Rational Numbers | FIO

Question 12

State true or false. Explain your reasoning.

(i) The cube of any odd number is even.

(ii) There is no perfect cube that ends with 8.

(iii) The cube of a 2-digit number may be a 3-digit number.

(iv) The cube of a 2-digit number may have seven or more digits.

(v) Cube numbers have an odd number of factors.

Solution

Understand the Question

A cube of a number is obtained by multiplying the number by itself three times: $n^3 = n \times n \times n$ .
Multiplying odd numbers always results in an odd number, while multiplying even numbers results in an even number.
The number of digits in the cube of a 2-digit number ( $10 \le n \le 99$ ) ranges between the digits of $10^3 = 1{,}000$ (4 digits) and $99^3 = 970{,}299$ (6 digits).
Only perfect squares have an odd number of factors; other numbers (including non-square perfect cubes) have an even number of factors.

(i) The cube of any odd number is even.

Step 1 · Evaluate cubes of odd numbers

Calculating the cubes of sample odd numbers:

\begin{aligned} 1^3 &= 1 \times 1 \times 1 = 1 \\ 3^3 &= 3 \times 3 \times 3 = 27 \\ 5^3 &= 5 \times 5 \times 5 = 125 \end{aligned}

The product of odd numbers is always odd. Therefore, the cube of an odd number is always odd.

Answer

(i) False

(ii) There is no perfect cube that ends with 8.

Step 1 · Check unit digits of cubes

Calculating perfect cubes of numbers ending in 2:

\begin{aligned} 2^3 &= 2 \times 2 \times 2 = 8 \\ 12^3 &= 12 \times 12 \times 12 = 1728 \end{aligned}

Any integer ending in 2 has a cube that ends in 8.

Answer

(ii) False

(iii) The cube of a 2-digit number may be a 3-digit number.

Step 1 · Find cube of smallest 2-digit number

The smallest 2-digit number is $10$ .

\begin{aligned} 10^3 &= 10 \times 10 \times 10 \\ &= 1000 \end{aligned}

$1000$ has $4$ digits. Any 2-digit number greater than $10$ has a cube greater than $1000$ , so the cube of any 2-digit number has at least $4$ digits.

Answer

(iii) False

(iv) The cube of a 2-digit number may have seven or more digits.

Step 1 · Find cube of largest 2-digit number

The largest 2-digit number is $99$ .

\begin{aligned} 99^3 &= (100 - 1)^3 \\ &= 100^3 - 3 \times 100^2 \times 1 + 3 \times 100 \times 1^2 - 1^3 \\ &= 1{,}000{,}000 - 3 \times 10{,}000 + 300 - 1 \\ &= 1{,}000{,}000 - 30{,}000 + 300 - 1 \\ &= 970{,}299 \end{aligned}

$970{,}299$ has $6$ digits. Since the cube of the largest 2-digit number has $6$ digits, no 2-digit number can have a cube with $7$ or more digits.

Answer

(iv) False

(v) Cube numbers have an odd number of factors.

Step 1 · Determine number of factors for cube numbers

A number has an odd number of factors if and only if it is a perfect square.

For example:

For $8 = 2^3$ , the factors are $1, 2, 4, 8$ (a total of $4$ factors, which is even).
For $27 = 3^3$ , the factors are $1, 3, 9, 27$ (a total of $4$ factors, which is even).

Since not all cube numbers are perfect squares, cube numbers do not necessarily have an odd number of factors.

Answer

(v) False

Common Mistakes

Squares vs. Cubes Factors: Assuming that because perfect squares have an odd number of factors, perfect cubes do as well. A number only has an odd number of factors if it is a square (where one factor pairs with itself, e.g. $\sqrt{n} \times \sqrt{n} = n$ ).
Digit Range of 2-digit Cubes: Overlooking that the minimum number of digits for a 2-digit cube is 4 (from $10^3 = 1000$ ) and the maximum is 6 (from $99^3 = 970{,}299$ ).

Question 12

More questions in FIO