Download (1 2 ) + 6

1
2
+6
1 2
2
+4
1 3
2
+1
1 4
2 ?
Express your
(1)
What is the value of 1 + 4
answer as a common fraction.
(2)
How many different positive three-digit integers can be formed using only
the digits in the set {2, 3, 5, 5, 5, 6, 6} if no digit may be used more times than it appears in
the given set of available digits?
(3)
A tennis coach divides her 9-player squad into three 3-player groups with
each player in only one group. How many different sets of three groups can be made?
(4)
Four red candies and three green candies can be combined to make many
different flavors. Flavors are different if the percent red is different, so 3 red / 0 green is
the same flavor as 2 red / 0 green; and likewise 4 red / 2 green is the same flavor as 2 red
/ 1 green. If a flavor is to be made using some or all of the seven candies, how many
different flavors are possible?
(5)
If x is an element of the set {−1, 1, 2} and y is an element of
{−2, −1, 0, 1, 2}, how many distinct values of x y are positive?
(6)
A jar contains two red marbles, three green marbles, ten white marbles and
no other marbles. Two marbles are randomly drawn from this jar without replacement.
What is the probability that these two marbles drawn will both be red? Express your answer
as a common fraction.
(7)
What is the probability of Jonah picking a vowel if he randomly chooses a
letter from the word ”CAT”? Express your answer as a common fraction.
(8)
Thad has an unlimited supply of 3-cent and 4-cent stamps. If he has to put
exactly 37 cents of postage on a letter, how many different combinations of 3-cent and/or
4-cent stamps could Thad use?
(9)
Julie baked cupcakes for her family at home and for a party at school. She
iced 4 cupcakes with red frosting, 2 cupcakes with orange frosting, 2 with yellow, 2 with
green, 3 with blue and the last 3 with violet frosting. Each cupcake is iced with exactly one
color of frosting. Julie plans to take exactly 10 of the cupcakes to her party, and will take
either all of the cupcakes of a particular color or none of the cupcakes of that color. How
many different combinations of cupcakes could she take to her party?
(10)
Tara rolls three standard dice once. What is the probability that the sum of
the numbers rolled will be three or more? Express your answer as a percent.
(11)
On the refrigerator, MATHCOUNTS is spelled out with 10 magnets, one
letter per magnet. Two vowels and three consonants fall off and are put away in a bag. If
the Ts are indistinguishable, how many distinct possible collections of letters could be put
in the bag?
(12)
Kevin will start with the integers 1, 2, 3 and 4 each used exactly once and
written in a row in any order. Then he will find the sum of the adjacent pairs of integers in
each row to make a new row, until one integer is left. For example, if he starts with 3, 2,
1, 4, and then takes sums to get 5, 3, 5, followed by 8, 8, he ends with the final sum 16.
Including all of Kevin’s possible starting arrangements of the integers 1, 2, 3 and 4, how
many possible final sums are there?
(13)
Mike and Ike each roll two standard six-sided dice. What is the probability
that the positive difference between their sums is greater than eight? Express your answer
as a common fraction.
(14)
Four packages are delivered to four houses, one to each house. If these
packages are randomly delivered, what is the probability that exactly two of them are
delivered to the correct houses? Express your answer as a common fraction.
(15)
How many integers from 100 through 999, inclusive, do not contain any of
the digits 2, 3, 4 or 5?
(16)
The digits 2, 3, 4, 7 and 8 will be put in random order to make a positive
five-digit integer. What is the probability that the resulting integer will be divisible by 11?
Express your answer as a common fraction.
(17)
When five standard six-sided dice are rolled sequentially there are 65 = 7776
possible outcomes. For how many outcomes is the sum of the five rolled numbers exactly
27?
(18)
Consider the sequence a1 = 1, a2 = 13, a3 = 135, . . . of positive integers.
The k term ak is defined by appending the digits of the k th odd integer to the preceding
term ak−1 . For example, since the 5th term is a5 = 13579, then the 6th term is
a6 = 1357911. Note that a3 is the first term of the sequence divisible by 9. What is the
value of m such that am is the 23rd term of the sequence that is divisible by 9?
th
(19)
If 1 ≤ a ≤ 10 and 1 ≤ b ≤ 36, for how many ordered pairs of integers (a, b)
p
√
is a + b an integer?
