TTP GRE - Chapter 3 - Incorrect Questions

The product of M and N is 24. M and N are positive integers and N is odd. - Quantity A: N - Quantity B: M A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

M×N = 24 M = 24/N since N is a positive odd integer. - If N = 1, M = 24, M>N - If N = 3, M = 8, M>N A) Quantity A is greater. Source: 3.7 - Factors

The product of M and N is 24 and N is a positive odd integer. - Quantity A: N - Quantity B: M A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

M×N = 24 M = 24/N since N is a positive odd integer. - If N = 1, M = 24, M>N - If N = 5, M = 24/5, M<N D) The relationship cannot be determined from the information given. Source: 3.7 - Factors

If N+12 is a positive odd integer, which of the following must be true about the sum of the next four integers? A) It is a multiple of 3 B) It is a multiple of 4 C) It is a prime number D) It is a multiple of 5 E) It is a multiple of 2

N+12 = Odd Next Integer (1) = Even Next Integer (2) = Odd Next Integer (3) = Even Next Integer (4) = Odd E) it is a multiple of 2. Source: 3.4.1 - Addition & Subtraction Rules for Even and Odd Numbers.

If X is a positive odd integer and Y is a negative even integer, which of the following must be true? A) X³ + Y is a positive idd integer B) X² + Y² is a negative odd integer C) X⁰ + Y¹¹ is a negative odd integer D) X + Y is a positive odd integer E) X + Y is a negative odd integer

Plugging in real numbers of X and Y... C) X⁰ + Y¹¹ is a negative odd integer Source: 3.4.1 - Addition & Subtraction Rules for Even and Odd Numbers.

At a certain high school, a student's popularity is determined by his or her locker number. Whoever has the locker number with the greatest number of distinct prime factors is the most popular student int he school. If Johanna, Jamal, Brianna, and Dyson get lockers with the numbers 300, 400, 150, 420, respectively, who is the most popular student? A) Johanna B) Jamal C) Brianna D) Dyson E) They have equal popularity

Prime Factorization of 300: 2²×3¹×5²; 3 distinct prime factors. Prime Factorization of 400: 2⁴×5²; 2 distinct prime factors. Prime Factorization of 150: 2×3¹×5²; 3 distinct prime factors Prime Factorization of 420: 2²×3¹×5¹×7¹; 4 distinct prime factors Dyson's locker number has the most distinct prime factors (4). Dyson is the most popular student. Source: 3.9.1 - Prime Factorization

Quantity A: The units digit of 7⁹⁵ Quantity B: The units digit of 3⁹⁵ A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

Quantity A: For the units digit of 7⁹⁵, a base of 7 produces the following pattern: - 7¹ = 7 - 7² = 9 - 7³ = 3 - 7⁴ = 1 As we can see from the above, the patterns of the units digit of any power of 7 repeat every 4 exponents. The pattern is 7-9-3-1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. Quantity B: The units digit of 3⁹⁵, a base of 3 produces the following pattern: - 3¹ = 3 - 3² = 9 - 3³ = 7 - 3⁴ = 1 As we can see from the above, the patterns of the units digit of any power of 3 repeat every 4 exponents. The pattern is 3-9-7-1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. B) Quantity B is greater. Source: 3.22.3 - Patterns in Units Digits:

What is the units digit of 18! + 4!? A) 0 B) 1 C) 2 D) 3 E) 4

Since 18! has at least one (5×2) pair, its units digit will be 0. With some multiplication we can see that 4! = 4×3×2×1 - 24. Thus, we are adding a number with a units digit of 0 to a number with a units digit of 4. Thus, the sum has units digits of 4 since 0 + 4 = 4. E) 4 Source: 3.15 - Determining the Number of Trailing Zeros in a number, 1.20 - Introduction to Factorials

On a certain plane, 2/5 of the passengers speak Farsi, and 3/4 speak Hebrew. If all of the passengers on the plane speak at least one of these languages, what is the smallest number of passengers that could be on the plane? A) 12 B) 15 C) 20 D) 24 E) 40

Since 2/5 of the passengers speak Farsi, and 3/4 speak Hebrew, the smallest number of passengers on the plane is the LCM of 5 and 4. Therefore 20 is the minimum number of passengers that could be on the plane. C) 20 Source: 3.13.7 - Word Problems Involving Divisibility

M¹¹P⁸N⁵ < 0 Quantity A: MN Quantity B: 0 A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

Since M¹¹P⁸N⁵ < 0, none of M¹¹, P⁸, N⁵ can be 0. Since P is raised to an even exponent, and P⁸ != 0, P⁸ must be positive. It follows that MN < 0. B) Quantity B is greater. Source: 3.6.8 - Even and Odd Exponents Versus Positive and Negative Answers.

