Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?

Question

Homi · Accepted Answer

Understand the Question

A perfect cube is obtained by multiplying a natural number by itself three times: $n^3 = n \times n \times n$ .
To find the number of cubes with $1$ $1$ , $2$ $2$ , or $3$ $3$ digits, we calculate consecutive cubes of natural numbers ( $1^3, 2^3, 3^3, \dots$ $1^{3}, 2^{3}, 3^{3}, \dots$ ) and count how many fall in each range:
- 1-digit range: $1$ to $9$
- 2-digit range: $10$ to $99$
- 3-digit range: $100$ to $999$

Step 1 · Find 1-Digit Cubes

For a 1-digit number, the range is $1 \le n^3 \le 9$ .

\begin{aligned} 1^3 &= 1 \\ 2^3 &= 8 \\ 3^3 &= 27 \end{aligned}

The 1-digit cubes are $1$ and $8$ .

Therefore, there are $2$ cubes with $1$ digit.

Step 2 · Find 2-Digit Cubes

For a 2-digit number, the range is $10 \le n^3 \le 99$ .

\begin{aligned} 3^3 &= 27 \\ 4^3 &= 64 \\ 5^3 &= 125 \end{aligned}

The 2-digit cubes are $27$ and $64$ .

Therefore, there are $2$ cubes with $2$ digits.

Step 3 · Find 3-Digit Cubes

For a 3-digit number, the range is $100 \le n^3 \le 999$ .

\begin{aligned} 5^3 &= 125 \\ 6^3 &= 216 \\ 7^3 &= 343 \\ 8^3 &= 512 \\ 9^3 &= 729 \\ 10^3 &= 1000 \end{aligned}

The 3-digit cubes are $125, 216, 343, 512,$ and $729$ .

Therefore, there are $5$ cubes with $3$ digits.

Step 4 · Observe the Pattern

Summarizing the number of cubes in each digit range:

We observe that the number of cubes does not follow a simple arithmetic pattern, but increases significantly in the 3-digit range.

Answer

Number of $1$ -digit cubes $= 2$ ( $1, 8$ )

Number of $2$ -digit cubes $= 2$ ( $27, 64$ )
Number of $3$ -digit cubes $= 5$ ( $125, 216, 343, 512, 729$ )
Observation: The number of cubes in each range does not increase at a constant rate, but the count expands as the range widens.

Common Mistakes

Boundary Inclusion Error: Counting $10^3 = 1000$ as a 3-digit cube when it actually has $4$ digits.
Confusing Cubes with Squares: Multiplying the base by itself only once ( $n^2$ ) instead of twice ( $n^3$ ).
Skipping Base Values: Missing intermediate values like $7^3 = 343$ or $8^3 = 512$ when testing the 3-digit range.

Question 25