AFCAT Numerical Ability · Squares, Cubes and Roots

Squares, Their Last Digits, and Perfect Squares

Square a number quickly with (a + b)² and (a − b)², read the last digit of a square, and test whether a number is a perfect square by its prime factors.

All Squares, Cubes and Roots notes

Why this matters

Most questions here are one step: square a number, read a last digit, or say whether a number is a perfect square. Marks go on the middle term of (a + b)², on numbers whose last digit only looks right, and on a prime factor left without a partner.

Concept 1 of 3: Squaring a number, and the square identities

Squaring multiplies a number by itself. For a number near a round one, split it as a round number plus or minus a small one and use the identities:

97 = 100 - 3

is quicker to square than

97 \times 97

Definition

$n^2 = n \times n$ . A negative number squared is positive: $(-6)^2 = 36$ .
$(a + b)^2 = a^2 + 2ab + b^2$ , so $41^2 = 1600 + 80 + 1 = 1681$ .
$(a - b)^2 = a^2 - 2ab + b^2$ , so $29^2 = 900 - 60 + 1 = 841$ .
$a^2 - b^2 = (a + b)(a - b)$ , so $53^2 - 47^2 = 100 \times 6 = 600$ .
A number ending in 5: write the tens part times the next number, then 25. $65^2$ is $6 \times 7 = 42$ followed by 25, that is $4225$ .

Square of a sum or a difference

(a \pm b)^2 = a^2 \pm 2ab + b^2

Worked example

Find

97^2

and

103^2

without long multiplication.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Squares, Cubes and Roots · Squares and Perfect Squares

Simplify:

(a - b)^2

with

a = 10, b = 2

The middle term

(4 + 5)^2 = 81

, not

16 + 25 = 41

. The square of a sum keeps the cross term

2ab

Brackets decide the sign

(-7)^2 = 49

, but

-7^2 = -49

: without brackets only the 7 is squared.

Drill 4 more on squaring a number, and the square identities

Concept 2 of 3: The last digit of a square

The last digit of a square depends only on the last digit of the number. So one glance at the last digit rules many numbers out as perfect squares.

Definition

Square the last digit and keep the last digit of that. Squares end only in 0, 1, 4, 5, 6 or 9, so a number ending in 2, 3, 7 or 8 is never a perfect square. Two more checks:

A square ending in 5 always ends in 25.
A square ending in 0 ends in an even number of zeros (400, 3600, 250000).

Last digit of n	Last digit of n²	Examples
0	0	$20^2 = 400$
1 or 9	1	$11^2 = 121$ , $19^2 = 361$
2 or 8	4	$12^2 = 144$ , $18^2 = 324$
3 or 7	9	$23^2 = 529$ , $17^2 = 289$
4 or 6	6	$14^2 = 196$ , $16^2 = 256$
5	5	$25^2 = 625$ The last two digits are always 25.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Squares, Cubes and Roots · Squares and Perfect Squares

Which of the following is a perfect square?

A good ending is not proof

2,231 ends in 1, a possible ending, but it lies between

47^2 = 2209

and

48^2 = 2304

. The last digit can rule a number out, never in.

Squares ending in 5

2,675 ends in 5, but in 75, not 25, so it is not a perfect square.

Drill 3 more on the last digit of a square

Concept 3 of 3: Testing a perfect square by prime factors

In a perfect square every prime factor comes in pairs, because the root's factors appear twice. Write the number in primes and look at the powers.

Definition

Write $n$ as a product of primes. It is a perfect square when every exponent is even.
The square root takes half of each exponent: $784 = 2^4 \times 7^2$ , so $\sqrt{784} = 2^2 \times 7 = 28$ .
To find the square in a range, bracket it with known squares: between 300 and 350 the only one is $18^2 = 324$ , since $17^2 = 289$ and $19^2 = 361$ .
Knowing the squares up to $30^2 = 900$ by heart makes most of these one step.

Perfect square test

n = p^{a} q^{b} \cdots \text{ is a square when } a, b, \ldots \text{ are all even}

Worked example

Is 3,136 a perfect square? If so, find its square root.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Squares, Cubes and Roots · Squares and Perfect Squares

180

a perfect square? (Using prime factorization)

One odd power is enough to fail

300 = 2^2 \times 3 \times 5^2

has two pairs, but the single 3 means it is not a perfect square.

Factor all the way down

72 = 8 \times 9

looks like two squares' worth, but

8 = 2^3

. So

72 = 2^3 \times 3^2

has an odd power and is not a square.

Drill 5 more on testing a perfect square by prime factors

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Squaring a number, and the square identities
Square of a sum or a difference
$(a \pm b)^2 = a^2 \pm 2ab + b^2$
Testing a perfect square by prime factors
Perfect square test
$n = p^{a} q^{b} \cdots \text{ is a square when } a, b, \ldots \text{ are all even}$

Reference tables (1)

The last digit of a square6 rows

Last digit of n	Last digit of n²	Examples
0	0	$20^2 = 400$
1 or 9	1	$11^2 = 121$ , $19^2 = 361$
2 or 8	4	$12^2 = 144$ , $18^2 = 324$
3 or 7	9	$23^2 = 529$ , $17^2 = 289$
4 or 6	6	$14^2 = 196$ , $16^2 = 256$
5	5	$25^2 = 625$ The last two digits are always 25.

Watch out for (6)

The middle term
→ Squaring a number, and the square identities
Brackets decide the sign
→ Squaring a number, and the square identities
A good ending is not proof
→ The last digit of a square
Squares ending in 5
→ The last digit of a square
One odd power is enough to fail
→ Testing a perfect square by prime factors
Factor all the way down
→ Testing a perfect square by prime factors

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