A Square and A Cube Class 8: Perfect Squares, Cubes, Square Roots and Cube Roots

What A Square and A Cube is about

Key concepts

Square numbers and the locker puzzle

NCERT section 1.1 Square Numbers

A square number is a number multiplied by itself: $n\times n=n^2$, read 'n squared'. Squares of natural numbers ($1,4,9,16,25,\dots$) are called perfect squares. The name comes from geometry: a square of side $n$ units holds $n^2$ unit squares.

Why squares have an odd number of factors. Factors come in partner pairs whose product is the number: for 12 the pairs are $1\times 12$, $2\times 6$, $3\times 4$, so 12 has 6 factors (even). For a square such as 36, one pair is $6\times 6$, and 6 has itself as its partner, so it is counted once. That leaves an odd number of factors. So:

• only square numbers have an odd number of factors;
• in the locker game, a locker is toggled once for each factor of its number, so only the square-numbered lockers end up open;
• a prime has exactly 2 factors, so prime-numbered lockers are toggled exactly twice.

Squares are not only for whole numbers: ${\left(\frac{2}{3}\right)}^2=\frac{4}{9}$ and $(1.5{)}^2=2.25$.

Worked example

Suppose the locker game is played with 150 lockers and 150 people. How many lockers stay open at the end, and which is the highest-numbered open locker?

Try it first, then show the solutionSolution

1. A locker stays open only if it is toggled an odd number of times, that is, if its number has an odd number of factors.
2. Only perfect squares have an odd number of factors, so the open lockers are $1,4,9,\dots$
3. $12^2=144\le 150$ but $13^2=169>150$, so the squares up to 150 are $1^2$ to $12^2$.

Answer: 12 lockers stay open; the highest is locker 144.

Patterns in perfect squares: units digit, zeros and parity

NCERT section 1.1 Square Numbers

Look at the units digit:

• a perfect square always ends in 0, 1, 4, 5, 6 or 9;
• a number ending in 2, 3, 7 or 8 is never a square;
• but ending in 0, 1, 4, 5, 6 or 9 does not prove a number is a square (26 ends in 6 and is not a square). The units digit can only rule numbers out.

Useful links between a number and its square:

• numbers ending in 1 or 9 have squares ending in 1;
• numbers ending in 4 or 6 have squares ending in 6;
• numbers ending in 2 or 8 have squares ending in 4; ending in 3 or 7 give 9; ending in 5 gives 5.

Zeros: if a number ends in $k$ zeros, its square ends in $2k$ zeros. So a perfect square can only end in an even number of zeros: 1,000 (three zeros) is not a square, while 4,90,000 (four zeros) is $700^2$.

Parity: the square of an even number is even; the square of an odd number is odd.

Worked example

Without calculating, decide which of these cannot be perfect squares: 7,396; 12,000; 5,483; 90,000. Then say what the last digit of $86^2$ must be.

Try it first, then show the solutionSolution

1. 5,483 ends in 3, so it cannot be a square.
2. 12,000 ends in three zeros, an odd number, so it cannot be a square.
3. 7,396 and 90,000 pass both tests, so they may be squares (in fact $86^2=7396$ and $300^2=90000$).
4. 86 ends in 6, and $6\times 6=36$, so $86^2$ ends in 6.

Answer: 5,483 and 12,000 cannot be squares; $86^2$ ends in 6.

Squares as sums of odd numbers

NCERT section 1.1 Square Numbers

The differences between consecutive squares are the odd numbers: $4-1=3$, $9-4=5$, $16-9=7$, and so on. Turning this round, **the sum of the first n odd numbers is $n^2$**, so $1+3+5+7=16=4^2$. A picture proof: each new inverted L-shaped strip of dots added to a square makes the next square, and each strip has 2 more dots than the last.

The $n$th odd number is $2n-1$. So to go from $n^2$ to $(n+1{)}^2$, add the $(n+1)$th odd number, $2n+1$:

• $(n+1{)}^2=n^2+2n+1$.

