The Last Digit Alone Can Rule Out a Perfect Square

It can. A perfect square can only end in 0, 1, 4, 5, 6 or 9. So $1078$ ends in 8 and is not a perfect square — no calculation needed.

The test rules numbers out instantly, though it can never prove one is a square. This page covers everything in the CBSE Class 8 Mathematics chapter's first part: squares and perfect squares, the digit tests, square roots by prime factorisation, and square-number patterns.

The square of a number is that number multiplied by itself, written $n^2$. A perfect square is a number that is the square of a natural number.

$$5^2=5\times 5=25,12^2=144,17^2=289$$

The perfect squares in order are

$$1,4,9,16,25,36,49,64,81,100,121,144,169,196,225$$

and they are worth knowing up to $20^2=400$ by heart, since they appear constantly.

Identifying them from a list: among $16,20,36,40,49$, the perfect squares are $16=4^2$, $36=6^2$ and $49=7^2$. The others are not, because no natural number squares to 20 or 40.

Squares appear in area problems, since a square of side 7 cm has area $7^2=49{\text{cm}}^2$ — which is where the name comes from.

The pattern worth noticing early is how the gaps grow. Between $1$ and $4$ is 3, between $4$ and $9$ is 5, between $9$ and $16$ is 7 — the odd numbers, in order. That observation returns as a full method later in this page.

Both let you reject a number quickly without any long calculation.

The units-digit test. Square the digits $0$ to $9$ and look at the last digit of each result:

$$0,1,4,9,16,25,36,49,64,81$$

The last digits are $0,1,4,9,6,5,6,9,4,1$ — so only 0, 1, 4, 5, 6 and 9 ever appear.

Therefore a perfect square never ends in 2, 3, 7 or 8.

Applying it: $1078$ ends in 8, $2023$ ends in 3, $567$ ends in 7 — none can be a perfect square.

The trailing-zero test. A perfect square always has an even number of trailing zeros.

- $100$ has two zeros and $100=10^2$. A perfect square.
- $1600$ has two zeros and $1600=40^2$. A perfect square.
- $1000$ has three zeros, an odd number, so it is not a perfect square.
- $40$ has one zero, so it is not either.

The reason is that squaring $10$ gives two zeros, squaring $100$ gives four, and so on — zeros always arrive in pairs.

The limitation must be stated, because it is what the question tests. These tests can only rule out. A number ending in 6 with two trailing zeros has passed both tests and may still not be a square — $136$ ends in 6 and is not a perfect square. To prove a number is one you must factorise it, which is the next section.

Break the number into prime factors, split them into two identical groups, and one group is the square root. The symbol is  ​.

Worked example. Find $\sqrt{324}$.

Factorise: $324=2\times 2\times 3\times 3\times 3\times 3$, so

$$324=2^2\times 3^4$$

Pair the factors: $(2\times 3\times 3)\times (2\times 3\times 3)$. Each group gives

$$\sqrt{324}=2\times 3^2=2\times 9=18$$

Check: $18^2=324$. Correct.

Worked example. Find $\sqrt{1764}$.

$$1764=2^2\times 3^2\times 7^2\Rightarrow \sqrt{1764}=2\times 3\times 7=42$$

Check: $42^2=1764$. Correct.

Worked example. Find $\sqrt{2304}$.

$$2304=2^8\times 3^2\Rightarrow \sqrt{2304}=2^4\times 3=16\times 3=48$$

The method also proves whether a number is a perfect square, which the digit tests could not do. If every prime appears an even number of times, the factors pair off exactly and the number is a perfect square. If any prime is left unpaired, it is not — so $200=2^3\times 5^2$ has a spare 2 and is not a perfect square.

Every perfect square $n^2$ is the sum of the **first $n$ odd numbers**, starting from 1.

$$1=1=1^2$$
$$1+3=4=2^2$$
$$1+3+5=9=3^2$$
$$1+3+5+7=16=4^2$$
$$1+3+5+7+9=25=5^2$$

So $49=7^2$ is the sum of the first seven odd numbers:

$$1+3+5+7+9+11+13=49$$

And running it backwards tests a number. Keep subtracting successive odd numbers from 36: $36-1=35$, $-3=32$, $-5=27$, $-7=20$, $-9=11$, $-11=0$. It reached exactly zero after six subtractions, so $36=6^2$. If the subtraction never lands on zero, the number is not a perfect square.

Other patterns for predicting the next term:

- The difference between consecutive squares is an odd number: $25-16=9$, $36-25=11$, $49-36=13$
- So the next square after $n^2$ is $n^2+(2n+1)$. From $49$, the next is $49+15=64$
- Squares of numbers ending in 5 end in 25: $15^2=225$, $25^2=625$
- The square of an even number is even; of an odd number, odd

The reason the odd numbers appear is visual. Adding a row and a column to an $n\times n$ square of dots needs $n+n+1=2n+1$ extra dots — always an odd number — which turns it into the next square.