(20)
Tamyra is making four cookies and has exactly four chocolate chips. If she
distributes the chips randomly into the four cookies, what is the probability that there are
no more than two chips in any one cookie? Express your answer as a common fraction.
(21)
Janelle generates a two-digit integer by rolling a six-sided die twice. The
result of her first roll is the tens digit, and the result of her second roll is the ones digit.
What is the probability that the resulting integer is divisible by 6? Express your answer as a
common fraction.
(22)
A set of nine cards is labeled with the following values:
1.25,
All the cards are then placed into three bags,
with cards placed in the same bag only if they show an equivalent value. One card is
randomly selected from each bag. What is the probability that the cards showing 1.25,
0.50 and 0.75 are selected? Express your answer as a common fraction.
1
10
1 9
2 , 125%, 0.75, 8 , 1 4 , 12 , 50%, 0.50.
(23)
How many combinations of pennies (1 cent), nickels (5 cents) and/or dimes
(10 cents) are there with a total value of 25 cents?
(24)
Nine Ping-Pong balls are numbered 1 through 9. How many different
combinations of three balls have a sum of 16?
(25)
A reference book lists a set of annual calendars. For any given year, there is
a calendar in the set that corresponds to it. How many annual calendars must be included
in the set in order to have a corresponding calendar for every possible year?
(26)
(27)
How many ordered pairs (x , y ) satisfy BOTH conditions below?
Condition I: x = 1 or y = 0 or y = 2
Condition II: x = 0 or x = 2 or y = 1
How many continuous paths from A to B, along segments of the figure, do
not revisit any of the six labeled points?
A
C
B
D
F
E
(28)
There are six bottles of soda, three bottles of juice and one bottle of water
in a cooler. If a bottle is randomly selected from the cooler, what is the probability that it
is the bottle of water? Express your answer as a common fraction.
(29)
Derek’s phone number, 336-7624, has the property that the three-digit
prefix, 336, equals the product of the last four digits, 7 × 6 × 2 × 4. How many seven-digit
phone numbers beginning with 336 have this property?
(30)
Nathan will roll two six-sided dice. What is the probability that he will roll a
number less than three on the first die and a number greater than three on the second die?
Express your answer as a common fraction.
Copyright MATHCOUNTS Inc. All rights reserved
Answer Sheet
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Answer
81/16
43 integers
280 sets
11 flavors
5 values
1/105
1/3
3 combos
5 combinations
100 %
75 collections
5 sums
5/648
1/4
180 integers
1/10
35 outcomes
69
10 ordered pairs
51/64
1/6
1/24
12 combinations
8 combos
14 calendars
5 ordered pairs
10 paths
1/10
84 phone numbers
1/6
Problem ID
3B21
1C322
4C322
341
A402
525B
BA11
A531
54D5
451
045
315C
3B322
0B51
2D33
05D5
34D5
BD02
C5212
A5D1
D301
AA01
5142
4341
C241
A242
5D02
4541
44C
D031
Copyright MATHCOUNTS Inc. All rights reserved
Solutions
(1) 81/16
ID: [3B21]
No solution is available at this time.
(2) 43 integers
ID: [1C322]
If all three digits are different, there are 4 choices for the first digit, 3 for the second, and 2
for the third, giving (4)(3)(2) = 24 integers. If two of them are the same, the repeated
digit digit is either 5 or 6. There are 2 choices for the repeated digit, then 3 choices for the
non-repeating digit, and 3 ways to arrange these digits (for example, if the repeating digit
is 5 and the non-repeating digit is 6, we can have 655, 565, and 556). This gives
(2)(3)(3) = 18 integers. Finally, if all three digits are the same, the number must be 555.
So there are 24 + 18 + 1 = 43 possible integers.
(3) 280 sets
ID: [4C322]
For now, call the groups group A, group B, and group C. We will count the number of
ways to fill up these groups. Since in reality, the groups are indistinguishable from each
other (i.e., once the groups have been chosen, we can rearrange the letters of the groups
without changing anything), we will end up overcounting. Since we can rearrange the
letters of the groups in 3! = 6 ways, we will count the same
configuration 6 times, so we
9
will end up overcounting by a factor
of 6. There are 3 ways to choose 3 players out of
6
the 9 to be in group A, and 3 ways to choose 3 players out of the
remaining 6 to be in
9 6
group B. The remaining 3 student go into group C. This gives 3 3 ways. Since we
overcounted by a factor of 6, our answer is 61 93 63 = 280 .
