Apply Number Properties - Praxis Core Math

1. When a positive integer $k$ is divided into 150, the remainder is 6. When $k$ is divided into 205, the remainder is 5. What is the greatest possible value of $k$?
   Answer: 8 If dividing 150 by $k$ gives a remainder of 6, it means that $150 - 6 = 144$ is perfectly divisible by $k$. If dividing 205 by $k$ gives a remainder of 5, it means that $205 - 5 = 200$ is perfectly divisible by $k$. So, $k$ must be a common divisor of 144 and 200. To find the greatest possible value of $k$, we find the greatest common divisor (GCD) of 144 and 200. $144 = 12^2 = (2^2 \cdot 3)^2 = 2^4 \cdot 3^2$ $200 = 2 \cdot 100 = 2 \cdot 10^2 = 2 \cdot (2 \cdot 5)^2 = 2^3 \cdot 5^2$ The GCD is $2^3 = 8$. A divisor must be larger than the remainder. Since $k=8$ is larger than both remainders (6 and 5), it is a valid answer.
2. Two positive integers have a least common multiple of 84 and a greatest common divisor of 2. If one of the integers is 12, what is the other integer?
   Answer: 14 For any two positive integers $a$ and $b$, $a \cdot b = \text{GCD}(a, b) \cdot \text{LCM}(a, b)$. Let the two integers be 12 and $x$. We are given LCM = 84 and GCD = 2. So, $12 \cdot x = 2 \cdot 84$, which is $12x = 168$. Dividing both sides by 12 gives $x = 14$. We can check our answer: GCD(12, 14) = 2, and LCM(12, 14) = LCM($2^2 \cdot 3$, $2 \cdot 7$) = $2^2 \cdot 3 \cdot 7 = 84$.
3. How many positive integers are common factors of 36 and 90?
   Answer: 6 The common factors of two numbers are the factors of their greatest common divisor (GCD). First, find the GCD of 36 and 90. $36 = 2^2 \cdot 3^2$ $90 = 2 \cdot 3^2 \cdot 5$ The GCD is $2^1 \cdot 3^2 = 2 \cdot 9 = 18$. Now, find how many factors the GCD (18) has. The factors of 18 are 1, 2, 3, 6, 9, and 18. There are 6 common factors.
4. The least common multiple of an integer $x$ and 18 is 72. The greatest common divisor of $x$ and 18 is 6. What is the value of $x$?
   Answer: 24 We can use the property that for any two positive integers $a$ and $b$, $a \cdot b = \text{GCD}(a, b) \cdot \text{LCM}(a, b)$. In this case, $a=x$ and $b=18$. So, $x \cdot 18 = \text{GCD}(x, 18) \cdot \text{LCM}(x, 18)$. We are given $\text{LCM}(x, 18) = 72$ and $\text{GCD}(x, 18) = 6$. Plugging in the values: $x \cdot 18 = 6 \cdot 72$, which simplifies to $x \cdot 18 = 432$. Solving for $x$: $x = 432 \div 18 = 24$.
5. A florist has 90 roses and 75 tulips to create identical bouquets for a wedding. If she must use all the flowers with none left over, what is the greatest number of identical bouquets she can create?
   Answer: 15 To find the greatest number of identical bouquets, we need to find the greatest common divisor (GCD) of the number of roses (90) and tulips (75). The number of bouquets must be a factor of both 90 and 75. Prime factorization of 90 is $2 \cdot 3^2 \cdot 5$. Prime factorization of 75 is $3 \cdot 5^2$. The GCD is the product of the lowest powers of common prime factors: $3^1 \cdot 5^1 = 15$. Therefore, the greatest number of identical bouquets she can create is 15.
6. If $x$ and $y$ are positive integers such that $x$ is a multiple of 4 and $y$ is a multiple of 5, then the product $xy$ must be a multiple of which of the following?
   Answer: 20 We are given that $x$ is a multiple of 4, so we can write $x = 4a$ for some integer $a$. We are also given that $y$ is a multiple of 5, so we can write $y = 5b$ for some integer $b$. The product $xy$ is $(4a)(5b) = 20ab$. Since $a$ and $b$ are integers, $ab$ is an integer. Therefore, $xy$ is always 20 times some integer, which means $xy$ must be a multiple of 20.
7. The number $N$ is defined by its prime factorization, $N = 2^3 \cdot 5^2 \cdot 11$. Which of the following integers is NOT a factor of $N$?
   Answer: 75 A number is a factor of $N$ if its prime factorization is a 'subset' of $N$'s; that is, it only contains prime factors of $N$ raised to powers less than or equal to their powers in $N$. The prime factors of $N$ are 2, 5, and 11. A) $40 = 8 \cdot 5 = 2^3 \cdot 5^1$. This is a factor of $N$. B) $55 = 5^1 \cdot 11^1$. This is a factor of $N$. C) $75 = 3 \cdot 25 = 3^1 \cdot 5^2$. This is NOT a factor of $N$ because it contains the prime factor 3, which is not a prime factor of $N$. D) $100 = 4 \cdot 25 = 2^2 \cdot 5^2$. This is a factor of $N$.