Test for a square: keep subtracting $1,3,5,\dots$ from the number. If you land exactly on 0 after $k$ steps, the number is $k^2$; if you jump past 0, it is not a square.

Between two squares: from $n^2$ to $(n+1{)}^2$ the gap is $2n+1$, so there are exactly $2n$ non-square numbers strictly between them.

Triangular numbers: two neighbouring triangular numbers add to a square, for example $6+10=16$.

Worked example

Given $47^2=2209$, find $48^2$ using odd numbers. How many whole numbers lie strictly between $30^2$ and $31^2$?

Try it first, then show the solutionSolution

1. $48^2=47^2+$ the 48th odd number $=2209+(2\times 48-1)=2209+95=2304$.
2. Between $n^2$ and $(n+1{)}^2$ there are $2n$ numbers; with $n=30$ that is 60.
3. Check: $31^2-30^2-1=961-900-1=60$.

Answer: $48^2=2304$; 60 numbers lie between $30^2$ and $31^2$.

Square roots by prime factorisation

NCERT section 1.1 Square Numbers

If $y=x^2$, then $x$ is a square root of $y$. Both $8^2$ and $(-8{)}^2$ equal 64, so 64 has two square roots, $+8$ and $-8$. The symbol  ​ in this chapter means the positive root: $\sqrt{64}=8$.

Prime factorisation method:
1. Write the number as a product of primes.
2. Try to put the primes into pairs (equivalently, split them into two identical groups).
3. If every prime pairs up, the number is a perfect square; take one prime from each pair and multiply to get the root.
4. If some prime is left unpaired, the number is not a square.

So $5184=2^6\times 3^4$: every exponent is even, and $\sqrt{5184}=2^3\times 3^2=72$.

Smallest multiplier or divisor: if a prime is unpaired, multiply by it to complete the pair, or divide by it to remove it. For example $2450=2\times 5^2\times 7^2$: the 2 is alone, so multiply by 2 to get $4900=70^2$ (or divide by 2 to get $1225=35^2$).

Worked example

Find the smallest number by which 2,450 must be multiplied to give a perfect square. Find the square root of the product.

Try it first, then show the solutionSolution

1. $2450=2\times 5\times 5\times 7\times 7$.
2. The 5s and 7s pair up; the single 2 does not.
3. Multiply by 2: $4900=(2\times 2)\times (5\times 5)\times (7\times 7)$.
4. $\sqrt{4900}=2\times 5\times 7=70$.

Answer: Multiply by 2; $\sqrt{4900}=70$.

Estimating square roots

NCERT section 1.1 Square Numbers

Most numbers are not perfect squares, but we can still trap their square roots between two whole numbers.

1. Find the perfect squares just below and just above the number.
2. The root lies between their roots.
3. Decide which end it is nearer by seeing which square is closer.

Example: $289=17^2<300<324=18^2$, so $17<\sqrt{300}<18$. 300 is 11 more than 289 but 24 less than 324, so $\sqrt{300}$ is nearer 17.

For larger perfect squares, combine the bracket with the units digit. For $\sqrt{3249}$: $50^2=2500$ and $60^2=3600$, so the root is in the 50s; it ends in 9, so the root ends in 3 or 7; $55^2=3025<3249$, so the root is above 55, giving 57. Check: $57^2=3249$.

To find $55^2$ quickly, expand: $(50+5{)}^2=2500+2\times 50\times 5+25=3025$.

Worked example

A square tablecloth must have an area of at most 500 cm², with a whole-number side in centimetres. What is the largest possible side? Between which two whole numbers does $\sqrt{500}$ lie, and which is it closer to?

Try it first, then show the solutionSolution

1. $22^2=484$ and $23^2=529$, so $22<\sqrt{500}<23$.
2. The largest whole-number side with area at most 500 cm² is 22 cm (484 cm²).
3. 500 is 16 above 484 and 29 below 529, so $\sqrt{500}$ is closer to 22 (it is about 22.36).