(4) 11 flavors
ID: [341]
Denote the ratio by x : y , where x is the number of red candies and y is the number of
green. We can have 0, 1, 2, 3, or 4 red candies and 0, 1, 2, or 3 green candies. Thus,
there are 5 · 4 = 20 potential ratios. However, a 0 : 0 ratio is not allowed (there would be
no candy!), so we subtract one for a total of 19 ratios possible. Now we must subtract the
ratios we’ve over-counted. In particular, 0 : 1 is the same as 0 : 2 and 0 : 3, and 1 : 0 is the
same as 2 : 0, 3 : 0, and 4 : 0. Also, 1 : 1 is the same as 2 : 2 and 3 : 3, and 2 : 1 is the
same as 4 : 2. Thus, we have over-counted by 8 ratios, so our final answer is 19 − 8 = 11 .
(5) 5 values
ID: [A402]
Raising −1 and 1 to integer powers will only result in answers of 1 or −1. Raising 2 to
each of the powers in {−2, −1, 0, 1, 2} gives 41 , 21 , 1, 2, and 4. Thus, there are 5 distinct
positive values of x y .
(6) 1/105
ID: [525B]
The total number of marbles is 2 + 3 + 10 = 15. The probability that the first marble
drawn will be red is 2/15. Then, there will be one red left, out of 14. Therefore, the
probability of drawing out two red marbles will be:
2 1
1
·
=
15 14
105
(7) 1/3
ID: [BA11]
One of the letters in ”CAT” is a vowel, and there are three total letters. Since probability
is the number of successful outcomes divided by the total number of outcomes, the
1
.
probability of picking a vowel is
3
(8) 3 combos
ID: [A531]
Thad must use at least one 3-cent stamp, because there is no way to make 37 cents from
4-cent stamps. So, the problem is now to make 34 cents from 3-cent and 4-cent stamps.
If we use one 4-cent stamp, we are left with 30 cents. This can be satisfied with ten
3-cent stamps.
If we use two 4-cent stamps, we are left with 26 cents. If we use three, we are left with
22 cents. If we use four, we are left with 18 cents. This can be satisfied with six 3-cent
stamps.
If we use five 4-cent stamps, we are left with 14 cents. If we use six, we are left with 10
cents. If we use seven, we are left with 6 cents, which can only be satisfied with two 3-cent
stamps.
Thus, there are 3 different combinations.
(9) 5 combinations
ID: [54D5]
If Julie includes one of the colors that cover three cupcakes, she must also include the
other color that covers three cupcakes. This is because she must make ten cupcakes total,
and all of the other colors cover an even number of cupcakes, so there is no way to make
ten with three and some combination of even numbers. Thus, if she includes blue and
violet, she has four cupcakes left to choose. There are three ways in which she can choose
four cupcakes if she chooses colors that cover two (green and orange, green and yellow, or
orange and yellow). Alternately, she can choose a color that covers four (red). Finally, if
she doesn’t include any colors that cover three cupcakes, she must choose all of the other
cupcakes in order to make ten. Thus, Julie has 5 different combinations of cupcakes.
(10) 100 %
ID: [451]
The smallest possible number that Tara can roll is a 3, by getting a 1 on each die. Thus,
she will always get a sum of three or more. Our answer is 100% .
(11) 75 collections
ID: [045]
Let’s divide the problem into two cases: one where 0 or 1 T’s fall off and one where both
T’s fall off:
0 or
1 T’s:
3 6
= 3 × 20 = 60
2 3
2
T’s:
3 5
= 3 × 5 = 15
2 1
Total: 60 + 15 = 75
(12) 5 sums
ID: [315C]
Let’s say the four integers were arranged as a, b, c, d across to start. We can then
compute each row in terms of these variables:
a
b
c
d
a+b
b+c
c+d
a+2b+c
b+2c+d
a+3b+3c+d
As we can see from the final sum, the order of a and d, as well as the order of b and c,
will not affect the final sum. So, we consider each of the six possible ways to choose two
a d b
1 2 3
1 3 2
numbers from possible starting values of 1, 2, 3, 4.