All of the digits of a three-digit number N are greater than 1. The digits of N are distinct, and the product of the digits of N is 80. Quantity A: The average of the three digits of N. Quantity B: 5 A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

There is only one combination of three distinct digits that multiply to 80: 2, 5, and 8. The average of the digits will always be 5. C) The two quantities are equal. Source: 3.8 - Multiples, 3.9.1 - Prime Factorization

How many trailing zeros (zeros to the right of the last non-zero digit of a number) are in 25! ? A) 1 B) 3 C) 5 D) 6 E) 9

To determine the number of trailing zeros in 25!, we need to determine the number of (5×2) pairs, as each (5×2) creates one trailing zero. Thus, we need to determine the total number of twos and total numbers of fives in 25!; as there are more twos than fives in the prime factorization, we can just determine just the number of fives. 25/5¹ = 5 25/5² = 1 We see that there are 5 + 1 = 6 fives in 25!, which means that there are 6 (5×2) pairs, which means there are 6 trailing zeros. D) 6 Source: 3.15 - Determining the Number of Trailing Zeros in a Number, 3.17.2 - Shortcut for Determining the Numbers of Primes in a Factorial.

For which of the following values of M will M! have more than 6 trailing zeros? I. 21 II. 26 III. 31 A) None B) III Only C) II and III D) I and III E) I, II, and III

To determine the number of trailing zeros, we need to determine the number of (5×2) pairs, as each (5×2) creates one trailing zero. Thus, we need to determine the total number of twos and total numbers of fives in 25!; as there are more twos than fives in the prime factorization, we can just determine just the number of fives. 21 21/5¹ = 4 4 fives are in 21, which means there are 4 trailing zeros. 26 26/5¹ = 5 26/5² = 1 6 fives are in 26, which means there are 6 trailing zeros. 31 31/5¹ = 6 31/5² = 1 7 fives are in 26, which means there are 7 trailing zeros. B) III Only Source: 3.17.1 - Division Properties of Factorials, 3.17.2 - Shortcut for Determining the Number of Primes in a Factorial, 3.15 - Determining the Number of Trailing Zeros in a Number.

Which of the following are divisible by 2, 3, and 8? A) 33,333,333,000 B) 2,334,455,921 C) 444,444,400

To solve this, we need to use our divisibility rules for 2, 3, and 9. 2: The number must be an even number to be divisible by 2. 3: The sum of the digits of the number must be divisible by 3 for the number to the divisible by 3. 8: The numbers must be even and the last three digits must be divisible by 8 for the entire number to be divisible by 8. The number is also divisible by 8 if the original number ends in three zeros (000). A) 33,333,333,000 Source: 3.10 - the Least Common Multiple (LCM)

Quantity A: The units digits of 300³ + 301³ + 302³ + 303³ + 304³ Quantity B: The units digits of 300⁴ + 301⁴ + 302⁴ + 303⁴ + 304⁴ A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

We start by evaluating Quantity A. Since we care only about the units digit of the sum, we can determine the value of each units digit raised to the 3rd power in 300³ + 301³ + 302³ + 303³ + 304³, and 4th power in 300⁴ + 301⁴ + 302⁴ + 303⁴ + 304⁴. Quantity A's Units Digits = 0+1+8+7+4 = 20, Quantity A = 0. Quantity B's Units Digits = 0+1+6+1+6 = 14, Quantity B = 4. B) Quantity B is greater. Source: 3.22.3 - Patterns in Units Digits

Sara is an avid lottery player. In the certain games she plays, she must pick one number between 30 and 39, inclusive, one number between 40 and 49, inclusive, and one number between 50 and 59, inclusive. She believes that she will have the best chance of winning if her three numbers, as a set, have the greatest number of distinct prime factors possible. According to Sara's theory, which of the following sets of three numbers should she use? A) 32-48-52 B) 33-42-56 C) 39-40-54 D) 38-49-51 E) 36-42-56

We'll find the factors of each set of numbers and then determine which set has the greatest number of distinct or different prime numbers. D) 38-49-51 has the most distinct prime numbers. Source: 3.9.1 - Prime Factorization

If X !=0, and (-Y²-XY)/X = Y+X, the value represented by which of the following expressions is equal to its own opposite? A) X² B) X² + Y² C) Y² D) (Y+X)² E) (X=Y)²

- (-Y²-XY)/X = Y+X - X[(-Y²-XY)/X = Y+X] - -Y²-XY = XY +X² - XY + x² + Y² + XY = 0 - Y² + 2XY + X² = 0 - (Y+X)² = 0 D) (Y+X)² Source: 3.3.1 - Properties of Zero

What is the sum of the greatest common factor and the lowest common multiple of 72 and 120?

