Solutions to Set 5

Problem #1: 720
Problem #2: 1
Problem #3: 25
Problem #4: 270
Problem #5: 10
Problem #6: 7
Problem #7: 9
Problem #8: 5/108
Problem #9: 16
Problem #10: √21

Explanations

Problem #1: At first, pretend that the 3 books that you have to keep together are just 1 big book. Then, instead of 7 books, you have 5 books to put in order. Therefore, there are 5!=5*4*3*2*1=120 possibilities.
However, you haven't taken into consideration that the 3 "stuck together" books can be put in any order. In fact, for each of the 120 possibilities we counted, those 3 "stuck together" books can be put in 3!=3*2*1=6 orders.
The total number of possibilities, therefore, is
120 * 6 = 720.
Problem #2: The number 5 itself is the only one. Any other number ending in 5 would be a multiple of 5, and therefore would not be prime.
Read more about Prime Numbers.
Problem #3: The scenario in which you would get the highest number of votes is where each of the other candidates gets only one vote. So assume that you received x votes, while John, Hillary, and Barack each received 1 vote. Since you know that each candidate received an average of 7 votes, you can write an equation based on the definition of the average (arithmetic mean):
(x + 1 + 1 + 1)/4 = 7
(x + 3)/4 = 7
x + 3 = 28
x = 25.
Therefore, you could have received a maximum of 25 votes.
Read more about The Arithmetic Mean in Equations.
Problem #4: Let a represent the measure of angle A. Since angle B is the complement of angle A, the measure of angle B can be expressed as 90-a.
The supplement of angle A is
180 - a.
The supplement of angle B is
180 - (90 - a)
= 180 - 90 + a
= 90 + a.
Therefore, the sum of the supplements is
(180 - a) + (90 + a)
= 270 degrees.
Alternatively, you could solve this problem without using variables. Just test a few angles and their complements, and you will see that the result is always 270 degrees. The algebraic method is generally reliable and tidy, but it isn't always the fastest!
Read more about Adding Angles.
Problem #5: Any number that is both a perfect square and a perfect cube must be an integer to the 6th power: 1⁶, 2⁶, 3⁶, and so on.
The problem gives the range 1 to 1,000,000, which can be written as 1⁶ to 10⁶. So, there are 10 "perfect sixths" in the range.
Problem #6: Let b, v, g represent the number of Bervokens, VonMergles, and Guzzalles, respectively. Using the given information, you can write three algebraic equations:
Tentacles: 16b + 8v + 20g = 112
Nostrils: 3b + 2v + g = 11
Horns: 5b + 14v + 3g = 45
If you add those three equations, you will get:
24b + 24v + 24g = 168
b + v + g = 7.
Therefore, the total number of monsters under my bed is 7.
Read more about Translating Word Problems.
Problem #7: The biggest 3-digit number that you can get by adding two 2-digit numbers is 99+99=198. Therefore, the three-digit number AAH must be between 100 and 198 inclusive. Therefore, its first digit must be 1, and you now know that A=1.
In the rightmost column of the alphanumeric, you see that A+A=H, so H is equal to 1+1=2.
Having found that A=1 and H=2, you know that HA is 21 and AAH is 112. Therefore, LA is equal to AAH-HA=112-21=91. It follows that L=9.
Problem #8: Since each roll has 6 different possible values, there are 6*6*6=216 different ways to roll a die 3 times. Of those 216 ways, there are 10 ways to roll a sum of 6:
[1] (1, 1, 4)
[2] (1, 2, 3)
[3] (1, 3, 2)
[4] (1, 4, 1)
[5] (2, 1, 3)
[6] (2, 2, 2)
[7] (2, 3, 1)
[8] (3, 1, 2)
[9] (3, 2, 1)
[10] (4, 1, 1)
Therefore, the probability of rolling a sum of 6 is 10/216=5/108.
Problem #9: Let a, b, and c represent the radii of the circles centered at points A, B, and C, respectively. Since the circles are externally tangent to one another,
AB = a + b
BC = b + c
AC = a + c
Therefore, the perimeter of triangle ABC is
32 = AB + BC + AC
32 = (a + b) + (b + c) + (a + c)
32 = 2a + 2b + 2c.
16 = a + b + c.
The sum of the radii is therefore 16.
Read more about Circles Tangent to Each Other.
Problem #10:
Use the Pythagorean Theorem with right triangle AEF to find AE = √24 = 2√6. Then use the Pythagorean Theorem with right triangle ADE to find AD = √23. Similarly, AC = √22 and AB = √21.
Read more about The Pythagorean Theorem.