USAMTS Problem 1/1/12

 

Determine the smallest five-digit positive integer N such that 2N is also a five-digit integer and all ten digits from 0 to 9 are found in N and 2N.

 

First, we assume that a five digit number is greater than 10000; it does not start with 0. 

 

We want to find the smallest such number, and for this we will start with the smallest possible numbers starting from the leftmost digits of N and 2N.  Let us start with 1 as the first digit of N.  This makes the first digit of 2N  2 or 3 and 2 is the smaller of these.

 

0 cannot be a digit of N because then its corresponding digit in 2N would either be 0 or 1, both of which are already taken. 

 

The next smallest number we can use for the second digit of N is 3.

 

The next smallest number that we can use for the third digit of N is 4.  Now that we have no more numbers less than 5 that we can use for N, there must be a carryover onto the third digit of 2N, making it be 9 instead of 8.  By now, our numbers look like this:

 

 N = 134??

2N = 269??

 

We still have not used numbers 5, 7, 8, and 0. (Remember, 0 cannot be used for a digit in N). The next smallest number we can use for the fourth digit of N is 5.  However, there will be a carryover because there are no more available digits less than five, and the fourth digit of 2N would be 1, which we have already used. 

 

The next smallest number we can use for the fourth digit of N is 7.  The fourth digit of 2N would then be 5.  However, this leaves us with the numbers 8 and 0 for the last digits of N and 2N, and 28 = 16, whose last digit is not 0. 

 

So, our next smallest number which we can use for the fourth digit of N is 8.  This would mean that the fourth digit of 2N is 7.   This leaves only 5 for the last digit of N.  52 = 10.  0 would be the last digit of 2N, and everything fits together perfectly.  So, the smallest numbers N and 2N such that the digits 0 through 9 are used exactly once are:

 

 N = 13485

2N = 26970

 

Note: I used cards numbered 0-9, and found one can systematically carry out this method in seconds.  Actually, there are many more numbers N and 2N that satisfy the given requirements.  In fact, if we consider numbers that begin with 0 five-digit numbers, then, by using the above procedure, we find that the smallest numbers N and 2N are:

 

 N = 06729

2N = 13459

USAMTS Problem 2/1/12

 

It was recently shown that 2²²⁴ + 1 is not a prime number.  Find the four rightmost digits of this number.

 

It is pretty easy to see that if a ≡ b (mod 10000), then a² ≡ b² (mod 10000).  In other words, to obtain the rightmost four digits of x², you don’t have to take the square of x.  Instead, you just have to take the rightmost four digits of the square of the rightmost four digits of x.  To prove this, let

 

x = 10000z + a,

 

where a is the last four digits of x, and z is a non-negative integer.  Then:

 

x² = (10000z + a)²

x² = 10000²z² + 210000a + a²

 

Since the first two terms of the right hand side are divisible by 10000, they are not part of the last four digits of x².  So, the last four digits of a² are the same as the last four digits of x².

 

With this in mind, we can repeatedly take the squares of the last four digits, starting with 2 as shown in the table below.  After 24 steps, we add 1, thus finding the last four digits of 2²²⁴ + 1.

 

 

22x ≡	... (mod 10000)
21 ≡	2
22 or 22 ≡	4
24 or 42 ≡	16
28 or 162 ≡	256
224 or 2562 ≡	5536
225 or 55362 ≡	7296
226 or 72962 ≡	1616
227 or 16162 ≡	1456
228 or 14562 ≡	9936
229 or 99362 ≡	4096
2210 or 40962 ≡	7216
2211 or 72162 ≡	656
2212 or 6562 ≡	336
2213 or 3362 ≡	2896
2214 or 28962 ≡	6816
2215 or 68162 ≡	7856
2216 or 78562 ≡	6736
2217 or 67362 ≡	3696
2218 or 36962 ≡	416
2219 or 4162 ≡	3056
2220 or 30562 ≡	9136
2221 or 91362 ≡	6496
2222 or 64962 ≡	8016
2223 or 80162 ≡	6256
2224 or 62562 ≡	7536

 

2²²⁴ ≡ 7536 (mod 10000), so 2²²⁴ + 1 ≡ 7537 (mod 10000).

USAMTS Problem 3/1/12

 

Determine the integers a, b, c, d, and e for which

(x² + ax + b)(x³  + cx² + dx + e) = x⁵  9x  27 .

 

Given (x² + ax + b)(x³  + cx² + dx + e) = x⁵  9x  27:

x⁵ + (a + c)x⁴ + (b + ca + d)x³ + (bc + ad + b)x² + (bd + ea)x + be

= x⁵  9x  27

 

By comparing the coeficients of both sides, we get five equations:

a + c = 0                (1

b + ca + d = 0        (2

bc + ad + b = 0      (3

bd + ea = -9                     (4

be = -27                 (5

 

1) gives:

c = -a                     (6

 

2) gives:

b  a² + d = 0

or d = a²  b                  (7

 

3) gives:

-ba + a(a²  b) + e = 0

or e = ba  a³ + ab

or e = 2ab  a³                         (8

 

4) gives:

b(a²  b) + a(2ab  a³) = -9

or a²b  b² + 2a²b  a⁴ + 9 = 0

or a⁴  3a²b + (b²  9) = 0                (9

 

Solving for a²

a² = (3b ± √(9b²  4b² + 36))/2

or a² = (3b ± √(5b² + 36))/2              (10

 

 

From 5) we get:

be = -27

, and since b and e are integers, b (and e) must be one of the following: ±1, ±3, ±9, or ±27.