- Prime Factorization of 72: 3²×2³. - Prime Factorization of 120: 2³×3¹×5¹. LCM: - Prime Factorization of 72: 3²×2³. - Prime Factorization of 120: 2³×3¹×5¹. LCF of 72 and 120 is 2³×3²×5¹ = 360 GCF: - Prime Factorization of 72: 3²×2³. - Prime Factorization of 120: 2³×3¹×5¹. GCF of 72 and 120 = 2³×3¹ = 24 Sum of GCF and LCM: 24 + 360 = 384. Sources: 3.10.1 - Finding the LCM, 3.11.1 - Finding the GCF

- Quantity A: The number of prime numbers between 40 and 50, inclusive. - Quantity B: The number of prime numbers between 50 and 60, inclusive. A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

- Prime Numbers between 40 and 50, inclusive: 41, 43, and 47 (so 3 total). - Prime Numbers between 50 and 60, inclusive: 53 and 59 (so 2 total). A) Quantity A is greater. Source: 3.9 - Prime Numbers

X, Y, A, and B are positive integers, X is a multiple of A, and B is a factor of Y. Quantity A: The average of X and Y Quantity B: The average of A and B A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

- We know that X is a multiple of A, so X >=A. - We also know that B is a factor of Y, so B <= Y. Since X is always greater than or equal to A, and Y is always greater than or equal to B, the average of X and Y will be always greater than or equal to the average of A and B. However, we don't know which case it must be: greater than or equal to. If the former is true, then Quantity A is greater, but if the latter case is true, then the two quantities are equal. D) The relationship cannot be determined from the information given. Source: 3.7 - Factors

What is the greatest value of a positive integer X such that (2^X) is a factor of 100⁸⁰? A) 40 B) 80 C) 100 D) 160 E) 240

100⁸⁰ = (10²)⁸⁰ = 10¹⁶⁰ - (2×5)¹⁶⁰ D) 160 Source: 3.13.4 - Divisibility with Exponents

How many integers are between 3 and 86/7, inclusive? A) 8 B) 9 C) 10 D) 11 E) 12

86/7 = 12 + (2/7) - Because the word inclusive is used, we include 3 in the list of integers. - Thus, the numbers are 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. C) 10 Source: 3.2 - Integers

If B and C are digits, and 9BC is a 3-digit number that is divisible by 3, which of the following could be a possible product of B and C? Indicate all such answers. A) 0 B) 3 C) 4 D) 9 E) 15 F) 16 G) 21

A) 0 D) 9 Source: 3.13.8 - Divisibility Rules

Which of the following is true? A) The least common multiple of 9 and 12 is greater than the greatest common factor of 36 and 72. B) The greatest common factor of 12 and 24 is greater than the least common multiple of 6 and 15. C) The least common multiple of 20 and 12 is greater than the greatest common factor of 96 and 64. D) The greatest common factor of 32 and 48 is greater than the least common multiple of 7 and 30. E) The least common multiple of 10 and 12 is less than the greatest common factor of 60 and 45.

C) The least common multiple of 20 and 12 is greater than the greatest common factor of 96 and 64. Source: 3.10.1 - Finding the LCM, 3.11.1 - Finding the GCF.

N is a one-digit number, N+4 is a two-digit number, and N has only two factors. - Quantity A: N - Quantity B: 7 A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

C) The two quantities are equal. If a one-digit number N is added to 4 and the sum is a two-digit number, then n must be 6, 7, 8, or 0. Now every number N has at least two factors -N (itself) and 1. An integer greater than 1 is a prime if and only if it has no other factors. Now N has only two factors, which means N is a prime number, and 7 is the only prime number among 6, 7, 8, and 9. So N is 7. Source: 3.7 - Factors

If X is an integer and (X)(X²)(X³) is positive, which of the following could be negative? A) X⁴ B) (X)(X³) C) (X)(X) D) X + X³ E) 3X²

It is important to understand that when a nonzero quantity is raised to an even exponent, the result will be positive regardless of the sign of the base. Since we are told that (X)(X²)(X³), which is really X⁶, is positive, X is either positive or negative (the only value X can't be is zero). The question wants to know which answer choice could be negative. By looking at the given answer choices, we can disregard that X is positive since if X is positive, then none of the answer choices could be negative. So X itself must be negative. A: X⁴ (-)⁴ → + B: (-)(-)³ → (-)(-) → + C) (X)(X) → (-)(-) → + D) X + X³ → (-) + (-)³ → (-) + (-) → - E) 3X² → 3 × (-)² → + D) X + X³ Source: 3.6.8 - Even and Odd Exponents Versus Positive and Negative Answers.

If N is a prime number between 0 and 100, how many positive divisors does n³ have? A) 1 B) 2 C) 3 D) 4 E) 5

Let's test a few numbers. The smallest prime number is 2, so let's start with it. 2³ =8, and the factors of 8 are 1, 2, 4, and 8. So there are 4 positive divisors. Let's test one other small prime number, 3. 3³ is 27 and it's divisible by 1, 3, 9, and 27. With any prime number that we pick, its cube will always have 4 positive divisors. D) 4 Source: 3.7 - Factors, 3.9 - Prime Numbers