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MO-ARML – 10/19/14
Solutions to Practice Questions
1. The interior angles of a convex polygon form an arithmetic progression with a common
difference of 4. It’s known that the largest interior angle is 172. Find the number n of sides of
this polygon.
n
 344  4n  4   180(n  2) .
2
 174n  2n 2  180n  360 , 2n2 + 6n – 360 = 0, n2 + 3n – 180 = 0, (n – 12)(n + 15) = 0
 n = 12
The interior angles are 172, 168, …., 172 – 4(n – 1). Hence,
2. Determine the number of positive integers n such that a regular convex polygons with n sides
whose interior angles, when measured in degrees, are integers.
180( n  2)
360
is an integer. That is, 180 
is an
n
n
integer, and thus n is a divisor of 360. Since 360 = 23325, it has (3 + 1)(2 + 1)(1 + 1) = 24
divisors. However, there are no polygons with 1 side or 2 sides, so there are 22 possible n.
We are looking for positive integer n such that
3. Find the unique ordered pair (x, y) such that (4x2 + 6x + 4)(4y2 – 12y + 25) = 28.
2

3 
9 
2
28 = (4x2 + 6x + 4)(4y2 – 12y + 25) =   2 x     4     2 y  3   25  9 


2 
4 

7
3
 3 3 
, .
 16  28  2 x   0 and 2y – 3 = 0  (x, y) = 
4
2
 4 2

4. If

cos 3 x 1
sin 3 x
 , determine
.
cos x 3
sin x
sin 3 x cos 3 x sin 3 x cos x  sin x cos x sin(3 x  x)



2
1
sin x cos x
sin x cos x
sin 2 x
2
sin 3 x
1 7
 2  .
Hence,
sin x
3 3
5. For any positive integer n, let s(n) be the sum of the digits of n in base 10 representation. Find the
largest n such that n = 7s(n).
Assuming n = ad ad-1ad-2a2a1a0 in base-10. Then,
ad 10d  ad 110d 1   a110  a0  7  ad  ad 1   a1  a0 
ad 10d  7   ad 1 10d 1  7  
 a2 100  7   a1 10  7   6a0  54  d 1
3a1 = 6a0  a1 = 2a0  the largest possible a1 is 8 and the largest n is 84.
1
6. Let f (x) be a polynomial of degree 14 such that f  k   for all k = 1, 2, …, 15. Determine
k
f 16 .
1
1
 0 has 15 roots, but f  x   is not a polynomial, so consider
x
x
p(x) = xf(x) – 1 instead. It’s a polynomial of degree 15 with roots x = 1, 2, 3, …, 15. Hence,
xf(x) – 1 = p(x) = a(x – 1)(x – 2)(x – 3)(x – 14)(x – 15) for some constant a.
1
1
15 14 13  2 1  1 .
Let x = 0  1 = a(1)15!  a =
. Then, 16f(16) – 1 =
15!
15!
2 1
 .
Hence, f 16  
16 8
Note that f  x  
7. The polynomial p(x) = x5 + x2 + 1 has roots r1, r2, r3, r4, and r5. Let q(x) = x2 – 2. Compute
q(r1)q(r2)q(r3)q(r4)q(r5).
Note that p(x) = (x  r1) (x  r2)(x  r3)(x  r4)(x  r5), and
q(r1)q(r2) q(r3) q(r4) q(r5) = (r12 – 2) (r12 – 2)(r12 – 2)(r12 – 2)(r12 – 2)












= r1  2 r1  2 r2  2 r2  2 r3  2 r3  2 r4  2 r4  2 r5  2 r5  2
=

2  r1
=p

2  r2

2  r3
 2  p  2   4

2  r4






2  r5  2  r1  2  r2  2  r3  2  r4  2  r5

2  3 4 2  3  9  32  23
8. Given two positive integers m and n, let (m n) denote the concatenation of m by n, which is
obtained by appending the digits of n to the end of m’s. For example, (123 45) = 12345. Find 3digit positive integers m and n such that 6(m n) = (n m).
Since both m and n are 3-digit, we have 6(1000m + n) = 1000n + m.
 5999m = 994n, 857m = 142n  m = 142, n = 857. (857142 = 6142857)
9. Let an = 6n + 8n. Determine the remainder upon dividing a99 by 49.
A99 = 699 + 899 = (7 – 1)99 + (7 + 1)99 = 49A + 2997  217 = 14 (mod 49)
9 x 2 sin 2 x  4
for 0 < x < /2.
x sin x
4
Let y = x sin x, so we are minimizing 9 y  . By the Arithmetic-Mean-Geometric-Mean
y
10. Find the minimum value of
4
4
4
 2 9 y   2 36  12 , and the equality holds when 9y  , that is, when
y
y
y
2
4
π 2
y  , implying the minimum value 9 y  is 12. Note that x sin x x0  0, x sin x x π   , so
3
y
2 3
2
2
π
x sin x  has a solution in 0  x  by the Intermediate Value Theorem.
3
2
Inequality, 9 y 
