AoPS:

AoPS: Introduction to Counting & Probability

Chapter 1 Counting is Arithmetic

Counting Lists of Numbers Problem1.1 How many #s are in the list 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18? Obviously there are 18 numbers. That was pretty easy. The counting was done for us!

Problem 1.2 How many #s are in the list 7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29? In other words, how many #s are there between 7 and 29 inclusive? (include 7 & 29 in the count)

Solution A clever way to approach this problem is to convert it to a problem like problem 1.1, by subtracting 6 from every # in the list:

A clever way to approach this problem is to convert it to a problem like problem 1.1, by subtracting 6 from every # in the list: 7 8 9 … 29 -6 -6 -6 -6 1 2 3 … 23

A clever way to approach this problem is to convert it to a problem like problem 1.1, by subtracting 6 from every # in the list: 7 8 9 … 29 -6 -6 -6 -6 1 2 3 … 23 You may notice that we found that there are 29-7+1 = 23 #s from 7 to 29, inclusive.

Concept: Given 2 positive #s, a and b, with b > a, find a formula for how many #s there are between a and b inclusive.

Concept: Given 2 positive #s, a and b, with b > a, find a formula for how many #s there are between a and b inclusive. We can subtract a-1 from our list of #s from a to b to get a list of #s starting at 1:

Concept: Given 2 positive #s, a and b, with b > a, find a formula for how many #s there are between a and b inclusive. We can subtract a-1 from our list of #s from a to b to get a list of #s starting at 1: aa+1 a+2 . . . b -(a-1) -(a-1) -(a-1) . . . -(a-1) 1 2 3 . . . b – a + 1

Concept: Given 2 positive #s, a and b, with b > a, find a formula for how many #s there are between a and b inclusive. We can subtract a-1 from our list of #s from a to b to get a list of #s starting at 1: aa+1 a+2 . . . b -(a-1) -(a-1) -(a-1) . . . -(a-1) 1 2 3 . . . b – a + 1 Our new list of #s has b – a + 1 numbers in it.

Problem 1.3 How many multiples of 3 are between 62 and 215?

Problem 1.3 How many multiples of 3 are between 62 and 215? We see that 62/3 = 20 2/3, so the simplest multiple of 3 is 3 X 21 = 63. Similarly, 215/3 = 71 2/3, so the largest multiple of 3 is 3 X 71 = 213.

Problem 1.3 How many multiples of 3 are between 62 and 215? We see that 62/3 = 20 2/3, so the simplest multiple of 3 is 3 X 21 = 63. Similarly, 215/3 = 71 2/3, so the largest multiple of 3 is 3 X 71 = 213. So our list is 63,66,69, …,213. Divide it by 3 to convert it to a list we know how to count:

Problem 1.3 How many multiples of 3 are between 62 and 215? We see that 62/3 = 20 2/3, so the simplest multiple of 3 is 3 X 21 = 63. Similarly, 215/3 = 71 2/3, so the largest multiple of 3 is 3 X 71 = 213. So our list is 63,66,69, …,213. Divide it by 3 to convert it to a list we know how to count: 21, 22, 23, . . . , 71.

We know how to count this list! Subtracting 20 from each number in the list gives 1, 2, 3, . . . , 51

We know how to count this list! Subtracting 20 from each number in the list gives 1, 2, 3, . . . , 51 DO NOT BE TEMPTED TO DO THIS: 215 – 62 = 153 = 51 3 3 IT DOESN’T ALWAYS WORK! See the next problem

Problem 1.4: How many multiples of 10 are between 9 & 101? How many multiples of 10 are between 11 & 103? We know that 101-9 = 103 – 11 = 92, so shouldn’t you get the same answers? Why aren’t they the same?

Problem 1.4: List 1: the multiples of 10 are 10, 20, 30, …, 100, so there are 10 multiples.

Problem 1.4: List 1: the multiples of 10 are 10, 20, 30, …, 100, so there are 10 multiples. List 2: the multiples of 10 are 20, 30, …, 100, so there are 9 multiples.

Problem 1.4: List 1: the multiples of 10 are 10, 20, 30, …, 100, so there are 10 multiples. List 2: the multiples of 10 are 20, 30, …, 100, so there are 9 multiples. The shortcut doesn’t work: 101 – 9 = 103 – 11 = 92 = 9.2 10 10 10 So how would you know whether the answer is 9 or 10?

Problem 1.5 How many 4-digit numbers are perfect cubes?

Solution: How many 4-digit numbers are perfect cubes? The smallest 4-digit cube is 1000 = 103

Solution: How many 4-digit numbers are perfect cubes? The smallest 4-digit cube is 1000 = 103 The largest 4-digit perfect cube is a little harder to find and requires a little experimentation. Start by noting 203 = 8000. By trial & error, 213 = 9261 223 = 10,648 So 9261 = 213 is the largest 4-digit cube & the list is 1000, . . . , 9261.

Solution: There is a much better way that we can write this list: 103, 113, 123, . . . , 203, 213 So the number of #s in the list is the same as 10, 11, 12, . . . , 20, 21 and that means there are 12 numbers in the list!

Now it’s your turn! 1. How many numbers are in the list 36, 37, 38, …, 92, 93? 2. How many numbers are in the list 4, 6, 8, . . . , 128, 130? 3. How many numbers are in the list -33, -28, -23, …, 52, 57?

