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Unit 4 – Number Patterns and Fractions
Lesson 2 – GCF and Equivalent Factorization
Black – GCF and Equivalent Factorization
Here is a set of mysteries that will help you sharpen your thinking skills. In each
exercise, use the clues to discover the identity of the mystery fraction.
1. My numerator is 6 less than my denominator.
I am equivalent to 3 .
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2. My denominator is 5 more than twice my numerator.
I am equivalent to 1 .
3
3. The GCF of my numerator and denominator is 3.
I am equivalent to 2 .
5
4. The GCF of my numerator and denominator is 5.
I am equivalent to 4 .
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5. My numerator and denominator are prime numbers.
My numerator is one less than my denominator.
6. My numerator and denominator are prime numbers.
The sum of my numerator and denominator is 24.
7. My numerator is divisible by 3.
My denominator is divisible by 5.
My denominator is 4 less than twice my numerator.
8. My numerator is divisible by 3.
My denominator is divisible by 5.
My denominator is 3 more than twice my numerator.
9. My numerator is a one-digit prime number.
My denominator is a one-digit composite number.
I am equivalent to 8 .
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10. My numerator is a prime number.
The GCF of my numerator and denominator is 2.
I am equivalent to 1 .
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11. Problem Solving. A store has 7 regular customers. Customer 1 shops at the store
everyday. Customer 2 shops at the store every second day. Customer 3 shops at the
store every third day etc. How many days from today will all 7 regular customers
shop at the store on the same day?
12. The number n is a prime number between 20 and 30. If you divide n by 8, the
remainder is 5. What is the value of n?
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Unit 4 – Number Patterns and Fractions
Lesson 2 – GCF and Equivalent Factorization
13. Five consecutive two-digit positive integers, each less than 30, are not prime. What
is the largest of these five integers?
14. What is the sum of the three distinct prime factors of 47,432?
15. What is the tens digit of the product of the first six prime numbers?
16. What is the greatest prime factor of 3105?
17. What is the positive difference between the two largest prime factors of 159,137?
18. What number can be subtracted from both the numerator and denominator of 19/24
so that the resulting fraction will be equivalent to 3/4?
19. The fraction x does not change its value when 3 is added to both its numerator and
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its denominator. What is the value of x?
20. A fraction is equivalent to 3/5. Its denominator is 60 more than its numerator. What
is the numerator of this fraction?
21. Find two pairs of variable expressions that have 24xy2 as their greatest common
factor.
22. The sum of the numerator and denominator of a fraction is 160. The fraction is
equivalent to 3 . What is the original fraction?
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23. What is the sum of the reciprocals of the positive integer factors of 28?
24. What is the sum of the positive integer factors of 225?
25. What is the smallest positive integer that has 2, 3, 4, 6, 7 and 12 as factors?
26. What is the number of positive factors of 648?
27. 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?
28. What is the greatest whole number less than 150 that has an odd number of distinct
positive factors?
29. What is the least positive integer with exactly 10 factors?
30. 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?
31. When the positive integers with exactly three factors are listed in ascending order,
what is the fifth number listed?
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Unit 4 – Number Patterns and Fractions
Lesson 2 – GCF and Equivalent Factorization
32. 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?
Solutions
1.
18
24
2.
5
15
3.
4.
5.
6
15
10
15
2
3
7.
12
20
8.
6
15
9.
2
8
10.
6. 2
10
11.LCM of all 7 numbers → 420 days
12. The primes between 20 and 30 are 23 and 29. If we divide 23 by 8, we get a
remainder of 7. If we divided 29 by 8, we get a remainder of 5. The value of n must
be 29.
13. This problem is asking about two-digit numbers that are not prime, in other words,
they will each be composite numbers. We are given an upper limit of 30, so all of
these numbers will be in the teens or twenties. Let’s go ahead and list them out,
circling the numbers that are not prime:
10 11 12 13 14
15
16 17 18
19 20 21
22 23 24 25 26 27 28 29
5 non-primes
The largest of the five consecutive non-prime integers is 28.
14. The prime factorization of 47,432 is 23 x 72 x 112. The sum of the three prime
factors is 2 + 7 + 11 = 20.
15. The first six prime numbers are 2, 3, 5, 7, 11 and 13. The product of the three
primes 7, 11 and 13 is the very special number 1001. This is helpful in determining the
product of all the numbers. Since 2 – 3 – 5 = 30 and 7 – 11 – 13 = 1001, then the
product of all six primes is 30 – 1001 = 30,030. The tens digit is 3.
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Unit 4 – Number Patterns and Fractions
Lesson 2 – GCF and Equivalent Factorization
16. Since the digits of 3105 have a sum of 3 + 1 + 0 + 5 = 9, 3105 is divisible by 9. In
fact, 3105 = 32 x 345. However, the digits of 345 have a sum of 3 + 4 + 5 = 12, so
345 must be divisible by 3. This implies that 3105 = 33 x 115. Finally, since 115 ends in
5, it is divisible by 5, giving 3105 = 33 x 5 x 23. The greatest prime factor of 3105 is
23.