8. If an integer is divisible by both 12 and 15, then it must also be divisible by which of the following?
   Answer: 30 An integer divisible by both 12 and 15 must be a multiple of their least common multiple (LCM). Let's find the LCM of 12 and 15. $12 = 2^2 \cdot 3$ $15 = 3 \cdot 5$ The LCM is $2^2 \cdot 3 \cdot 5 = 60$. Any number that is a multiple of 60 must be divisible by any factor of 60. Let's check the options to see which one is a factor of 60. A) 18 is not a factor of 60. B) 24 is not a factor of 60. C) 30 is a factor of 60. Therefore, any multiple of 60 must also be a multiple of 30. D) 180 is a multiple of 60, not a factor, so not every multiple of 60 is a multiple of 180.
9. At a bus station, Bus A departs every 12 minutes and Bus B departs every 20 minutes. If both buses depart together at 1:00 PM, what time will it be the third time they depart together (including the 1:00 PM departure)?
   Answer: 3:00 PM First, find the time interval at which the buses depart together. This is the least common multiple (LCM) of 12 and 20. $12 = 2^2 \cdot 3$ $20 = 2^2 \cdot 5$ LCM = $2^2 \cdot 3 \cdot 5 = 60$. The buses depart together every 60 minutes, or every hour. The first departure together is at 1:00 PM. The second will be 60 minutes later at 2:00 PM. The third departure will be another 60 minutes later, at 3:00 PM.
10. A positive integer $n$ has exactly three positive factors. Which of the following must be true about $n$?
    Answer: $n$ is the square of a prime number. The number of factors of an integer is determined by the exponents in its prime factorization. If the factorization is $p_1^{a_1} p_2^{a_2} \dots$, the number of factors is $(a_1+1)(a_2+1)\dots$. For a number to have exactly 3 factors, this product must equal 3. Since 3 is a prime number, there can only be one term in the product, and it must be 3. This means $a_1+1=3$, so $a_1=2$. The prime factorization must be of the form $p^2$ for some prime number $p$. Therefore, $n$ must be the square of a prime number. For example, $9=3^2$ has factors 1, 3, 9. $4=2^2$ has factors 1, 2, 4.
11. The integer $k$ is a multiple of 6 and a factor of 120. Which of the following is NOT a possible value for $k$?
    Answer: 48 We need to check which of the given options satisfies both conditions. The conditions are: (1) $k$ is a multiple of 6, and (2) $k$ is a factor of 120. Let's test each option. A) 12 is a multiple of 6 ($6 \cdot 2 = 12$) and a factor of 120 ($120 \div 12 = 10$). So, 12 is a possible value. B) 30 is a multiple of 6 ($6 \cdot 5 = 30$) and a factor of 120 ($120 \div 30 = 4$). So, 30 is a possible value. C) 48 is a multiple of 6 ($6 \cdot 8 = 48$), but it is not a factor of 120 ($120 \div 48 = 2.5$). So, 48 is NOT a possible value. D) 60 is a multiple of 6 ($6 \cdot 10 = 60$) and a factor of 120 ($120 \div 60 = 2$). So, 60 is a possible value.
12. A rectangular garden measuring 42 feet by 54 feet is to be divided into identical square plots with no area left over. What is the minimum number of square plots required?
    Answer: 63 This is a two-step problem. First, to minimize the number of square plots, we must maximize the size of each square plot. The side length of the largest possible square plot is the greatest common divisor (GCD) of the garden's dimensions, 42 and 54. $42 = 2 \cdot 3 \cdot 7$ $54 = 2 \cdot 3^3$ The GCD is $2 \cdot 3 = 6$. So, the largest possible square plots are 6 feet by 6 feet. Second, find how many of these plots fit into the garden. The number of plots along the 42-foot side is $42 \div 6 = 7$. The number of plots along the 54-foot side is $54 \div 6 = 9$. The total number of plots is $7 \cdot 9 = 63$.
13. Let $M$ be the least common multiple of 18, 24, and 30. Which of the following is a prime factor of $M$?
    Answer: 5 To find the least common multiple (LCM), we first find the prime factorization of each number. $18 = 2 \cdot 9 = 2 \cdot 3^2$ $24 = 8 \cdot 3 = 2^3 \cdot 3$ $30 = 3 \cdot 10 = 2 \cdot 3 \cdot 5$ The LCM is the product of the highest powers of all prime factors involved (2, 3, and 5). $M = \text{LCM} = 2^3 \cdot 3^2 \cdot 5^1 = 8 \cdot 9 \cdot 5 = 360$. The prime factors of $M$ are 2, 3, and 5. Of the choices given, only 5 is a prime factor of $M$.
