Putnam Mathematical Competition

Description: The Putnam Mathematical Competition is an annual mathematics competition for undergraduate students in the United States and Canada. It is one of the most prestigious mathematics competitions in the world, and is known for its challenging problems.
Number of Questions: 15
Created by:
Tags: mathematics mathematical competitions putnam mathematical competition
Attempted 0/15 Correct 0 Score 0

Let $f(x)$ be a continuous function on the interval $[0, 1]$. If $f(0) = 0$ and $f(1) = 1$, then there exists a point $c$ in the interval $(0, 1)$ such that $f(c) = c$.

  1. True

  2. False


Correct Option: A
Explanation:

This is a consequence of the Intermediate Value Theorem.

Let $A$ be a $3 imes 3$ matrix with real entries. If the determinant of $A$ is zero, then $A$ is not invertible.

  1. True

  2. False


Correct Option: A
Explanation:

If the determinant of $A$ is zero, then the columns of $A$ are linearly dependent, which means that $A$ is not invertible.

Let $p$ be a prime number. If $a$ and $b$ are integers such that $a^p + b^p = c^p$, then $a + b = c$.

  1. True

  2. False


Correct Option: B
Explanation:

This is false. For example, $3^2 + 4^2 = 5^2$, but $3 + 4 eq 5$.

Let $f(x)$ be a continuous function on the interval $[0, 1]$. If $f(x) > 0$ for all $x$ in $[0, 1]$, then there exists a point $c$ in the interval $(0, 1)$ such that $f(c) = 1$.

  1. True

  2. False


Correct Option: B
Explanation:

This is false. For example, $f(x) = x$ is continuous on $[0, 1]$, $f(x) > 0$ for all $x$ in $[0, 1]$, but there is no point $c$ in $(0, 1)$ such that $f(c) = 1$.

Let $A$ be a $3 imes 3$ matrix with real entries. If the eigenvalues of $A$ are all real, then $A$ is diagonalizable.

  1. True

  2. False


Correct Option: A
Explanation:

This is a consequence of the Spectral Theorem.

Let $p$ be a prime number. If $a$ and $b$ are integers such that $a^p + b^p = c^p$, then $a = b = c$.

  1. True

  2. False


Correct Option: A
Explanation:

This is a consequence of Fermat's Last Theorem.

Let $f(x)$ be a continuous function on the interval $[0, 1]$. If $f(x) > 0$ for all $x$ in $[0, 1]$, then there exists a point $c$ in the interval $(0, 1)$ such that $f(c) > 1$.

  1. True

  2. False


Correct Option: A
Explanation:

This is a consequence of the Intermediate Value Theorem.

Let $A$ be a $3 imes 3$ matrix with real entries. If the determinant of $A$ is nonzero, then $A$ is invertible.

  1. True

  2. False


Correct Option: A
Explanation:

This is a consequence of the fact that the determinant of a matrix is zero if and only if the matrix is not invertible.

Let $p$ be a prime number. If $a$ and $b$ are integers such that $a^p + b^p = c^p$, then $a + b + c = 0$.

  1. True

  2. False


Correct Option: A
Explanation:

This is a consequence of Fermat's Last Theorem.

Let $f(x)$ be a continuous function on the interval $[0, 1]$. If $f(x) > 0$ for all $x$ in $[0, 1]$, then there exists a point $c$ in the interval $(0, 1)$ such that $f(c) = 2$.

  1. True

  2. False


Correct Option: B
Explanation:

This is false. For example, $f(x) = x^2$ is continuous on $[0, 1]$, $f(x) > 0$ for all $x$ in $[0, 1]$, but there is no point $c$ in $(0, 1)$ such that $f(c) = 2$.

Let $A$ be a $3 imes 3$ matrix with real entries. If the eigenvalues of $A$ are all distinct, then $A$ is diagonalizable.

  1. True

  2. False


Correct Option: A
Explanation:

This is a consequence of the fact that a matrix is diagonalizable if and only if its eigenvalues are all distinct.

Let $p$ be a prime number. If $a$ and $b$ are integers such that $a^p + b^p = c^p$, then $a^2 + b^2 = c^2$.

  1. True

  2. False


Correct Option: B
Explanation:

This is false. For example, $3^2 + 4^2 = 5^2$, but $3^4 + 4^4 eq 5^4$.

Let $f(x)$ be a continuous function on the interval $[0, 1]$. If $f(x) > 0$ for all $x$ in $[0, 1]$, then there exists a point $c$ in the interval $(0, 1)$ such that $f(c) = 3$.

  1. True

  2. False


Correct Option: B
Explanation:

This is false. For example, $f(x) = x^3$ is continuous on $[0, 1]$, $f(x) > 0$ for all $x$ in $[0, 1]$, but there is no point $c$ in $(0, 1)$ such that $f(c) = 3$.

Let $A$ be a $3 imes 3$ matrix with real entries. If the determinant of $A$ is equal to the product of its eigenvalues, then $A$ is diagonalizable.

  1. True

  2. False


Correct Option: A
Explanation:

This is a consequence of the fact that the determinant of a matrix is equal to the product of its eigenvalues if and only if the matrix is diagonalizable.

Let $p$ be a prime number. If $a$ and $b$ are integers such that $a^p + b^p = c^p$, then $a + b + c = p$.

  1. True

  2. False


Correct Option: B
Explanation:

This is false. For example, $3^2 + 4^2 = 5^2$, but $3 + 4 + 5 eq 2$.

- Hide questions