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MO-ARML – 9/14/14
Solutions to Practice Questions
1. In a triangle ABC, a = 12, b = 17, and c = 13. Compute b cos C – c cos B
Solution. Use the Law of Cosines.
a 2  b2  c 2
a 2  c2  b2 2b2  2c2 b2  c2
b cos C – c cos B = b 
c


2ab
2ac
2a
a
2
2
17  13
289  169 120
=


 10 .
12
12
12
2. There exists a digit Y such that for any digit X, the seven-digit number 1 2 3 X 5 Y 7 is
not a multiple of 11. Compute Y.
Solution. Consider the ordered pairs of digits (X; Y) for which 1 2 3 X 5 Y 7 is a multiple of 11.
Recall that a number is a multiple of 11 if and only if the alternating sum of the digits is a
multiple of 11. Since 1 + 3 + 5 + 7 = 16, the sum of the remaining digits, namely 2 + X + Y,
must equal 5 or 16. Thus X + Y must be either 3 or 14, making X = 3  Y (if Y = 0, 1, 2, or 3) or
14  Y (if Y = 5, 6, 7, 8, or 9). Thus a solution (X; Y) exists unless Y = 4.
3. Compute the sum of all values of k for which there exist positive real numbers x and y
2
5

 2k  1
 log x y  log y x
satisfying the following system of equations: 
.
log x2 y 5  log y 2 x3  k  3


1
1
Solution. Let a = log x y , and notice that log y x 
, log x2 y  log x y , and
2
log x y
5

 2a  a  2k  1
.

5
3
 a
 k 3
 2
2a
Multiply the second equation and subtract he first equation (to eliminate k) will give us
8
8
a   5 , or equivalently, 3a2 + 5a – 8 = 0, or, (3a +8)(a – 1) = 0. Hence, a = 1 or a 
.
a
3
These are the two cases when the system has solutions for x and y. When a = 1, the first
8
equation yields 7 = 2k – 1, that is, k = 4. When a 
, the first equation yields
3
16 15
173
149
149
  2k  1 , or
 2k  1 ,
 2k , implying k 
. Hence, the sum is
3
8
24
24
48
149 43
149 43
4


= 4
.
48 48
48 48
1
1
log y 2 x  log y x 
. Then, the systems of equation becomes
2
2 log x y
4. For {1, 2, 3, …, n} and each of its non-empty subsets, an alternating sum is defined as follows.
Arrange the number in the subset in decreasing order and then, beginning with the largest,
alternately add and subtract successive numbers. For example, the alternating sum for
{1, 2, 4, 6, 9} is 0 – 6 + 4 – 2 + 1 = 6 and for {5} it is simply 5. Find the sum of all such
alternating sums for n = 7.
Solution. The non-empty subsets of {1, 2, 3, …, 7}, except {7}, can be divided into 26 – 1 = 63
pairs of one not containing 7 and one does, such as {1, 2, 3} and {1, 2, 3, 7}. The alternating
sums of each pair would add up to exactly 7. For example, (3 – 2 + 1) + (7 – 3 + 2 – 1) = 7.
Hence the sum of all alternating sums for n = 7 is equal to 763 + 7 = 764 = 448 (the second
summand is for the singleton set {7}).
5. Let N be the least integer greater than 20 that is a palindrome in both base 20 and base 14.
For example, the three-digit base-14 numeral (13)5(13)14 (representing 13142 + 5 14 + 13
Is a palindrome in base 14, but not in base 20, and the three-digit base-14 numeral (13)3114 is
not a palindrome in base 14. Compute the base-10 representation of N.
Solution. Because N is greater than 20, the base-20 and base-14 representations of N must be at
least two digits long. The smallest possible case is that N is a two-digit palindrome in both
bases. Then N = 20a + a = 21a, where 1  a  19. Similarly, in order to be a two-digit
palindrome in base 14, N = 14b + b = 15b, with 1  b  13. So N would have to be a multiple
of both 21 and 15. The least common multiple of 21 and 15 is 105, which has the base 20
representation of 105 = 5520 and the base-14 representation of 105 = 7714, both of which are
palindromes. Thus the answer is 105.
6. Compute sin24 + sin28 + sin212 +  + sin2176
1  cos 2 x
, sin24 + sin28 + sin212 +  + sin2176
2
1  cos8 1  cos16 1  cos 24
1  cos352
=


