SMT 2013
Advanced Topics Test Solutions
February 2, 2013
1. How many positive threedigit integers
a
b
c
can represent a valid date in 2013, where either
a
corresponds to a month and
b
c
corresponds to the day in that month, or
a
b
corresponds to a
month and
c
corresponds to the day? For example, 202 is a valid representation for February
2nd, and 121 could represent either January 21st or December 1st.
Answer: 273
Solution:
The integers which are valid have a 11 correspondence to days in the first 9 months
– this is straightforward to see for all positive integers that do not have a 1 in the hundreds
place and just requires careful inspection of the case where 1 is in the hundreds place. There
are 365

31

30

31 = 273
such days.
2. Consider the numbers
{
24
,
27
,
55
,
64
, x
}
.
Given that the mean of these five numbers is prime
and the median is a multiple of 3, compute the sum of all possible positive integral values of
x
.
Answer: 60
Solution:
The restriction on the median means either
x
≤
27 or 3

x
and
x <
55. Hence, the
sum of all five numbers is 24 + 27 + 55 + 64 +
x
= 170 +
x <
225, so the average is
>
170
5
= 34
and
<
225
5
= 45. The only prime numbers in this range are 37, 41, and 43, which yield
x
= 15,
x
= 35, or
x
= 45. 35 is greater than 27 but not a multiple of 3, so it doesn’t work. Hence, the
answer is 15 + 45 = 60
.
3. Nick has a terrible sleep schedule. He randomly picks a time between 4 AM and 6 AM to fall
asleep, and wakes up at a random time between 11 AM and 1 PM of the same day. What is the
probability that Nick gets between 6 and 7 hours of sleep?
Answer:
3
8
Solution:
Consider the rectangle with lowerleft corner at (4
,
11) and upperright corner at
(6
,
13). We want to compute the probability that a randomly generated point inside the rectangle
falls above the line
y
=
x
+ 6 and below the line
y
=
x
+ 7. These lines cut out a trapezoid
of area
3
2
, so therefore the probability that a randomly generated point in the rectangle falls in
this trapezoid is
3
2
4
=
3
8
.
4. Given the digits 1 through 7, one can form 7! = 5040 numbers by forming different permutations
of the 7 digits (for example, 1234567 and 6321475 are two such permutations).
If the 5040
numbers obtained are then placed in ascending order, what is the 2013
th
number?
Answer: 3657214
Solution:
When the numbers are ordered, the first 6! = 720 numbers all have 1 in the millions
place value. The next 720 numbers all have 2 in the millions place value. The 2013
th
number
must then lie in the next batch, with 3 as the millions place value digit. Within the third batch
of 720 numbers, the first 5! = 120 have a 1 in the hundred thousands place value, the next 120
have a 2 in the hundred thousands place, the next 120 have a 4, and so on. Continuing in the
same manner, we can deduce that the 2013
th
number is 3657214
.
5. An unfair coin lands heads with probability
1
17
and tails with probability
16
17
. Matt flips the coin
repeatedly until he flips at least one head and at least one tail. What is the expected number
of times that Matt flips the coin?