Problem of the Week Archive

Problem of the Week Archive
Best of 2014 – December 29, 2014
Problems & Solutions
As the year comes to a close, let’s take a look back at some of our favorite problems of 2014.
2014 Chapter Team #6
Abigail, Bartholomew and Cromwell play a game in which they take turns adding 1, 2, 3 or 4 to a
sum in order to create an increasing sequence of primes. For example, Abigail must start with either
2 or 3. If she chooses 2, then Bartholomew can add 1 to make 3, or he can add 3 to make 5. If
Bartholomew makes 3, then Cromwell can add 2 to make 5, or he can add 4 to make 7. Abigail,
Bartholomew and Cromwell take turns, in that order, until no more primes can be made, and the
game ends. The player who makes the last prime wins. If Bartholomew wins, how many primes were
made?
We'll let A, B and C stand for Abigail, Cromwell and Bartholomew, respectively. Let's suppose the game proceeds as follows:
A chooses 2
B adds 1 to make 3
C adds 2 to make 5
A chooses 2 to make 7
B chooses 4 to make 11
C chooses 2 to make 13
A chooses 4 to make 17
B chooses 2 to make 19
C chooses 4 to make 23
A chooses...
Well, A can’t choose, since there are no primes between 24 and 27. That means Cromwell chose the last prime. But we want
Bartholomew to end up with the last prime.
So let's suppose, instead, that the game proceeds this way:
A chooses 2
B adds 3 to make 5
C adds 2 to make 7
A chooses 4 to make 11
B chooses 2 to make 13
C chooses 4 to make 17
A chooses 2 to make 19
B chooses 4 to make 23
Bartholomew has chosen the last prime and the total number of primes made was 8 primes.
2014 State Target #4
The sum of a list of seven positive integers is 42. The mean, median and mode are consecutive
integers, in some order. What is the largest possible integer in the list?
We are told that the sum of the seven integers is 42, so it follows that these numbers have a mean of 42/7 = 6. Thus, there
are three possibilities for the consecutive measures of central tendency 4, 5 and 6; 5, 6 and 7; 6, 7 and 8. But the values 4, 5
and 6 will yield the largest possible integer in such a list. Since the mean is 6, the mode and median could be 4 and 5,
respectively, or vice versa.
Case 1: Let’s suppose the median is 4 and the mode is 5. We have __, __, __, 4, 5, 5, __. Since we are looking for the
largest possible integer in such a list, we need to make the first three integers as small as possible. Doing so, we have
1, 2, 3, 4, 5, 5, __. The largest number in this particular list would be 42 – (1 + 2 + 3 + 4 + 5 + 5) = 42 – 20 = 22.
Case 2: Now, suppose the median is 5 and the mode is 4. We have __, 4, 4, 5, __, __, __. Again since we are looking for
the largest possible number in such a list, we’ll make three of the missing integers as small as possible. We would have
1, 4, 4, 5, 6, 7, __. The largest number in this particular list would be 42 – (1 + 4 + 4 + 5 + 6 + 7) = 42 – 27 = 15.
Therefore, the largest possible integer in the list is 22.
2014 National Sprint #17
For the following system of equations, what is the value of c?
a + b + c + d = 88
a + b + c + e = 84
a + b + d + e = 82
a + c + d + e = 78
b + c + d + e = 72
If we add the five equations the result is 4a +4b + 4c + 4d + 4e = 404 → a + b + c + d + e = 101. Since we are told that
a + b + d + e = 82, we can substitute to get 82 + c = 101, so c = 19.
2014 National Countdown #23
Of six MP3 players, two are defective and four are not. If Cecil randomly chooses two of these MP3
players, without replacement, what is the probability that the two he chooses are not defective?
Express your answer as a common fraction.
Suppose players A and B are defective, and players C, D, E and F are not defective. There are 6C2 = 6!/(4! × 2!) =
720/(24 × 2) = 15 ways Cecil can select a pair of MP3 players from these six players. Six of those pairs(C-D, C-E, C-F, D-E,
D-F, E-F) are the result of Cecil selecting two players that are not defective. Therefore, the probability that the two MP3 players
Cecil selects are not defective is 6/15 = 2/5.
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