Titu_Andreescu,_Jonathan_Kane_2_006_15_easy_problems_Middle_School

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UW-Whitewater / UT-Dallas Mathematics Meet – April 2006
MIDDLE SCHOOL
© Copyright Titu Andreescu and Jonathan Kane
Problem 1 
Michael is celebrating his fifteenth birthday today. How many Sundays have there been in his
lifetime?
Answer: 783
Michael was born in 2006 – 15 = 1991. Each year since then has had 52 full weeks plus one
extra day except for the four leap years (1992, 1996, 2000, 2004) which had 2 extra days. So,
there have been 52(7)(15) + 15 + 4 + 1 = 52(7)(15) + 20 days in Michael’s lifetime. This gives
52(15) + 2 full weeks plus 6 extra days which accounts for 783 Sundays.
Problem 2 
Evaluate .
Answer: 10
.
Problem 3
Find the sum of all the prime numbers less than 100 which are one more than a multiple of six.
Answer: 527 
The numbers which are one more than a multiple of six are 1, 7, 13, 19, 25, 31, 37, 43, 49, 55,
61, 67, 73, 79, 85, 91, 97. The sum of the prime numbers is 7 + 13 + 19 + 31 + 37 + 43 + 61 +
67 + 73 + 79 + 97 = 527.
Problem 4 
At the beginning of each hour from 1 o’clock AM to 12 NOON and from 
1 o’clock PM to 12 MIDNIGHT a coo-coo clock’s coo-coo bird coo-coos the number of times
equal to the number of the hour. In addition, the coo-coo clock’s coo-coo bird coo-coos a single
time at 30 minutes past each hour. How many times does the coo-coo bird coo-coo from 12:42
PM on Monday until 3:42 AM on Wednesday?
Answer: 279
In one twelve hour period, the coo-coo coo-coos 1 + 2 + 3 + 4 + 5 + þ+ 12 times on the hour and
a total of 12 times 30 minutes past each hour for a total of times. So for the
three twelve hour periods from 12:42 PM on Monday until 12:42 AM on Wednesday, the coo-
coo coo-coos 270 times. Then it will coo-coo 1 time at 1:00, 1 time at 1:30, 2 times at 2:00, 1
time at 2:30, 3 times at 3:00, and 1 time at 3:30 for a total of 9 more times. This gives a final
answer of 279.
Problem 5 
The sizes of the freshmen class and the sophomore class are in the ratio 5:4. The sizes of the
sophomore class and the junior class are in the ratio 7:8. The sizes of the junior class and the
senior class are in the ratio 9:7. If these four classes together have a total of 2158 students, how
many of the students are freshmen?
Answer: 630
This is the same as High School Problem 1.
Problem 6 
We draw a radius of a circle. We draw a second radius 23 degrees clockwise from the first
radius. We draw a third radius 23 degrees clockwise from the second. This continues until we
have drawn 40 radii each 23 degrees clockwise from the one before it. What is the measure in
degrees of the smallest angle between any two of these 40 radii?
Answer: 7
15 times 23 is 345, so the 16 radius drawn is 345 degrees from the first. Thus, the 17 radiusth th
drawn is 345 + 23 – 360 = 8 degrees from the first radius drawn. The following 15 radii drawn
will also be 8 degrees from another radius drawn. Similarly, the 33 radius drawn will be 8rd
degrees from the 17 radius and, therefore, 16 degrees from the first radius. This places it 7th
degrees from the second radius drawn. The remaining seven radii are similarly spaced 7 degrees
from the nearest drawn radii.
Problem 7 
At a movie theater tickets for adults cost 4 dollars more than tickets for children. One afternoon
the theater sold 100 more child tickets than adult tickets for a total sales amount of 1475 dollars. 
How much money would the theater have taken in if the same tickets were sold, but the costs of
the child tickets and adult tickets were reversed?
Answer: 1875
This is the same as High School Problem 2.
Problem 8 
A rogue spaceship escapes. 54 minutes later the police leave in a spaceship in hot pursuit. If the
police spaceship travels 12% faster than the rogue spaceship along the same route, how many
minutes will it take for the police to catch up with the rogues?
Answer: 450
This is the same as High School Problem 4.
Problem 9 
Moving horizontally and vertically from point to point along the lines in the diagram below, how
many routes are there from point A to point B which consist of six horizontal moves and six
vertical moves?
Answer: 132
Label point A and the other points on the bottom row with
the number 1. For each other point, label the point with a
number which is the sum of the numbers on the labels of the
points immediately to the left and immediately below the
point since this will give a count of the number of paths from
A to that point. Continuing this until point B is labeled,
gives the number of paths from A to B which is 132.
The answer is also given by the Catalan number
.
Problem 10 
How many rectangles are there in the diagram below such that the sum of the numbers within the
rectangle is a multiple of 7?
43 44 45 46 47 48 49
36 37 38 39 40 41 42
29 30 31 32 33 34 35
22 23 24 25 26 27 28
15 16 17 18 19 20 21
8 9 10 11 12 13 14
1 2 3 4 5 6 7
Answer: 163
This is the same as High School Problem 9.
Problem 11 
Consider the polynomials 
Find the coefficient of in .
Answer: 0
 
 which contains no x term.4
Problem 12 
How many positive integers divide the number 10! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 ?
Answer: 270
. Thus a number divides 10! if and only
if its prime factorization has no more than eight factors of 2, four factors of 3, two factors of five,
and one factor of 7. Therefore to get a number that divides 10!, select from 0 to 8 factors of two
(8+1 ways), select from 0 to 4 factors of 3 (3+1 ways), select 0 to 2 factors of 5 (2+1 ways), and
select 0 or one factor or 7 (1+1 ways) and take the product of all the selected factors. Thus the
number of ways to select such a positive integer divisor of 10! is given by 
(8 + 1)@(4 + 1)@(2 + 1)@(1 + 1) = 270.
Problem 13 
An equilateral triangle with side length 6 has a square of side length 6
attached to each of its edges as shown. The distance between the two
farthest vertices of this figure (marked A and B in the figure) can be
written as where m and n are positive integers. 
Find m + n.
Answer: 114
This is the same as High School Problem 10.
Problem 14 
The rodent control task force went into the woods one day and caught 200 rabbits and 18
squirrels. The next day they went into the woods and caught 3 fewer rabbits and two more
squirrels than the day before. Each day they went into the woods and caught 3 fewer rabbits and
two more squirrels than the day before. This continued through the day when they caught more
squirrels than rabbits. Up through that day how many rabbits did they catch in all?
Answer: 5491
The number of rabbits caught on the n day is 203 – 3n. The number of squirrels caught on theth
n day is 16 + 2n. The task force kept on catching rodents up through the first day when th
203 – 3n < 16 + 2n or 187 < 5n. This happens when n = 38. The number of rabbits caught is
then the sum of the arithmetic series (203 – 3) + (203 – 6) + (203 – 9) + þ + (203 – 38@3) =
. 
Alternatively, (203 – 3) + (203 – 6) + (203 – 9) + þ + (203 – 38@3) = 
203@38 – (3 + 6 + 9 + þ + 38@3) = 203@38 – 3@(1 + 2 + 3 + þ + 38) = .
Problem 15 
A concrete sewer pipe fitting is shaped like a cylinder with diameter 48 with a cone on top. A
cylindrical hole of diameter 30 is bored all the way through the center of the fitting as shown. 
The cylindrical portion has height 60 while the conical top portion has height 20. Find N such
that the volume of the concrete is NB.
Answer: 24,300
This is the same as High School Problem 17.
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