Question

Which one of the following is an odd number?
$\begin{align}
  &A.\;{2001^2} + 3 \\
  &B.\;{2002^2} + 10 \\
  &C.\;{2003^2} + 7 \\
  &D.\;{2004^3} + 1 \\
\end{align} $

Answer 1

Hint: This problem requires the concept of odd and even numbers. An odd number is the one which gives a remainder of 1 when divided by 2. Any number which ends with 1, 3, 5, 7 or 9 is an odd number. We will try to determine the final digit of each of the four options, and then identify which one is an odd number.

Complete step-by-step answer:
In the four given options, we do not need to find the squares of such large numbers, we just need to determine the final digits, and add the respective number so that we can determine if they are odd or even.

In option A, we have been given the number ${2001^2} + 3$. We know that the square of 1 is 1 itself. This means that any number ending with 1 will have its square which is ending with 1 as well. So we can write that-
$\begin{align}
  &1 \times 1 = 1 \\
  &2001 \times 2001 = {2001^2} = \_\_\_1 \\
  &{2001^2} + 3 = \_\_\_1 + 3 = \_\_\_4\left( {number\;ending\;with\;4} \right) \\
\end{align} $
A number ending with 4 is divisible by 2, so this number will be an even number.

In option B, we have been given the number ${2002^2} + 10$. We know that the square of 2 is 4. This means that any number ending with 2 will have its square which is ending with 4. So we can write that-
$\begin{align}
  &2 \times 2 = 4 \\
  &2002 \times 2002 = {2002^2} = \_\_\_4 \\
  &{2002^2} + 10 = \_\_\_4 + 10 = \_\_\_4\left( {number\;ending\;with\;4} \right) \\
\end{align} $
A number ending with 4 is divisible by 2, so this number will be an even number.

In option C, we have been given the number ${2003^2} + 7$. We know that the square of 3 is 9. This means that any number ending with 3 will have its square which is ending with 9. So we can write that-
$\begin{align}
  &3 \times 3 = 9 \\
  &2003 \times 2003 = {2003^2} = \_\_\_9 \\
  &{2003^2} + 7 = \_\_\_9 + 7 = \_\_\_6\left( {number\;ending\;with\;6} \right) \\
\end{align} $
A number ending with 6 is divisible by 2, so this number will be an even number.

In option D, we have been given the number ${2004^3} + 1$. We know that the cube of 4 is 64. This means that any number ending with 4 will have its cube which is ending with 4 as well. So we can write that-
$\begin{align}
  &4 \times 4 \times 4 = 64 \\
  &2004 \times 2004 \times 2004 = {2004^3} = \_\_\_4 \\
  &{2004^3} + 1 = \_\_\_4 + 1 = \_\_\_5\left( {number\;ending\;with\;5} \right) \\
\end{align} $
A number ending with 5 is not divisible by 2, so this number will be an odd number.
From the given options only the last option is an odd number, hence the correct option is D.

Note: A common mistake in this question is that the students start calculating the squares and cubes of such large numbers. This is not only time consuming but also leads to calculation mistakes. We have to just find the final digit of each of the four numbers, as that will be the deciding factor that the number is odd or even.