1 4 2
2 3 1
2 4 1
3 4 1
see from the table, there are a total of 5 different possible
c a+3b+3c+d
4
24
4
22
As we can
3
20
4
20
3
18
2
16
sums.
(13) 5/648
ID: [3B322]
The pairs of sums which result in differences greater than 8 are {12, 2}, {12, 3}, and
{11, 2}. There are 64 total outcomes for the roll of the four dice. Of these, only (6, 6, 1, 1)
and (1, 1, 6, 6) yield the pair of sums {12, 2}. The four rolls (1, 2, 6, 6), (2, 1, 6, 6),
(6, 6, 1, 2), and (6, 6, 2, 1) give the pair of sums {12, 3}. Finally, the four rolls (1, 1, 6, 5),
(1, 1, 5, 6), (6, 5, 1, 1), and (5, 6, 1, 1) give the pair of sums {11, 2}. Altogether, 4 + 4 + 2
of the 64 rolls result in a difference greater than eight, so the probability is
(4 + 4 + 2)/64 = 6/648 .
(14) 1/4
ID: [0B51]
Since there are 4 houses and 4 packages, we can choose 42 = 6 pairs of houses to be the
pair that will receive the correct package. In that case, the other two houses must have
one another’s package. The probability of this occuring for any arrangement is 14 · 13 · 21 , as
the first fraction represents the probability of a given house getting the correct package,
and the second fraction the subsequent probability that the other given house gets the
correct package, and the final fraction the probability that the last two houses have each
1
1
.
other’s packages. So, the probability is 6 · 2·3·4
=
4
(15) 180 integers
ID: [2D33]
The first digit can be 1, 6, 7, 8 or 9, offering 5 choices. The next two digits can be any of
those 5, or also 0, for 6 choices total. So the total number of integers possible is
5 · 6 · 6 = 180 .
(16) 1/10
ID: [05D5]
If the resulting integer is divisible by 11 then the sum of the first, third, and fifth digits has
the same remainder when divided by 11 as the sum of the second and fourth digits. This
only occurs when the first, third, and fifth digits are 2, 3, and 7 (in some order) and the
second and fourth
digits are 4 and 8 (in some order).
5
There are 2 total ways to partition these five digits into a group of 3 and a group of 2.
From above, only one of these partitions will result in five-digit integers that are divisible by
11. Therefore, our answer is 1/10 .
(17) 35 outcomes
ID: [34D5]
There are three ways for the sum of the five rolled numbers to equal 27: we can have two
6s and three 5s; three 6s, one 5, and one 4; or four 6s and one 3.
5!
In the first case, we rolled the numbers 6, 6, 5, 5, 5 in some order. There are
= 10
2!3!
distinct ways that we could have rolled these numbers.
In the second case, we rolled the numbers 6, 6, 6, 5, 4 in some order. There are
5!
= 20 distinct ways that we could have rolled these numbers.
3!1!1!
5!
=5
In the third case, we rolled the numbers 6, 6, 6, 6, 3 in some order. There are
4!1!
distinct ways that we could have rolled these numbers.
Adding all these outcomes, we see that there are 10 + 20 + 5 = 35 possible outcomes.
(18) 69
ID: [BD02]
First, recall that if the sum of the digits of a number is divisible by 9, then the number is
also divisible by 9. So, every time we hit an integer in this sequence that is divisible by 9,
we can simply disregard all of its digits as we continue building to the next integer. For
instance, after a3 = 135, we start again, considering 7, 79, 7911. 7911 is divisible by 9, so
we move on to 13, 1315, 131517. 131517 is divisible by 9, so we start again with 19,
1921, 192123. Notice that every third term is divisible by 9. Continuing in such a manner
will reveal that m must be 69 .
(19) 10pordered pairs
ID: [C5212]
√
√
√
b is an
If a + b is an integer, then its square a + b is also an integer. Therefore,
√
integer. In other words, b must be a perfect square. If we define c = b, then the problem
has asked us to find the number of ordered pairs (a, c) for which 1 ≤ a ≤ 10, 1 ≤ c ≤ 6,
and a + c is a perfect square. We check the 6 possibilities for c separately. If c = 1, then a
is 3 or 8. If c = 2, then a is 2 or 7. If c = 3, then a is 1 or 6. If c = 4, then a = 5, and if
c = 5, then a = 4. Finally, if c = 6, then a is either 10 or 3. Altogether, there are
2 + 2 + 2 + 1 + 1 + 2 = 10 ordered pairs (a, c) satisfying the given conditions.