Solutions: • 58 • Dividing each member of the list by 2, we get 2, 3, 4, …, 64,65, and then subtracting 1, we get 1,2,3, …,63,64, so there are 64 numbers. • We could add 3 to each member in the list to get -30,-25,-20,…,55,60, and divide by 5 to get -6,-5,-4,…,11,12. Then using the integer formula, we get 12 – (-6) + 1 = 19.

Try some more! 4. How many numbers are in the list 147, 144, 141, . . . , 42, 39? 5. How many numbers are in the list 3 2/3, 4 1/3, 5, 5 2/3, …, 26 1/3, 27? 6. How many positive multiples of 7 are less than 150?

Solutions: 4. First, reverse the list, then divide by 3 to get 13, 14, …, 48, 49, so 49 – 13 + 1 = 37. 5. First multiply each number by 3 to get 11, 13, 15, …, 79, 81. Then we can subtract 1 and divide by 2 to get 5, 6, 7, …, 39, 40. So we get 40 – 5 + 1 = 36. 6. 7 x 21 = 147 < 150 < 154 = 7 x 22, so 21 positive multiples of 7 are less than 150.

OK, one last time! 7. How many perfect squares are between 50 and 250? 8. How many odd perfect squares are between 5 and 211? 9. How many sets of four consecutive positive integers are there such that the product of the four integers is less than 100,000?

Solutions: 7. Since 72 < 50 < 82 and 152 < 250 < 162, the squares between 50 & 250 are 82,92,102,…,152. So there are 15 – 8 + 1 = 8. 8. Since 12 < 5 < 32 and 132 < 211 < 152, we have the list 32,52,72,…,132, which has the same # of members as 3,5,7,…,13 which = 6. 9. Note that 174 = 83521 < 100,000 < 104,976 =184. Since 17.54 ≈ 16 x 17 x 18 x 19, we check 16 x 17 x 18 x 19 = 93,024. Also 17 x 18 x 19 x 20 = 116,280, so 16 x 17 x 18 x 19 is the largest product of 4 consec. pos. integers which is less than 100,000. So there are 16 sets.

Counting with Addition and Subtraction Problem 1.6 At Northshore High School there are 12 players on the basketball team. All of the players are taking at least one foreign language class. The school offers only Spanish & French as its foreign language classes. 8 of the players are taking Spanish and 5 of the players are taking both languages. How many players are taking French?

Solution: The players taking French fall into 2 categories: those who take Spanish and those who don’t. The # of players taking French and Spanish = 5 (given in the problem).

Solution: The players taking French fall into 2 categories: those who take Spanish and those who don’t. The # of players taking French and Spanish = 5 (given in the problem). Next count the # of players taking French but not Spanish. There are 12 players on the team in total, and 8 of them take Spanish, so there are 12 – 8 = 4 not taking Spanish. Since every player must take at least one language, there are 4 taking French.

Solution: So the # of players taking French is the sum of the # of players in each of the two categories, 5 + 4 = 9. There is another way to solve this problem: Draw a Venn Diagram. Use a Venn Diagram whenever you wish to count things or people which occur in two or three overlapping groups. See the next page.

Place points in the circles to represent the players. A point that is in the French circle that is not in the Spanish circle represents one player taking French but not Spanish. Spanish French

A point in the region that is in both circles represents a player taking both languages. Spanish French

A player taking Spanish but not French is represented by a point inside the Spanish circle but not French one. Spanish French

Finally a point placed outside both circles represents a player who is in neither class. Spanish French

Now we can use the diagram to solve the problem. Put 5 points in the intersection of both circles because there are 5 players in both classes. Spanish French

Now, since there are 8 players taking Spanish, and 5 points are already inside the Spanish circle on the right, there must be 3 more points inside the Spanish circle not in the French circle. Add 3. Spanish French

Since we have 12 total points and we know there aren’t any outside both circles, there must be 4 left inside the French circle but not inside the Spanish circle so add 4 points. Spanish French

So now we can just read off the answer – there are 9 points inside the French circle on the left. Spanish French 4 5 3

Problem 1.7 There are 27 cats at the pound. 14 of them are short-haired. 11 of them are kittens. 5 of them are long-haired adult cats. How many of them are short-haired kittens?

Solution: Draw a Venn Diagram, with one circle for cats with short hair and one circle for cats which are kittens.

Which #s do we want to place in the regions? Since 5 cats don’t have short hair & are not kittens, we know there are 5 cats outside both circles. Kittens Short Hair 5

At this point we can’t immediately fill any of the other numbers, because none of our #s corresponds exactly to a region of the diagram. For example, we know there are 11 kittens, but there’s no single region of the diagram that corresponds to “kittens”: there’s a region for “short-haired kittens” and a region for “long-haired kittens.” So were going to have to use a little bit of thought. (Be very careful here)

The part of the right circle that does not intersect with “short hair” must represent “long-haired kittens.” Short Hair Kittens Kittens Short Hair This is the region that represents long-haired kittens. This region represents short-hair kittens. 5 5

Introduce a variable x. Call the # of cats in one of the regions inside the circles x and try to find other regions in terms of x. Let the # of “short-haired kittens” be x. Short Hair Kittens Kittens Short Hair x 5 5

Since there are a total of 14 short-haired cats, and x of them are kittens, we know that 14 – x of them are not kittens. Then we have 11 – x kittens that are not short-haired. Short Hair Kittens Kittens Short Hair 14 - x x 11 - x 5 5

AoPS:

AoPS:

Presentation Transcript

AoPS

AoPS

AoPS