Download Potpourri - Blaine School District

2011 Northwest High School Math Championship
Potpourri Test Solutions
1. Since 32 · 31 = 992, the answer is a multiple of 992. The only multiple of 992 in the interval
from 1500 to 2500 is 1984. The sum of the digits of 1984 is 22 .
2. Note that n is odd and is a multiple of 5. Therefore the units digit of n is 5 .
3. Since x2 − y 2 = (x + y) (x − y) = 100, we get x − y =
(x + y)2 − (x − y)2 = 252 − 42 = 609 .
100
25
= 4. Thus we have 4xy =
4. Adding together the areas of the squares, the total area of the rectangle is 22 + 52 + . . . +
332 + 362 = 4209 = 3 · 23 · 61. Since the side-length of one of the squares is 36, each dimension
of the rectangle must be at least 36. Thus the dimensions of the rectangle must be 69 by 61,
making its perimeter 260 .
5. Every factor n ≤ 5 can be paired with the factor m = 30
so that mn = 30 in each pair. Thus
n
the product of all eight factors is 304 = (1 · 30) (2 · 15) (3 · 10) (5 · 6), so k = 4 .
6. Since x2 − 9 = (x − 3) (x + 3), either x + 3 = ±1 or x − 3 = ±1. Thus the four values of x,
each of which yields a prime, are ±2, ±4. The product of these values of x is 64 .
7. The sum of the ten numbers is 630, and the sum of six of these numbers is 57 · 6 = 342. Thus
the sum of the other four numbers is 288, and therefore their average is 72 .
8. The sum of the lengths of any two sides of a triangle must be greater than the length of
the third side. Consequently, the longest side must be less than the sum of the other sides.
Starting with the longest side, and writing the other sides in decreasing order, the only possible
triples of integers are (7, 7, 1) , (7, 6, 2) , (7, 5, 3) , (7, 4, 4) , (6, 6, 3) , (6, 5, 4), and (5, 5, 5). Thus
the desired number of triangles is 7 .
9. Let a = f (4) and b = f ( 14 ). Setting x = 4 yields a + 5b = 7. Setting x = 14 yields b + 5a = 13
.
4
37
Solving these two equations yields a = 96 , which means p + q = 133 .
!
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10. Let a = x − x1 and b = 1 − x1 . Then a + b = x and (a + b) (a − b) = a2 − b2 = x − 1.
Therefore a − b = x−1
. Adding yields 2a = x + √x−1
= 1 + a√2 , which says that (a − 1)2 = 0.
x
x
Therefore a = 1, so x− x1 = 1. This yields x = 1±2 5 . Only 1+2 5 satisfies the original equation.
Thus the number of real solutions to the given equation is 1 .