 
2
2
2
2
44 1
  cos8  cos16  cos 24   cos 352  . Using complex number
=
2 2
z = cos 8 + i sin 8, we know z45 = 1. That is, z45 – 1 = 0, or (z – 1)(z44 + z43 +  + z2 + z + 1)
= 0. Hence, z44 + z43 +  + z2 + z + 1 = 0. Since, zk = cos 8k + i sin 8k by DeMovire’s
Theorem, we know the real part of z44 + z43 +  + z2 + z + 1, which is 0, is equal to
cos8  cos16  cos 24   cos 352 + 1. Therefore, cos8  cos16  cos 24   cos 352
44 1
45
  1 
= 1, and sin24 + sin28 + sin212 +  + sin2176 =
.
2 2
2
Solution. Since sin2x =
7. Find the area of the region defined by x2  y 2  x  y .
Solution. By symmetry we only have to calculate the area of this region within the first
quadrant, which is now bounded by x 2  y 2  x  y; x  0, y  0 . The first inequality is the
2
2
1 
1 1

same as x  x  y  y  0 or  x     y    .
2 
2
2

Hence we are look for the part of the circle centered
1
1 1
at  ,  with radius
that is in the first quadrant.
2
2 2
2
2
It can be broken into a right isosceles triangle with side
1 1
1
length 1 and a half-circle with radius
. The area together = + π .
2 4
2
1 1 
Hence, the total area of the original region = 4   π   2  π
2 4 
8. Compute the smallest positive integer n such that 214n and 2014n have the same number
of divisors.
Solution. Assume Since 214 = 2107 and 2014 = 21953, they don’t have the same number of
divisors. Multiplying 214 and 2014 by anything relatively prime to both 214 and 2014 will not
make their numbers of divisors the same. Also, 107n and 1953n should have the same number
of divisors. Hence, all prime divisors of n must be from 107’s and/or 1953’s. For convenience,
write n  19a 153b 1107c 1 , where a, b, c  1. Then, 107n  19b 153c 1107 d has ab(c + 1) divisors
and 19  53n  19a 53b107c 1 has (a + 1)(b + 1)c divisors, and these two products are equal.
Hence, ab(c + 1) = (a + 1)(b + 1)c, or equivalently, abc + ab = abc + ac + bc + c,
ab  ac  bc  c = 0.
The goal is to minimize n, so try c = 1. Then, ab – a – b – 1 = 0, (a – 1)(b – 1) = 2, implying
a = 2, b = 3, or a = 3, b = 2. Again, to minimize n, we use a = 3, b = 2, and then
n  19253  19133 . Note that it can be verified that, if c  2, then the possible n’s will be
greater than 19133.
 1000 
9. Compute the greatest integer k  1000 such that 
 is a multiple of 7.
 k 
1000  1000  999  998
Solution. 
. The largest multiple of 7 less than 1000 is 994, but

1 2  3
 k 
1000  1000  1000  999  998 994
is not divisible by 7. The next largest multiple of 7



1 2  3 7
 993   7 
1000  1000  1000  999  998 994 987
is 987, but 
is still not divisible by 7. This


1 2  3 7 14
 986   14 
process produce a factor of 7 until the numerator is a multiple of 49. Since 980 is divisible by
1000  1000  1000  999  998 994 987 980
49, 
is divisible by 7. The greatest


1 2  3 7 14 21
 979   21 
 1000 
integer k  1000 such that 
 is a multiple of 7 is 979.
 k 
10. The equation z6 + z3 + 1 = 0 has complex roots z = r(cos  + i sin ) with argument  between
90 and 180 . Determine the degree measure of .
Solution.
Since (z3 – 1)(z6 + z3 + 1) = z9 – 1, the roots of z6 + z3 + 1 = 0 are among the roots of z9 – 1 = 0
that are not roots of z3 – 1 = 0.
The arguments of the roots of z9 – 1 = 0 in 90 ,180  are 120 ,160 , but 120 is the argument
of a root of z3 – 1 = 0, so the only available  is 160 .
1.