14. The product of two positive integers is 600, and their least common multiple is 120. What is their greatest common divisor?
    Answer: 5 For any two positive integers $a$ and $b$, the product of the numbers is equal to the product of their greatest common divisor (GCD) and least common multiple (LCM). The formula is $a \cdot b = \text{GCD}(a, b) \cdot \text{LCM}(a, b)$. We are given that $a \cdot b = 600$ and $\text{LCM}(a, b) = 120$. Plugging these values into the formula gives $600 = \text{GCD}(a, b) \cdot 120$. To find the GCD, divide the product by the LCM: $\text{GCD} = 600 \div 120 = 5$.
15. How many distinct positive integers are factors of the number 180?
    Answer: 18 To find the number of factors of an integer, first find its prime factorization. $180 = 18 \cdot 10 = (2 \cdot 3^2) \cdot (2 \cdot 5) = 2^2 \cdot 3^2 \cdot 5^1$. The number of factors is found by taking each exponent in the prime factorization, adding 1 to it, and then multiplying these new numbers together. The exponents are 2, 2, and 1. So, the number of factors is $(2+1)(2+1)(1+1) = 3 \cdot 3 \cdot 2 = 18$.
16. Two positive integers have a sum of 28 and a greatest common divisor of 4. If one of the integers is 12, what is their least common multiple?
    Answer: 48 Let the two integers be 12 and $b$. We are given their sum is 28, so $12 + b = 28$, which gives us $b = 16$. We need to verify that GCD(12, 16) = 4. Finding the prime factorizations: $12 = 2^2 \cdot 3$ and $16 = 2^4$. The GCD is $2^2 = 4$, which matches the given condition. Now we find the LCM using the formula: LCM(12, 16) = $\frac{12 \times 16}{\text{GCD}(12, 16)} = \frac{192}{4} = 48$.
17. For a school event, a teacher wants to arrange chairs in rows. Whether the teacher places 5, 6, or 8 chairs in a row, there are always 2 chairs left over. What is the smallest possible number of chairs greater than 2 that the teacher could have?
    Answer: 122 Let $C$ be the total number of chairs. The problem states that if you subtract 2 from $C$, the result is a multiple of 5, 6, and 8. So, $C-2$ must be the least common multiple (LCM) of 5, 6, and 8. $5 = 5^1$ $6 = 2 \cdot 3$ $8 = 2^3$ The LCM is $2^3 \cdot 3 \cdot 5 = 8 \cdot 3 \cdot 5 = 120$. So, $C-2 = 120$. This means $C = 122$. This is the smallest number of chairs greater than 2 that satisfies the conditions.
18. Let $G$ be the greatest common divisor of 48 and 72, and let $L$ be the least common multiple of 10 and 12. What is the value of $L - G$?
    Answer: 36 This problem requires two separate calculations. First, find $G$, the GCD of 48 and 72. $48 = 16 \cdot 3 = 2^4 \cdot 3$ $72 = 8 \cdot 9 = 2^3 \cdot 3^2$ GCD(48, 72) = $2^3 \cdot 3^1 = 8 \cdot 3 = 24$. So, $G = 24$. Second, find $L$, the LCM of 10 and 12. $10 = 2 \cdot 5$ $12 = 2^2 \cdot 3$ LCM(10, 12) = $2^2 \cdot 3 \cdot 5 = 4 \cdot 3 \cdot 5 = 60$. So, $L = 60$. Finally, calculate $L - G = 60 - 24 = 36$.
19. If the greatest common divisor of two distinct positive integers $x$ and $y$ is $x$, which of the following statements must be true?
    Answer: $y$ is a multiple of $x$. The greatest common divisor (GCD) of two integers is the largest integer that divides both of them without a remainder. If the GCD of $x$ and $y$ is $x$ itself, it means that $x$ must divide both $x$ and $y$. Since $x$ always divides $x$, the condition implies that $x$ must also be a factor of $y$. If $x$ is a factor of $y$, then $y$ is a multiple of $x$. For example, GCD(4, 12) = 4; here, 12 is a multiple of 4.
20. The number $K$ is the product of three distinct prime numbers. Which of the following must be the number of positive factors of $K^2$?
    Answer: 27 Let the three distinct prime numbers be $p$, $q$, and $r$. Then $K = p \cdot q \cdot r = p^1 \cdot q^1 \cdot r^1$. Now, we find $K^2$: $K^2 = (p^1 \cdot q^1 \cdot r^1)^2 = p^2 \cdot q^2 \cdot r^2$. To find the number of factors of $K^2$, we add 1 to each exponent in its prime factorization and multiply the results. The exponents are 2, 2, and 2. So, the number of factors is $(2+1)(2+1)(2+1) = 3 \cdot 3 \cdot 3 = 27$.

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