(20) 51/64
ID: [A5D1]
No solution is available at this time.
(21) 1/6
ID: [D301]
No solution is available at this time.
(22) 1/24
ID: [AA01]
No solution is available at this time.
(23) 12 combinations
ID: [5142]
First, we count the number of combinations that include pennies: we could have all
pennies, all pennies and one nickel, all pennies and two nickels, all pennies and three nickels,
all pennies and four nickels, all pennies and one dime, all pennies and two dimes, all pennies
and one dime and one nickel, all pennies and one dime and two nickels. As for the no
pennies case, we can have five nickels, one dime and three nickels, two dimes and one
nickel. So, there are 9 + 3 = 12 combinations.
(24) 8 combos
ID: [4341]
We proceed by casework.
Case I: One of the balls is the 9.
The other two balls must sum to 7. There are three ways to do this: 1+6, 2+5, or 3+4.
Case II: One of the balls is the 8, and the 9 is not chosen. (we’ve counted those cases
already)
The other two balls must sum to 8. There are three ways to do this: 1+7, 2+6, or 3+5.
Case III: One of the balls is the 7, and the 8 and the 9 are not chosen.
The other two balls must sum to 9. There are two ways to do this: 3+6 or 4+5.
If we proceed in this manner, we see that there are no further possibilities that we haven’t
counted (if we choose the 6 and can’t choose the 7, 8, or 9, there is no way to make the
necessary sum of 10 with the remaining two balls). Thus, there are 8 such combinations.
(25) 14 calendars
ID: [C241]
No solution is available at this time.
(26) 5 ordered pairs
ID: [A242]
Proceed case-by-case in condition I. If x = 1, then by condition II, y = 1 since the first two
possibilities are excluded. If y = 0, then either x = 0 or x = 2. If y = 2, then likewise,
either x = 0 or x = 2. This gives 5 possible ordered pairs.
(27) 10 paths
ID: [5D02]
We denote a path from A to B by writing the labeled points visited, such as A-C-B (first
going to C then to B).
Case 1: Path ends in C-B. There are clearly four such paths, which we can determine
systematically; A-C-B, A-D-C-B, A-D-F -C-B, and A-D-E-F -C-B.
Case 2: Path ends in F -B. The possible paths are easy to determine systematically as
A-C-F -B, A-C-D-F -B, A-C-D-E-F -B, A-D-C-F -B, A-D-F -B, A-D-E-F -B, yielding 6
possible paths.
Therefore there are a total of 10 such paths.
(28) 1/10
ID: [4541]
No solution is available at this time.
(29) 84 phone numbers
ID: [44C]
We begin by factoring 336. 336 = 24 · 3 · 7. Because we are looking for phone numbers, we
want four single digits that will multiply to equal 336. Notice that 7 cannot be multiplied
by anything, because 7 · 2 is 14, which is already two digits. So, one of our digits is
necessarily 7. 3 can be multiplied by at most 2, and the highest power of 2 that we can
have is 23 = 8. Using these observations, it is fairly simple to come up with the following
list of groups of digits whose product is 336:
1, 6, 7, 8
2, 4, 6, 7
2, 3, 7, 8
3, 4, 4, 7
For the first three groups, there are 4! = 24 possible rearrangements of the digits. For
the last group, 4 is repeated twice, so we must divide by 2 to avoid overcounting, so there
are 4!2 = 12 possible rearrangements of the digits. Thus, there are 3 · 24 + 12 = 84
possible phone numbers that can be constructed to have this property.
(30) 1/6
ID: [D031]
For the first die to be less than three, it must be a 1 or a 2, which occurs with probability
1
3 . For the second die to be greater than 3, it must be a 4 or a 5 or a 6, which occurs with
probability 21 . The probability of both of these events occuring, as they are independent, is
1
1 1
.
3 · 2 =
6
Copyright MATHCOUNTS Inc. All rights reserved