Answer: 22 cm; $\sqrt{500}$ lies between 22 and 23, closer to 22.

Cube numbers and their patterns

NCERT section 1.2 Cubic Numbers

A cube number is a number multiplied by itself three times: $n\times n\times n=n^3$. A cube of edge 4 units is made of 4 layers of $4\times 4$ unit cubes, so it has $4^3=64$ unit cubes. The perfect cubes begin $1,8,27,64,125,216,343,512,729,1000$.

Units digits of cubes (unlike squares, every digit 0 to 9 can occur):

• digits 0, 1, 4, 5, 6, 9 keep their own units digit when cubed ($4^3=64$, $9^3=729$);
• 2 and 8 swap ($2^3=8$, $8^3=512$);
• 3 and 7 swap ($3^3=27$, $7^3=343$).

Zeros: a cube ends in a number of zeros that is a multiple of 3 ($10^3=1000$, $100^3=10^6$). So no cube ends in exactly 00.

Parity: the cube of an odd number is odd; of an even number, even. Cubes of fractions and negatives work as expected: $(-6{)}^3=-216$, ${\left(\frac{2}{3}\right)}^3=\frac{8}{27}$.

Taxicab numbers: $1729=1^3+12^3=9^3+10^3$ is the smallest number that is a sum of two positive cubes in two different ways (the Hardy-Ramanujan number). The next two are 4104 and 13832.

Worked example

Without calculating the full cube, give the last digit of $37^3$ and of $58^3$. Can 4,000 be a perfect cube?

Try it first, then show the solutionSolution

1. 37 ends in 7, and cubes of numbers ending in 7 end in 3 ($7^3=343$). So $37^3$ ends in 3.
2. 58 ends in 8, and $8^3=512$ ends in 2. So $58^3$ ends in 2.
3. 4,000 ends in exactly 3 zeros, so it passes the zero test, but $4000=4\times 10^3$ and 4 is not a cube, so 4,000 is not a perfect cube.

Answer: $37^3$ ends in 3; $58^3$ ends in 2; 4,000 is not a perfect cube.

Cube roots by prime factorisation

NCERT section 1.2 Cubic Numbers

If $y=x^3$, then $x$ is the cube root of $y$, written $x=\sqrt[3]{y}$. So $\sqrt[3]{8}=2$ and $\sqrt[3]{1000}=10$.

Method: write the prime factorisation and group equal primes in triplets (or split all the primes into three identical groups). If every prime forms complete triplets, the number is a perfect cube, and the cube root is the product of one prime from each triplet.

• $3375=3^3\times 5^3$, so $\sqrt[3]{3375}=3\times 5=15$.
• $500=2^2\times 5^3$: the 2s do not make a triplet, so 500 is not a cube.

In the factorisation of $n^3$, every prime of $n$ appears exactly three times as often: $12=2^2\times 3$ and $12^3=2^6\times 3^3$.

Smallest multiplier or divisor: complete each unfinished triplet (multiply) or remove the leftover primes (divide). For $1372=2^2\times 7^3$, multiply by 2 to get $2744=14^3$, or divide by 4 to get $343=7^3$.

Guessing cube roots of perfect cubes: the last digit of the root comes from the last digit of the cube, and the tens digit from the part before the last three digits. For 74,088: the last digit 8 means the root ends in 2; the leading part 74 lies between $4^3=64$ and $5^3=125$, so the tens digit is 4. Root: 42.

Worked example

Find $\sqrt[3]{74088}$ by prime factorisation and check it by the guessing method.

Try it first, then show the solutionSolution

1. $74088=2\times 2\times 2\times 3\times 3\times 3\times 7\times 7\times 7=2^3\times 3^3\times 7^3$.
2. One prime from each triplet: $2\times 3\times 7=42$.
3. Check by guessing: the last digit 8 gives a root ending in 2; 74 lies between $4^3=64$ and $5^3=125$, so the root is 42.

