Download Divisibility Rules – Blue Problems

Name _____________________________ Date _________Factors – Blue Problems
___
1. Find the factors of 23.
2. Find the factors of 24.
As the above examples show, two numbers may have greatly different numbers of factors, even though the
numbers are nearly equal.
3. How many factors does 23 have?
4. How many factors does 24 have?
You can use prime factorizations to find out how many factors a number has.
Example
Solution
How many factors does 40 have?
40 = 23  51
3+1=4
1+1=2
4x2=8
1. Write the prime factorization.
2. Take each exponent and add 1.
3. Multiply the sums.
40 has 8 factors
Complete the table. The first one is done for you.
Number
Prime
Factorization
Number of
Factors
Factors
6
23
4
1, 2, 3, 6
5.
44
6.
91
7.
125
8.
54
9.
664
10.
369
11.
475
12.
222
13. What is the sum of the reciprocals of the positive integer factors of 28?
14. What is the sum of the positive integer factors of 225?
15. What is the smallest positive integer that has 2, 3, 4, 6, 7 and 12 as factors?
16. What is the number of positive factors of 648?
17. What is the sum of the three greatest consecutive integers less than 200 for which the least number
has 4 as a factor, the second number has 5 as a factor and the greatest number has 6 as a factor?
18. What is the greatest whole number less than 150 that has an odd number of distinct positive factors?
19. What is the least positive integer with exactly 10 factors?
20. The product of the first 101 positive integers is divided by the product of the first 100 positive
integers. How many positive integer factors does the quotient have?
21. When the positive integers with exactly three factors are listed in ascending order, what is the fifth
number listed?
22. Tom’s graduating class has 288 students. At the graduation ceremony, the students will sit in rows
with the same number of students in each row. If there must be at least 10 rows and at least 15
students in each row, then there can be x students in each row. What is the sum of all possible values
of x?
23. Fences for Fido and Friends. You've decided to figure out the size of four rectangular dog pens.
They are arranged right next to each other, with their sides lining up evenly. The pictures below show
two possibilities for their relative sizes. You know the following:
1.
2.
3.
4.
The area of the yard is 169 square meters.
The entire yard is a square.
The sides of each pen are whole numbers of meters.
The area of the pen labeled A4 is 24 square meters.
What is one possible set of dimensions for the pens that satisfies the statements above?
Extra: Now that you have found one possible solution set, find another.
24. Tiling Table. Your grandfather gives each of his grandchildren a set of tiles to use to make a table.
Grandfather tells you, your brother, Avery, and your cousin, Geavonnie, that he has selected special
tiles for each of you!
In Grandfather's shop he gives Avery 120 large tiles (6" by 6"), he gives Geavonnie 256 medium sized
tiles (4" by 4"), and he gives you 1056 small tiles (2" by 2").
Then he reminds the three of you that before you leave, he
needs to know how much wood each of you will need for the trim
around the edge of your rectangular table.
*He hands you the catalog of exotic wood. As you look through the catalog you are amazed at the
prices.
*You say, "I think we should figure out the least amount of wood that we'll need for the trim on our
tables."
*Geavonnie agrees, "It's really expensive, and if we figure out how to use the tiles to make a
rectangular table with the smallest perimeter, we can save Grandfather some money."
Avery says, "Let's start with my table. If I have 10 rows of 12 tiles each, that should make a table
with the smallest perimeter."
You ask, "How did you do that so quickly?"
Avery says that he knows a shortcut. He explains, "I didn't figure it out by arranging the tiles,
figuring out all the possible perimeters, and then comparing them to find the smallest. Instead of that,
I thought about the numbers."
What's Avery's shortcut? How did he quickly find the dimensions of the rectangle with the smallest
perimeter?
Extra: Whose table will have the largest area?
Factors – Blue Solutions
1. 1. 23
2. 1, 2, 3, 4, 6, 8, 12, 24
3. 2
4. 8
5. 22  11; 6; 1, 2, 4, 11, 22, 44
6. 7  13; 4; 1, 7, 13, 91
7. 53; 4; 1, 5, 25, 125
8. 2  33; 8; 1, 2, 3, 6, 9 ,18,27,54
13. The positive integer factors of 28 are 1, 2, 4, 7, 14 and 28. The sum of their reciprocals is (1/1) + (1/2)
+ (1/4) + (1/7) + (1/14) + (1 /28) = (28 + 14 + + 4 + 2 + 1) ÷ 28 = 56 ÷ 28 = 2.
