SW-ARML
9-25-10
Problem (I) 1. Compute the number of positive integers less than 25 that cannot be written
as the difference of two squares of integers.
Solution. Suppose n = a2 – b2 = (a + b)(a – b), where a and b are integers. Because a + b and
a – b differ by an even number, they have the same parity. Thus n must be expressible as the
product of two even integers or two odd integers. This condition is sufficient for n, since
knowing a + b and a – b would allow us to solve for a and b. Hence the desired integers in the
given range are 2, 6, 10, 14, 18, and 22, for a total of 6 values.
Problem (I) 2. For digits A, B, and C, (AB)2 + (AC)2 = 1313. Compute A + B + C.
Solution. Note that (AB)2 + (AC)2 > 1313 if the tens digit A ≥ 3 and (AB)2 + (AC)2 < 1313 if the
tens digit A = 1. So, A = 2. Note that B and C must have opposite parity, so without loss of
generality, assume that B is even. Consider the numbers modulo 10: for any integer n, n2  0, 1,
4, 5, 6, or 9 mod 10. The only combination whose sum is congruent to 3 mod 10 is 4 + 9. So B =
2 or 8 and C = 3 or 7. Checking cases shows that 282 + 232 = 1313, so B = 8;C = 3, and A + B +
C = 13.
Problem (I) 3. For real numbers , B, and C, the zeros of T(x) = x3 +x2 + Bx + C are sin2,
cos2, and –csc2. Compute T(5).
Solution. Use the sum of the roots formula to obtain sin2 + cos2 + (–csc2) = –1, so csc2 = 2
1
1
and sin2 = . Therefore cos2 = . T(x) has leading coefficient 1, so by the factor theorem,
2
2
1
1
1 
1
567




T(x) =  x   x    x  2  and T(5) =  5    5    5  2  
.
2 
2
4
2 
2


Problem (T) 4. Compute all ordered pairs of real numbers (x, y) that satisfy both of the
equations: x2 + y2 = 6y – 4x + 12 and 4y = x2 + 4x + 12.
Solution. Rearrange the terms in the first equation to yield x2 + 4x + 12 = 6y – y2 + 24, so that
the two equations together yield 4y = 6y – y2 + 24, or y2 – 2y – 24 = 0, from which y = 6 or y =
–4. If y = 6, then x2 + 4x + 12 = 24, from which x = –6 or x = 2. If y = –4, then x2 + 4x + 12 =
–16, which has no real solutions because x2 + 4x + 12 = (x + 2)2 + 8 ≥ 8 for all real x. So there
are two ordered pairs satisfying the system, namely (–6; 6) and (2; 6).
1
Problem (T) 5. Define log*(n) to be the smallest number of times the log function must be
iteratively applied to n to get a result less than or equal to 1. For example, log*(1000) = 2 since
log 1000 = 3 and log(log 1000) = log 3 = 0.477…  1. Let a be the smallest integer such that
log*(a) = 3. Compute the number of zeros in the base 10 representation of a.
Solution. If log*(a) = 3, then log(log(log(a)))  1 and log(log(a)) > 1. If log(log(a)) > 1, then
log(a) > 10 and a > 1010. Because the problem asks for the smallest such a that is an integer,
choose a = 1010 + 1 = 10,000,000,001, which has 9 zeros.
Problem (T) 6. The six sides of convex hexagon A1A2A3A4A5A6 are colored red. Each of the
diagonals of the hexagon is colored either red or blue. Compute the number of colorings
such that every triangle AiAjAk has at least one red side.
Solution. Only two triangles have no sides that are sides of the original hexagon: A1A3A5 and
A2A4A6. For each of these triangles, there are 23 – 1 = 7 colorings in which at least one side is red,
for a total of 7·7 = 49 colorings of those six diagonals. The colorings of the three central
diagonals A1 A4 ; A2 A5 ; A3 A6 are irrelevant because the only triangles they can form include sides
of the original hexagon, so they can be colored in 23 = 8 ways, for a total of 8 · 49 = 392
colorings.
Problem (T) 7. Given an arbitrary finite sequence of letters (represented as a word), a
subsequence is a sequence of one or more letters that appear in the same order as in the
original sequence. For example, N, CT, OTT, and CONTEST are subsequences of the word
CONTEST, but NOT, ONSET, and TESS are not. Assuming the standard English alphabet
{A, B, …, Z}, compute the number of distinct four-letter “words” for which EE is a
subsequence.
Solution. Divide into cases according to the number of E's in the word. If there are only two E's,
 4
then the word must have two non-E letters, represented by ?'s. There are    6 arrangements of
 2
two E's and two ?'s, and each of the ?'s can be any of 25 letters, so there are 6 · 252 = 3750
possible words. If there are three E's, then the word has exactly one non-E letter, and so there are
4 arrangements times 25 choices for the letter, or 100 possible words. There is one word with
four E's, hence a total of 3851 words.
Problem (R) 8. If A, R, M, and L are positive integers such that A2 +R2 = 20 and M2 +L2 = 10,
compute the product A · R ·M · L.
Solution. The only positive integers whose squares sum to 20 are 2 and 4. The only positive
integers whose squares sum to 10 are 1 and 3. Thus A · R = 8 and M · L = 3, so A · R ·M · L = 24.
2
Problem (R) 9. Compute the last digit, in base 10, of the sum
T2 + (2T)2 + (3T)2 + … + (T2)2 and write the answer in terms of T.
Solution. Let S be the required sum.
T (T  1)(2T  1) T 3 (T  1)(2T  1)
S  T 2 12  22   T 2   T 2

.
6
6
Problem (R) 10. A fair coin is flipped n times. Compute the smallest positive integer n for
which the probability that the coin has the same result every time is less than 10%.
Solution. After the first throw, the probability that the succeeding n –1 throws have the same
1
1
1
result is n1 . Thus n1 
 2n-1 > 10  n – 1 ≥ 4, so n = 5 is the smallest value.
2
10
2
Problem (R) 11. Let T = TNYWR. Compute the smallest positive integer n such that there are
at least T positive integers in the domain of f(x) =
 x 2  2 x  n and write answer in terms of T.
Solution. Completing the square under the radical yields
n  1  ( x  1) 2 . The larger zero of
the radicand is 1  n  1 > 0 and the smaller zero is 1  n  1  0 , so the T positive integers
in the domain of f must be 1, 2, 3, …, T. Therefore, 1  n  1  T . Hence
n + 1 ≥ (T + 1)2. Therefore, n ≥ T2 + 2T and the smallest is T2 + 2T.
n  1  T  1 , and
Problem (R) 12. Let T = TNYWR which is positive. Compute the smallest positive real number
 x
x such that    T
x   x 
Solution. If
 x 
 T , the equation can be rewritten as follows:
x   x 
x   x  1
x
1 x
1
T 1
 ,
1  ,
 1 
1. Now that 0 < x < 1 is impossible, since it
T
T  x  T
T
 x 
 x 
would make the numerator of the original expression 0. x = 1 is also impossible because it
would make the denominator of the original expression 0. To make x as small as possible, place
T 1
it in the interval 1 < x < 2, so that  x  = 1. Then x =
.
T
3