Answer: $\sqrt[3]{74088}=42$.

Cubes as sums of odd numbers, successive differences and history

NCERT section 1.3 A Pinch of History

Odd numbers build cubes too, in blocks:

• $1=1^3$, $3+5=2^3$, $7+9+11=3^3$, $13+15+17+19=4^3$, ...
• $n^3$ is the sum of $n$ consecutive odd numbers, the first of which is $n^2-n+1$. Their average is $n^2$, and $n$ numbers with average $n^2$ add to $n^3$.

Successive differences: for squares $1,4,9,16,25$ the first differences are $3,5,7,9$ and the second differences are all 2. For cubes $1,8,27,64,125,216$ the differences become constant only at the third level (all equal to 6).

A pinch of history: the Babylonians kept tables of squares and cubes on clay tablets about 1700 BCE. In Sanskrit, *varga* meant a square (the figure and the power), *ghana* a cube, and *varga-varga* the fourth power. The word *mula*, the root of a plant, gave *varga-mula* (square root) and *ghana-mula* (cube root); the idea passed into Arabic (*jidhr*) and Latin (*radix*), which is why we say 'root' today.

Worked example

Find the 7 consecutive odd numbers that add up to $7^3$, and check the total.

Try it first, then show the solutionSolution

1. The first odd number is $n^2-n+1=49-7+1=43$.
2. The seven numbers are 43, 45, 47, 49, 51, 53, 55.
3. Their average is the middle one, 49, so the sum is $7\times 49=343=7^3$.

Answer: $43+45+47+49+51+53+55=343=7^3$.

Maths in real life: A Square and A Cube

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Perfect Cubes and Cube Roots
• Perfect Squares and Square Roots

A Square and A Cube formula sheet

Square and cube

$n^2=n\times n,n^3=n\times n\times n$

Sum of the first n odd numbers

$1+3+5+\cdots +(2n-1)=n^2$

The nth odd number is $2n-1$.

Next square

$(n+1{)}^2=n^2+2n+1$

Add the next odd number.

Numbers between consecutive squares

$(n+1{)}^2-n^2-1=2n$

Square root by factors

$\sqrt{p^{2a}{q}^{2b}}=p^a{q}^b$

Every exponent must be even.

Cube root by factors

$\sqrt[3]{p^{3a}{q}^{3b}}=p^a{q}^b$

Every exponent must be a multiple of 3.

Square of a sum (for estimating)

$(a+b{)}^2=a^2+2ab+b^2$

For example $45^2=1600+400+25=2025$.

Cube as odd numbers

$n^3=(n^2-n+1)+(n^2-n+3)+\cdots +(n^2+n-1)$

n terms with average $n^2$.

Hardy-Ramanujan number

$1729=1^3+12^3=9^3+10^3$

How the school exam asks A Square and A Cube

The question patterns to expect, and the method that works for each.

Tricks to remember A Square and A Cube

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Squares never end in 2, 3, 7, 8

  Think '2-3-7-8, never great': a number ending in any of these can never be a square.

  Use it when: Ruling out squares in one glance.

• Pairs for squares, triplets for cubes

  Square = 2 friends per group; cube = 3 friends per group. Anyone left without a full group must be multiplied in (or sent away by dividing).

  Use it when: Prime factorisation questions and smallest multiplier questions.

• Cube endings: 2 and 8 swap, 3 and 7 swap

  Only the 'swap pairs' 2-8 and 3-7 change their last digit when cubed; every other digit stays the same.

  Use it when: Guessing cube roots of large perfect cubes.

• Zeros: double for squares, triple for cubes

  k zeros in the number become 2k zeros in its square and 3k zeros in its cube.

  Use it when: Spotting numbers that cannot be squares or cubes.

• Mula means root

  Varga-mula = square root, ghana-mula = cube root; mula is the root of a plant.

  Use it when: History questions in the chapter.