 

To find out which one will produce integer a, we will test each possibillity by seeing if (3b ±  /2 is a perfect square

and for that we check first (5b² + 36) is a perfect square.

 

When b = ±1, 5b² + 36 = 41, which is not a perfect square.

 

When b = ±3, 5b² + 36 = 81, which is a perfect square, a². (3b ±  /2 = ±9 or 0, both of which are perfect squares

 

When b = ±9, 5b² + 36 = 441, which is a perfect square.  However,

(3b ±  /2 = ±24 or ±3, neither of which are perfect squares.

 

When b = ±27, 5b² + 36 = 3681, which is not a perfect square.

 

The only possible values for b are 3 and -3.

 

If b = -3, a² = -9 or 0

a² = -9 is impossible because a is a real number.

a² = 0 is also impossible because:

8) gives e = 2ab  a³ = 0

or be = 0

but from 5) we know that be = -27.

So b cannot equal -3

 

 

Our only possibility is b = 3.

If b = 3, a² = 9 or 0

We know a² = 0 is impossible.

a² = 9 is perfectly valid.

a = 3 or -3

 

 

 

b = 3

 

7) gives:

d = a² - b

or d = 6

 

5) gives:

e = -27/b

or e = -9

 

4) gives:

bd + ea = -9

or 18 + -9a = -9

or a = 3

 

1) gives:

c = -a

or c = -3

 

Checking with our original five equations:

(1) 3 + (-3) = 0  true

(2) 3 + + 6 = 0  true

(3) (-9) + 18 + (-9) = 0  true

(4) 18 + (-27) = -9  true

(5) 3(-9) = -27  true

 

a = 3

b = 3

c = -3

d = 6

e = -9

 

Thus: (x² + 3x + 3)(x³   3x² + 6x  9) = x⁵  9x  27

USAMTS Problem 4/1/12

 

A sequence of real numbers s₀, s₁, s₂, ... has the property that

s_is_j = s_{i + j} + s_{i  j}   for all nonnegative integers i and j with i ≥ j,

s_i = s_{i + 12}               for all nonnegative integers i, and

s₀ > s₁ > s₂ > 0.

Find the three numbers s₀, s₁, and s₂.

 

Given:

s_is_j = s_{i + j} + s_{i  j}    (1)

s_i = s_{i + 12}               (2)

s₀ > s₁ > s₂ > 0       (3)

 

When i = j = 0, (1) gives

s₀s₀ = s₀ + s₀

or (s₀)² = 2(s₀)

Because s₀ > 0, we can divide both sides by s₀, getting

s₀ = 2          (4)

 

We know that s₁₂ = s₀  (from (2))

 

When i = j = 6

(s₆)² = s₁₂ + s₀ = 2 + 2 = 4

s₆ = ±2

 

Now     (s₃)² = s₆ + s₀ = ±2 + 2 =  {4 or 0}

Thus      s₃ = ±2 or 0                 (5)

 

Now      s₂s₁ = s₃ + s₁

or s₃ = s₂s₁  s₁

s₃ = s₁(s₂  1)      (6)

 

Since        0 < s₂ < s₁ < 2

|s₁| < 2

|s₂  1| < 1

so |s₃| < 2

 

Thus, for s₃ ,   0 is the only possible value from (5),

s₃ = 0           (7)

 

  Since s₁ ≠ 0,  (6) gives

s₂ = 1

 

Now, putting i = j = 1 in (1), we get

(s₁)² = s₂ + s₀ = 1 + 2 = 3

s₁ =  as s₁ > 0

 

s₀ = 2

s₁ =  

s₂ = 1

 

Note: s₃ = 0.  We can easily calculate all values for s_i

eg: s₁ = s₁₁ = , s₂ = s₁₀ = 1, s₃ = s₉ = 0, s₄ = s₈ = -1, s₅ = s₇ = - , and s₆ = -2. 

Since s_i = s_{i + 12}, we can find all values of s_i where i is an integer.

USAMTS Problem 5/1/12

 

 

In the octahedron shown to the right, the base and top faces are equilateral triangles with sides measuring 9 and 5 units, and the lateral edges are all of length 6 units.  Determine the height of the octahedron; i.e., the distance between the base and the top face.

 

 

The given octahedron ABCDEF contains two equilateral triangles: equilateral ΔABC with sides measuring 5 units and equilateral ΔDEF with sides measuring 9 units.  Lateral edges AE, AF, BD, BF, CD, and CE are congruent and have a measure of 6 units.

 

Let the midpoint of BC be G, the center of ΔABC be O (since it is an equilateral triangle, the circumcenter is the same as the incenter), and the center of ΔDEF be P.

 

In an equilateral triangle (say, ΔABC) with center O and sides measuring a, an altitude AG = .

 

Since AO = BO = CO = 2OG

OG =                                (1)

OB = OA = OC =              (2)

 

We see that GD' = OD'  OG

 

and we also know that ΔCGD' is a right triangle,

so CD'² = CG² + GD'²

 

Also, ΔCDD' is a right triangle with side CD measuring 6,

so the height DD' =  

 

 

Now:

OG =                               [Using (1)]

 

(for ΔD'E'F') OD' =         [Using (2)]

 

Thus GD' = OD'  OG =      

 

We know that CG = 5/2, so 

CD'² = CG² + GD'² = (5/2)² +  = 5²/2² + (13²3)/6²

= 61/3

 

Thus, height DD' =  

=