17. Prime factoring a number is arguably the most difficult thing to do in mathematics.
There’s no quick way to do it. But, the divisibility rules will show that no prime less
than 11 will divide the number in the problem. Using the divisibility rules is a good way
to get started.
The divisibility rule for 2 says that a number is divisible by 2 if the last digit is even.
The last digit is 7, so the number is not divisible by 2.
The divisibility rule for 3 says that a number is divisible by 3 if the sum of the digits
is divisible by 3. Since 1 + 5 + 9 + 1 + 3 + 7 = 26 is not a multiple of 3, then 159,137 is
not divisible by 3, either.
A number divisible by 5 has units digit 0 or 5, so 159,137 is not divisible by 5.
The rule for divisibility by 7 is tricky. In short, remove the last digit double it and
subtract that from the number remaining; repeat until a number is reached that can
identified as divisible or not divisible by 7. For 159,137; 15,913 – 2(7) = 15,899; then
1589 – 2(9) = 1571; then 157 – 2(1) = 155; and, finally, 15 – 2(5) = 5, which is not
divisible by 7, so 159,137 is not divisible by either.
I determine of a number is divisible by 11, alternately add and subtract the digits.
For this number, 1 – 5 + 9 – 1 + 3 – 7 = 0, which is a multiple of 11, so the number is
divisible by 11.
In fact, 159,137 = 11 – 14,467. Then, a calculator can be used to show that 14,467 =
17 – 23 – 37. The difference between the two greatest factors is 37 – 23 = 14.
18. Generate fractions by subtracting 1, 2, 3, …, from the numerator and denominator of
given fraction 19/24. Stop when the result is a fraction equivalent to ¾. Subtracting
1 gives 18/23. Subtracting 2 gives 17/22, subtracting 3 gives 16/21, subtracting 4
gives 15/20 = 3/4. The number 4 was subtracted from the numerator and
denominator.
19. If x is 5, we have 5/5 = 1 before adding 3’s to numerator and denominator and 8/8 =
1 afterward. To find the algebraically, we could set up the equation x = x + 3 and
5
solve for x.
5+ 3
20. In the simplified fraction 3 , the numerator is 2 less than the denominator. In the
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equivalent fraction, the numerator 60 less than the denominator. This means that 2
units in the simplified fraction equal 60 units in the equivalent fraction, so 3 units
would equal 90. The equivalent fraction must be 90 , and the numerator is 90.
150
21. Sample answers: 48xy2z and 72x2y3, 24x5y3z and 96xy2.
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Unit 4 – Number Patterns and Fractions
Lesson 2 – GCF and Equivalent Factorization
22. Let’s assume we do not have any idea where to start. We know we need a fraction
equivalent to 3 . Multiplying the numerator and denominator by 2 gives us the
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equivalent fraction 6 , but the sum of the numerator and denominator is only 20.
14
The next fraction is 9 , but again its sum of 30 falls way short of 160. The next
21
fraction is 12 , and gives a sum of 40. We are still far from the sum of 160 we need,
28
but notice that our sum is increasing by 10 each time we increase the factor by which
we are multiplying the numerator and denominator. We just multiplied both by 4. We
need the sum to go up another 160 – 40 = 120, which is 12 more 10s. Raising our
factor of 4 by 12, we get 16. Let’s try multiplying the numerator and denominator by
16. We get 48 , which has the correct sum of 160.
112
Rather than using the above Guess, Check & Revise method, we could set up the
situation algebraically. We know we have to multiply the numerator and denominator
each by the same number, x, in order to get an equivalent fraction. We also know the
hew numerator and denominator must add to 160, so 3x + 7x = 160 or 10x = 160 or x =
16. We have determined we must multiply the numerator and denominator by 16, and
this yields 48 . Can we see now why the sums in our first solution kept increasing by
112
10?
23. 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.
24. The positive integer factors of 225 are: 1, 3, 5, 9, 15, 25, 45, 75, and 225. Their sum
is 403.
25. 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.
26. 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.
27. 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.
28. 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.
3
Unit 4 – Number Patterns and Fractions
Lesson 2 – GCF and Equivalent Factorization
29. 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.
30. 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! =
31. 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 ,….
The fifth number on this list is 112 = 121.
32. 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.
4
Unit 4 – Number Patterns and Fractions
Lesson 2 – GCF and Equivalent Factorization
Bibliography Information
Teachers attempted to cite the sources for the problems included in this problem set. In some cases,
sources may not have been known.
Problems
Bibliography Information
13-33
Math Counts (http://mathcounts.org)
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