must be congruent to e modulo 5.

The digit e is the remainder when n is divided by 5; because a is 1, e must be odd.  Furthermore, divisibility by 6 is dependent upon divisibility by 2 and 3; thus e must be congruent to (a)(b), that is, b modulo 3.  e can be 1, 3, or 5.

a is 1.

If e is 1, then we have

1bcd1

d and b must also be 1.  Now de = 11, so c = 3.

n = 11311.

If e is 3, then d must be 3, and so c must be 1.  In this case, b must be 2.  But b must equal e mod 3, or 0. Thus, e cannot be 3.

If e is 5, then d must be 0, and so c must be 1.  In this case, b must be 1.  But b must equal e mod 3, or 2.  Thus, e cannot be 5.

Thus, e = 1, and n = 11311.

USAMTS Problem 2/3/14

Given positive integers p, u, and v such that u2 + 2v2 = p, determine, in terms of u and v, integers m and n such that 3m2  2mn + 3n2 = 24p.  (It is known that if p is any prime number congruent to 1 or 3 modulo 8, then we can find integers u and v such that
u
2 + 2v2 = p.)

We know that, if

x = a2 + 2b2

y = c2 + 2d2

then xy can be written in the form k2 + 2l2:

We can write this simply as

(The above method, and many other interesting properties of sums of squares can be found in Mathematical Recreations by Maurice Kraitchik.)

p can be written as u2 + 2v2, and 24 can be written as 42 + 2(2)2

Thus,

We can write  as

Now,

So,

or

or

is symmetrical with respect to m and n, so m and n can be interchanged.

So,  and n = 3u      or         and n = 3u.

USAMTS Problem 3/3/14

Determine, with proof, the rational number  that equals

.

The nth term of this series can be written as

=

=

=

=

=

If we sum consecutive terms

we notice that all of the “middle” terms of our sum cancel each other out.

Thus, the sum

simplifies to

The sum of the series is .

USAMTS Problem 4/3/14

The vertices of a cube have coordinates (0, 0, 0), (0, 0, 4), (0, 4, 0), (0, 4, 4), (4, 0, 0),
(4, 0, 4), (4, 4, 0), and (4, 4, 4).  A plane cuts the edges of this cube at the points (0, 2, 0), (1, 0, 0), (1, 4, 4), and two other points.  Find the coordinates of the other two points.

We can write the equation of a plane as Ax + By + Cz + D = 0.  The plane that passes through points (0, 2, 0), (1, 0, 0), and (1, 4, 4) is not parallel to the x axis, so A ≠ 0.

So, we can write the equation of this plane as  x + ay + bz + c = 0 , where the coefficient of x is 1.  We know three points of this plane, so we can find the values of a, b, and c:

(0) + a(2) + b(0) + c = 0

(1) + a(0) + b(0) + c = 0

(1) + a(4) + b(4) + c = 0

2a + c = 0

1 + c = 0

1 + 4a + 4b + c = 0

c = -1

a =

b = (1 + 2  1) / 4 = -

So,

x + y  z  1 = 0 , or

2x + y  z  2 = 0

is the equation of the plane that intercepts points (0, 2, 0), (1, 0, 0), and (1, 4, 4).

The edges of the given cube are defined by the lines {x = 0, y = 0}, {x = 0, y = 4},
{
x = 4, y = 0}, {x = 4, y = 4}, {x = 0, z = 0}, {x = 0, z = 4}, {x = 4, z = 0}, {x = 4, z = 4},
{
y = 0, z = 0}, {y = 0, z = 4}, {y = 4, z = 0}, and {y = 4, z = 4}, where 0 < x, y, z < 4.

The plane  2x + y  z  2 = 0  intercepts:

2 = 0  intercepts:

Line {x = 0, z = 0} at (0, 2, 0),

Line {y = 0, z = 0} at (1, 0, 0),

Line {y = 4, z = 4} at (1, 4, 4)

It intercepts the line {x = 0, y = 4} at (0, 4, 2),

and the line {y = 0, z = 4} at (3, 0, 4)

The plane does not intercept lines {x = 0, y = 0}, {x = 4, y = 0}, {x = 4, y = 4}, {x = 0, z = 4}, {x = 4, z = 0}, {x = 4, z = 4}, and {y = 4, z = 0} within the domain of the cube.

USAMTS Problem 5/3/14

A fudgeflake is a planar fractal figure with 120º rotational symmetry such that three identical fudgeflakes in the same orientation fit together without gaps to forma  larger fudgeflake with its orientation 30º clockwise of the smaller fudgeflakes’ orientation, as shown on the right.  If the distance between the centers of the original three fudgeflakes is 1, what is the area of one of these three fudgeflakes?  Justify your answer.

Fudgeflakes have 120º rotational symmetry, and can tessellate together in a triangular pattern.  If we take a large number of fudgeflakes and arrange them in such a tessellation, we can approximate a triangular area.

Suppose we create such an array, and place markers at the center of each fudgeflake.  If we remove the fudgeflakes, we will have a triangular pattern of markers spaced 1 unit apart from one another.

On each of these markers we can replace a fudgeflake with another tessellating figure with 120º rotational symmetry, with centers spaced 1 unit apart from each other.

If our triangular region is large, the ratio of the area of the fudgeflakes to the total area of the triangular region is approximately 1, and can be brought within any desired accuracy close to 1 by increasing the size of the triangular region.  Similarly, the ratio of the area of our new, symmetrical, tessellating figures to the area of the total triangular region approximates 1.  This means that, the area of a tessellating figure with the above properties is independent of its shape.

A regular hexagon has 120º rotational symmetry and can tessellate with centers forming a triangular array, like a fudgeflake.  Thus, a regular hexagon with its center 1 unit away from a neighboring hexagon will have the same area as a fudgeflake which has center 1 unit away from a neighboring fudgeflake.

Such a hexagon has distance from center to side as , and area:

Thus, the area of a fudgeflake is .