Question 6

We will calculate the squares of the first 30 natural numbers and then observe the patterns in their properties.

Step 1 — Calculating the squares of natural numbers

Let us find the square of each natural number from 1 to 30. The square of a number $n$ is found by multiplying $n$ by itself, which is $n \times n$.

$1^2 = 1 \times 1 = \mathbf{1}$ $2^2 = 2 \times 2 = \mathbf{4}$ $3^2 = 3 \times 3 = \mathbf{9}$ $4^2 = 4 \times 4 = \mathbf{16}$ $5^2 = 5 \times 5 = \mathbf{25}$ $6^2 = 6 \times 6 = \mathbf{36}$ $7^2 = 7 \times 7 = \mathbf{49}$ $8^2 = 8 \times 8 = \mathbf{64}$ $9^2 = 9 \times 9 = \mathbf{81}$ $10^2 = 10 \times 10 = \mathbf{100}$ $11^2 = 11 \times 11 = \mathbf{121}$ $12^2 = 12 \times 12 = \mathbf{144}$ $13^2 = 13 \times 13 = \mathbf{169}$ $14^2 = 14 \times 14 = \mathbf{196}$ $15^2 = 15 \times 15 = \mathbf{225}$ $16^2 = 16 \times 16 = \mathbf{256}$ $17^2 = 17 \times 17 = \mathbf{289}$ $18^2 = 18 \times 18 = \mathbf{324}$ $19^2 = 19 \times 19 = \mathbf{361}$ $20^2 = 20 \times 20 = \mathbf{400}$ $21^2 = 21 \times 21 = \mathbf{441}$ $22^2 = 22 \times 22 = \mathbf{484}$ $23^2 = 23 \times 23 = \mathbf{529}$ $24^2 = 24 \times 24 = \mathbf{576}$ $25^2 = 25 \times 25 = \mathbf{625}$ $26^2 = 26 \times 26 = \mathbf{676}$ $27^2 = 27 \times 27 = \mathbf{729}$ $28^2 = 28 \times 28 = \mathbf{784}$ $29^2 = 29 \times 29 = \mathbf{841}$ $30^2 = 30 \times 30 = \mathbf{900}$

Step 2 — Observing unit digits of perfect squares

Let us look at the last digit (unit digit) of each square we calculated. The unit digits are: 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0. We can notice a pattern in these unit digits.

• Numbers ending in 1 or 9 have squares that end in 1.
• Numbers ending in 2 or 8 have squares that end in 4.
• Numbers ending in 3 or 7 have squares that end in 9.
• Numbers ending in 4 or 6 have squares that end in 6.
• Numbers ending in 5 have squares that end in 5.
• Numbers ending in 0 have squares that end in 0.

$\boxed{\text{Conjecture 1: A perfect square can only end in 0, 1, 4, 5, 6, or 9.}}$ $\boxed{\text{Conjecture 2: A perfect square can never end in 2, 3, 7, or 8.}}$

Step 3 — Observing differences between consecutive squares

Let us find the difference between the squares of consecutive natural numbers.

$2^2 - 1^2 = 4 - 1 = \mathbf{3}$ $3^2 - 2^2 = 9 - 4 = \mathbf{5}$ $4^2 - 3^2 = 16 - 9 = \mathbf{7}$ $5^2 - 4^2 = 25 - 16 = \mathbf{9}$ $6^2 - 5^2 = 36 - 25 = \mathbf{11}$

We observe that the difference between the square of a natural number $n$ and the square of the previous natural number $(n-1)$ is always an odd number. This difference is equal to $2n-1$.

$\boxed{\text{Conjecture 3: The difference between } n^2 \text{ and } (n-1)^2 \text{ is } 2n-1.}$ $\boxed{\text{This difference is always an odd number.}}$

Step 4 — Observing the sum of consecutive odd numbers

Let us look at how squares relate to the sum of the first few odd natural numbers.

$1 = \mathbf{1^2}$ $1 + 3 = \mathbf{4 = 2^2}$ $1 + 3 + 5 = \mathbf{9 = 3^2}$ $1 + 3 + 5 + 7 = \mathbf{16 = 4^2}$ $1 + 3 + 5 + 7 + 9 = \mathbf{25 = 5^2}$

We see that the sum of the first $n$ odd natural numbers is equal to $n^2$.

$\boxed{\text{Conjecture 4: The sum of the first } n \text{ odd natural numbers is } n^2.}$

Step 5 — Observing even and odd squares

Let us check if the original number is even or odd and what its square is.

• Squares of even numbers (like $2, 4, 6, \dots$) are $4, 16, 36, \dots$, which are all even.
• Squares of odd numbers (like $1, 3, 5, \dots$) are $1, 9, 25, \dots$, which are all odd.

$\boxed{\text{Conjecture 5: The square of an even number is always even.}}$ $\boxed{\text{Conjecture 6: The square of an odd number is always odd.}}$

Step 6 — Observing trailing zeros in squares

Let us look at the squares of numbers that end in zero.

$10^2 = \mathbf{100}$ $20^2 = \mathbf{400}$ $30^2 = \mathbf{900}$

We observe that if a number ends with one zero, its square ends with two zeros. In general, if a number ends with $n$ zeros, its square will end with $2n$ zeros.

$\boxed{\text{Conjecture 7: If a number ends with } n \text{ zeros, its square ends with } 2n \text{ zeros.}}$

Answer

(i) The squares of the first 30 natural numbers are: $1^2 = 1$, $2^2 = 4$, $3^2 = 9$, $4^2 = 16$, $5^2 = 25$, $6^2 = 36$, $7^2 = 49$, $8^2 = 64$, $9^2 = 81$, $10^2 = 100$, $11^2 = 121$, $12^2 = 144$, $13^2 = 169$, $14^2 = 196$, $15^2 = 225$, $16^2 = 256$, $17^2 = 289$, $18^2 = 324$, $19^2 = 361$, $20^2 = 400$, $21^2 = 441$, $22^2 = 484$, $23^2 = 529$, $24^2 = 576$, $25^2 = 625$, $26^2 = 676$, $27^2 = 729$, $28^2 = 784$, $29^2 = 841$, $30^2 = 900$. (ii) The patterns observed and conjectures made are:

1. Unit Digits: A perfect square can only end in 0, 1, 4, 5, 6, or 9. It can never end in 2, 3, 7, or 8.
2. Difference of Consecutive Squares: The difference between the square of a natural number $n$ and the square of $(n-1)$ is $2n-1$, which is always an odd number.
3. Sum of Odd Numbers: The sum of the first $n$ odd natural numbers is equal to $n^2$.
4. Even/Odd Squares: The square of an even number is always even, and the square of an odd number is always odd.
5. Trailing Zeros: If a number ends with $n$ zeros, its square ends with $2n$ zeros.