A Square and A Cube: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in A Square and A Cube

The slips examiners see most often, and what to write instead.

• ✗Mistake: Treating a units-digit check as proof that a number is a square

  Students learn that squares end in 0, 1, 4, 5, 6 or 9 and reverse it.

  Wrong working

  2,026 ends in 6, so it is a perfect square.

  Right working

  Ending in 6 only means it might be a square. $45^2=2025$ and $46^2=2116$, so 2,026 is not a square.

  ✓Do this: Use digit tests only to rule numbers out; prove squareness by factorisation or by squaring.

  Usually costs: The whole 1-2 mark answer.

• ✗Mistake: Multiplying by the whole unpaired part instead of only what is missing

  Confusion between completing a pair and repeating the factor.

  Wrong working

  $1350=2\times 3^3\times 5^2$; to make a cube multiply by $2\times 5^2=50$.

  Right working

  A triplet needs 3 of each prime. 2 needs two more 2s and $5^2$ needs one more 5, so multiply by $2^2\times 5=20$: $1350\times 20=27000=30^3$.

  ✓Do this: For each prime, write how many are present and how many are needed to reach the next multiple of 2 (or 3).

  Usually costs: Method mark and answer mark.

• ✗Mistake: Taking one copy of every prime instead of one from each group

  Pairs are formed but the root is read off wrongly.

  Wrong working

  $7056=2^4\times 3^2\times 7^2$, so $\sqrt{7056}=2\times 3\times 7=42$.

  Right working

  $2^4$ is two pairs of 2s, so take $2\times 2$: $\sqrt{7056}=2^2\times 3\times 7=84$.

  ✓Do this: Halve each exponent for a square root (divide by 3 for a cube root).

  Usually costs: The answer mark.

• ✗Mistake: Counting the end numbers when asked how many numbers lie between two squares

  'Between' is read as 'from ... to'.

  Wrong working

  Between $12^2=144$ and $13^2=169$ there are $169-144=25$ numbers.

  Right working

  Strictly between them: $169-144-1=24=2\times 12$.

  ✓Do this: Use $2n$ for the numbers strictly between $n^2$ and $(n+1{)}^2$.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting that a negative number also squares to the same value

  Only the positive root is used in calculations.

  Wrong working

  The only number whose square is 81 is 9.

  Right working

  $9^2=81$ and $(-9{)}^2=81$; 81 has two square roots, $\pm 9$, and $\sqrt{81}$ means the positive one, 9.

  ✓Do this: Say 'square roots are $\pm 9$' when asked for all roots; use  ​ for the positive one.

  Usually costs: 1 mark in a true/false or reasoning item.

NCERT exercise guide: A Square and A Cube

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on A Square and A Cube

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The units digit of $268^2$ is
   (a) 2
   (b) 4
   (c) 6
   (d) 8

   Hint

   Only the last digit matters.

   Answer

   (b) 4, since $8^2=64$.

2. Q2 · 1 mark · MCQExam level

   How many zeros are there at the end of $4000^2$?
   (a) 3
   (b) 4
   (c) 6
   (d) 9

   Hint

   Zeros double when squared.

   Answer

   (c) 6

3. Q3 · 1 mark · Very short answerExam level

   How many non-square numbers lie between $45^2$ and $46^2$?

   Hint

   Use $2n$.

   Answer

   90

4. Q4 · 2 marks · Very short answerExam level

   Find $\sqrt[3]{13824}$ by prime factorisation.

   Hint

   $13824=2^9\times 3^3$.

   Answer

   $2^3\times 3=24$

5. Q5 · 2 marks · Very short answerHarder exam level

   Given $64^2=4096$, find $65^2$ using the odd-number pattern.

   Hint

   Add the 65th odd number.

   Answer

   $4096+129=4225$

6. Q6 · 3 marks · Short answerHarder exam level

   Find the smallest number by which 8,820 must be divided so that the quotient is a perfect square, and find the square root of the quotient.

