Start learning →

GRE Real Numbers and Polynomials — practice questions

Explore 73 free Real Numbers and Polynomials questions in GRE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

Practise Real Numbers and Polynomials with Quill →

Real Numbers and Polynomials questions and exercises with answers

Showing questions 1–30 of 73.

For any integer $n$, the expression $n^2 - 1$ is always divisible by 8 when $n$ isRead the answer and worked solution →The product of any three consecutive positive integers is always divisible byRead the answer and worked solution →The number of trailing zeros at the end of $50!$ isRead the answer and worked solution →If $\text{HCF}(a, b) = 12$ and $\text{LCM}(a, b) = 180$, then the number of possible ordered pairs $(a, b)$ isRead the answer and worked solution →The largest number that divides 245 and 1029, leaving remainders 5 and 5 respectively, isRead the answer and worked solution →For some integer $m$, every odd integer can be written in the formRead the answer and worked solution →The square of any odd integer, when divided by 8, leaves remainderRead the answer and worked solution →The decimal expansion of $\dfrac{257}{2^3 \cdot 5^4}$ terminates after how many decimal places?Read the answer and worked solution →How many positive integers less than 1000 are divisible by neither 2 nor 5?Read the answer and worked solution →If $p$ is a prime and $p \mid a^2$, then which must be true?Read the answer and worked solution →The unit digit of $7^{2026}$ isRead the answer and worked solution →The smallest positive integer that is divisible by every integer from 1 to 10 isRead the answer and worked solution →If $a = 2^3 \cdot 3^4 \cdot 5^2$ and $b = 2^4 \cdot 3^2 \cdot 7$, then $\dfrac{\text{LCM}(a,b)}{\text{HCF}(a,b)}$ equalsRead the answer and worked solution →Which of the following is an irrational number?Read the answer and worked solution →The remainder when $2^{100}$ is divided by 7 isRead the answer and worked solution →Two positive integers have HCF 9 and LCM 90. If one number is 18, the other isRead the answer and worked solution →The HCF of two consecutive positive integers isRead the answer and worked solution →If $3^x = 4^y = 12^z$ with $x,y,z$ nonzero, then $\dfrac{1}{z}$ equalsRead the answer and worked solution →The degree of a polynomial is:Read the answer and worked solution →A polynomial of degree 1 is called a:Read the answer and worked solution →The general form of a quadratic polynomial is:Read the answer and worked solution →For $p(x) = x^2 - 3x - 4$, the value of $p(2)$ is:Read the answer and worked solution →The zeroes of a polynomial $p(x)$ are the values of $x$ for which:Read the answer and worked solution →For a quadratic $ax^2 + bx + c$, the sum of its zeroes is:Read the answer and worked solution →For a quadratic $ax^2 + bx + c$, the product of its zeroes is:Read the answer and worked solution →A quadratic polynomial can have at most how many zeroes?Read the answer and worked solution →The zeroes of the polynomial $x^2 - 3x - 4$ are:Read the answer and worked solution →Which of the following is NOT a polynomial?Read the answer and worked solution →A cubic polynomial can have at most how many zeroes?Read the answer and worked solution →A polynomial of degree 3 is called a:Read the answer and worked solution →

Worked practice questions

For any integer $n$, the expression $n^2 - 1$ is always divisible by 8 when $n$ is

  1. any integer
  2. an even integer
  3. a prime number
  4. an odd integer
Show answer and solution

Correct answer: D. an odd integer

If $n$ is odd, write $n = 2k+1$. Then $n^2 - 1 = 4k(k+1)$. Since $k(k+1)$ is a product of two consecutive integers it is even so $4k(k+1)$ is divisible by 8. For even $n$, $n^2-1$ is odd, so not divisible by 8.

Open this question →

The product of any three consecutive positive integers is always divisible by

  1. 4
  2. 12
  3. 9
  4. 6
Show answer and solution

Correct answer: D. 6

Among any three consecutive integers, at least one is divisible by 2 and exactly one is divisible by 3. Hence the product is divisible by $2 \times 3 = 6$. It need not be divisible by 4 (e.g. $1\cdot2\cdot3 = 6$).

Open this question →

The number of trailing zeros at the end of $50!$ is

  1. 12
  2. 13
  3. 11
  4. 10
Show answer and solution

Correct answer: A. 12

Trailing zeros are governed by the power of 5 in $50!$: $\lfloor 50/5 \rfloor + \lfloor 50/25 \rfloor = 10 + 2 = 12$. (Powers of 2 are more plentiful, so 5 is the limiting factor.)

Open this question →

How to practise

  1. Choose a Real Numbers and Polynomials question and try to work out your answer.
  2. Open the question to check the correct answer and read its worked solution.
  3. Continue in the app for bite-sized lessons, explanations and review with Quill. You can start without creating an account.

Put it into practice

Continue with Real Numbers and Polynomials practice for GRE. Read explanations as you go. No signup form needed to start.

Practise Real Numbers and Polynomials →

More Mathematics topics

MathematicsStatistics and ProbabilityArithmetic ProgressionSets and RelationsQuadratic EquationsUnit 1A1: Relations, Functions and GraphsUnit 6B: Definite Integration & Riemann SumsUnit 1B1: Limit Concepts and Graphical Limits