14. The positive integer factors of 225 are: 1, 3, 5, 9, 15, 25, 45, 75, and 225. Their sum is 403.
15. The 2, 3, 4 and 6 are all factors of 12, so we don’t have to worry about them. The 7, however, is
relatively prime with 12, so the LCM of 7 and 12, which is 84, is the smallest positive integer that is
divisible by 2, 3, 4, 6, 7 and 12.
16. The prime factorization of 648 is 23 x 34. Since the exponents on these prime factors are 3 and 4,
there are (3 + 1)(4 + 1) = 20 factors.
17. Multiples of 5 have a units digit of 0 or 5. No multiple of 4 ends in a 9, so our middle number must end
in 5. Working down from 200, we find that 194 is not a multiple of 4, but 184 is. 185 is divisible by 5
and 186 is divisible by 6, so we found the three consecutive integers. Their sum is 184 + 185 + 186 =
555.
18. Since factors generally come in pairs, the only numbers that have an odd number of distinct positive
factors are prefect-square numbers where one of the factors times itself makes number. The
greatest square number less than 150 is 144, which has 15 factors.
19. First, eliminate square number, because they have an odd number of factors, and therefore couldn’t
possibly have 10 factors. Then, list numbers with many factors. The answer is 48, which has the
factors 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
20. As shown below, there is tremendous cancellation, and fraction is equal to 101. Since 101 is a prime
number, it has exactly 2 factors.
101! 101 x 100! 101
100! =
100! =
21. Only square numbers can have an odd number of factors. Factors occur in pairs, so an odd number of
factors occurs only when the numbers in one pair are equal. For instance, the factors of 36 occur in
pairs: 1 x 36, 2 x 18, 3 x 12, 4 x 9, 6 x 6. But the last pair uses 6 twice, and when counting the factors
of 36, we’d only count 6 once. Hence, 36 has an odd number of factors, namely 9.
It helps to know that the number of factors for any prime to a power is one more than the power. For
example, 35 has 6 factors, and 23 has 4 factors. The only numbers with factors are squares of prime
numbers. Now, list the squares of primes in order, starting with 22:
22, 32, 52, 72, 112, 132, 172 ,….
2
The fifth number on this list is 11 = 121.
22. The product of the number of students in each row and the number of rows must be 288. The only
factor pairs of 288 where both factors are greater than 10 are 12 & 24 and 16 & 18. This tells us
there can be 12 rows with 24 students each, 16 rows with 18 students each and 18 rows with 16
students each. (Notice, there can’t be 24 rows of 12 students each.) The sum of possible numbers of
students per row is 24 + 16 + 18 = 58.
23. Fences for Fido and Friends. One possible set of dimensions is
A1= 9m x 6m A2= 7m x 9m A3= 7m x 4m A4= 6m x 4m
The area of the entire yard equaled 169 square meters. The figure was a square. The rule for area is L
x W. Therefore the the sides must be 13 and 13. Since they told me that the area of A4= 24 square
meters, the sides could be 6m & 4m. By subtracting 4 from 13 I got 9 for 1 side of A1. To get the
other side of A1 you have to subtract 6 from 13 and get 7. And since 9 times 6 equals 54 that is the
area of A1. To get the dimensions for A2 you have to take 6 from 13 and get 7 for one of the
dimensions. For the other dimension you take 4 away from 13 and get 9 for the dimensions for A2. To
get the dimensions for A3 you have to subtract 9 from 13 and get 4. And for the other side you have
to subtract 6 from 13 and get 7 for the dimensions of A3.
24. Tiling Table. Avery's short cut was to find a rectangular shape that was closest to a square. He came
up with two factors of 120 (number of tiles) that were closest to each other - 10 and 12.
Avery knew that the longer he made his table the larger the perimeter would be. If he laid his tiles
out next to each other, every tile would have at least two sides that would require the frame. The
closer he laid the tiles together and made the table closer to square shape, the fewer exposed sides
of tiles there would be and he would
need less of the expensive frame.
Extra Credit - Avery's table would have the largest area. Each tile is 36 sq.in. each (6in. x 6in.). He
has 120 tiles for a total of 4320 sq.in. (120 tiles x 36 sq.in). Geavonnie's tiles are 16 sq.in. each (4in. x
4in.). She has 256 tiles for a total of 4096 sq. in.
Bibliography Information
Teachers attempted to cite the sources for the problems included in this problem set. In some cases,
sources were not known.
Problems
Bibliography Information
23 , 24
The Math Forum @ Drexel
(http://mathforum.org/)
13 - 22
Math Counts (http://mathcounts.org)
1 - 12
Davison, David M. Prentice Hall Pre-Algebra
Tools for a Changing World. Needham, Mass:
Prentice Hall, 2001. Print.