Comparing and Ordering Fractions – Blue Problems
Enrichment 5-1 A Rational Maze
1. Find the correct path from start to finish through the maze. Proceed from one
circle to the next only if the second number is greater than the first number.
2. When the least common multiple of 8 and 20 is multiplied by the greatest common
factor of 8 and 20, what is the result?
3. The sum of three consecutive integers is a multiple of 89 and is a value between
600 and 900. What is the sum of the three integers?
4. How many positive integers less than or equal to 50 are multiples of 3 or 4, but not
5?
5. The house numbers in a housing development are the multiples of four starting at 4
and ending at 120. If metal digits cost 50 cents each, what is the total cost of all
of the metal digits that are needed to number the houses in this development?
6. Explain the following statement: When rewriting fractions with a common
denominator, the common denominator can be any common multiple of the two
original denominators, however, using the least common multiple is the most
convenient choice.
7. Multiple Days. For how many days during the year is the day a multiple of the
month?
8. What is the least positive multiple of 72 that has exactly 16 positive factors?
9. Count Off. At a school, the children lined up in a long row and counted off. A hat
was given to the sixteenth child in line and every sixteenth child after that. A
noisemaker was given to the twenty-fourth child and to every twenty-fourth child
after that. What were the positions in line of the first three children who received
both a hat and a noisemaker?
Comparing and Ordering Fractions – Blue Solutions
1.
2. The greatest common factor (GCF) of 8 and 20 is 4, and the least common multiple
(LCM) of 8 and 20 is 40. Their product is 4 x 40 = 160. Note that 8 x 20 = 160, also.
In general, the GCF(a, b) x LCM(a, b) = a x b.
3. There are four multiples of 89 between 600 and 900: 7 x 89 = 623, 8 x 89 = 712, 9
x 89 = 801, and 10 x 89 = 890. The sum of any three consecutive integers is always
a multiple of 3. Since 89 is prime and 9 is the only multiple of 3 in our list, the sum
of the three consecutive integers can only be 801. Incidentally, the three
consecutive integers are 266, 267 and 268.
4. There are 16 multiples of 3 less than 50 and 12 multiples of 4. We have counted the
4 multiples of 12 twice, however, so we must subtract 4. That’s 16 + 12 – 4 = 24 so
far. Now we need to exclude any multiples of five we might have counted. Those
would be the 3 multiples of 15 (15, 30 and 45) and the 2 multiples of 20 ( 20 and
40). That leaves 24 – 3 - 2 = 19 integers less than 50 that are multiples of 3 or 4,
but not 5.
5. We know that we are numbering houses from 4 through 120, but using only the
multiples of four. We could list these out, but let’s see if we can count them quickly.
Since 4 = 4 x 1 and 120 = 4 x 30, we are working with the first 30 multiples of 4.
Unfortunately, some are one-digit integers, others are two-digit integers and still
others have three digits. Let’s group them accordingly. (It helps to know that 100 is
the first three-digit multiple of 4.)
1-digit
4x1 = 4
4x2 = 8
2-digit
4x3 = 12
•
•
•
4x24 = 96
3-digit
4x25 = 100
•
•
•
4x30 = 120
2 mult.
24 – 2 = 22 mult.
30 – 24 = 6 mult.
The two one-digit multiples account for a total of two digits; there are 22 two-digit
multiples, resulting in 22x2 = 44 digits; and there are six three-digit multiples
yielding 6x3 = 18 digits. This is a total of 2 + 44 + 18 = 64 digits, which will cost 64
x $0.50 = $32.
6. Any common multiple will work, however, the LCM is smallest. This makes
multiplication easier.
7. Multiple Days. Ninety times. Consider this on a month-by-month basis.
Month Number
Number of Multiples
1
31
2
14
3
10
4
7
5
6
6
5
7
4
8
3
9
3
10
3
11
2
12
2
8. We know that 72 = 23 x 32, which means that there are 4 x 3 = 12 factors. If the
multiple of 72 included just one other prime factor different from 2 and 3, it would
follow that this multiple m could be represented by m = (23 x 32) x p, and it would
have 4 x 3 x 2 = 24 factors. Bringing a new prime factor into the picture creates
too many factors for the multiple. Therefore, let’s try to multiple (23 x 32) x 2 = 24
x 32, which has 5 x 3 = 15 factors. This doesn’t satisfy our condition, so we’re left
with trying (23 x 32) x 3 = 23 x 33, which has 4 x 4 = 16 factors. Our multiple is 23 x
33 = 216.
9. Count Off. Number 48, 96, and 144 are the first three common multiples of 16 and
24.
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
1
Davison, David M. Prentice Hall
Pre-Algebra Tools for a
Changing World. Needham,
Mass: Prentice Hall, 2001.
Print.
2–5,8
Math Counts
(http://mathcounts.org)
6
Larson, Boswell, Kanold, and
Stiff. Mathematics Concepts
and Skills Course 2 Math Log.
McDougal Littell, 2001.
7,9
Collier, C. Patrick. Menu
Collection Problems Adapted
from Mathematics Teaching in
the Middle School. New York:
National Council of Teachers of
Mathematics, 2000. Print.