   Hint

   $8820=2^2\times 3^2\times 5\times 7^2$.

   Answer

   Divide by 5; $8820÷5=1764=42^2$.

7. Q7 · 3 marks · Short answerHarder exam level

   State true or false and give a reason:
   (i) A perfect cube can end in exactly four zeros.
   (ii) The cube of an even number is always even.
   (iii) Every number ending in 9 is a perfect square.

   Hint

   Think of zeros in threes and counterexamples.

   Answer

   (i) False: cubes end in a multiple of 3 zeros. (ii) True: an even number has 2 as a factor, so its cube has $2^3$. (iii) False: 19 ends in 9 and is not a square.

8. Q8 · 3 marks · Short answerHarder exam level

   Find the smallest number by which 3,087 must be divided to make it a perfect cube. Find the cube root of the quotient.

   Hint

   Factorise; look for the incomplete triplet.

   Answer

   $3087=3^2\times 7^3$; divide by 9 to get $343=7^3$; cube root 7.

9. Q9 · 3 marks · Short answerHarder exam level

   You are told that 59,319 and 1,57,464 are perfect cubes. Find their cube roots without factorising, explaining each step.

   Hint

   Last digit, then the leading digits.

   Answer

   59,319 ends in 9, so the root ends in 9; 59 lies between $3^3=27$ and $4^3=64$, so the root is 39. 1,57,464 ends in 4, so the root ends in 4; 157 lies between $5^3=125$ and $6^3=216$, so the root is 54.

10. Q10 · 5 marks · Long answerHarder exam level

    A square garden has an area of 1,764 m². (a) Find its side. (b) Fence posts are placed every 3 m along the boundary, starting at a corner. How many posts are needed? (c) A path 2 m wide is laid all round the outside. Find the area of the path.

    Hint

    Find the side first; the outer square has side 4 m more.

    Answer

    (a) $\sqrt{1764}=42$ m. (b) Perimeter $=168$ m, so $168÷3=56$ posts. (c) Outer side 46 m: $46^2-42^2=2116-1764=352$ m².

11. Q11 · 3 marks · Short answerHOTS / competency

    The squares of two consecutive whole numbers differ by 151. Find the numbers, and explain why the difference of squares of consecutive numbers is always odd.

    Hint

    $(n+1{)}^2-n^2=n+(n+1)$.

    Answer

    $(n+1{)}^2-n^2=2n+1=151$, so $n=75$: the numbers are 75 and 76 ($5776-5625=151$). The difference $2n+1$ is always odd.

12. Q12 · 3 marks · Short answerHOTS / competency

    Suppose the locker game is played with 500 lockers and 500 people. How many lockers end up open, and which open locker has the largest number? Which lockers are touched exactly twice?

    Hint

    Count factors.

    Answer

    Squares up to 500 stay open: $22^2=484\le 500<529$, so 22 lockers, the largest being 484. Lockers touched exactly twice are the prime-numbered ones, which have exactly two factors.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Which numbers have an odd number of factors?Show answer

   Only perfect squares, because one factor pairs with itself.

2. 2.Which digits can never be the units digit of a perfect square?Show answer

   2, 3, 7 and 8.

3. 3.What is the sum of the first n odd numbers?Show answer

   $n^2$.

4. 4.How many numbers lie strictly between $n^2$ and $(n+1{)}^2$?Show answer

   $2n$.

5. 5.How do you test a number for being a perfect square by prime factorisation?Show answer

   Check that every prime can be paired, that is, every exponent is even.

6. 6.How many zeros can a perfect cube end with?Show answer

   A multiple of 3: 0, 3, 6, and so on.

7. 7.Which digits swap when cubed?Show answer

   2 and 8, and 3 and 7.

8. 8.What is the Hardy-Ramanujan number and why is it special?Show answer

   1729, the smallest number that is a sum of two positive cubes in two ways: $1^3+12^3$ and $9^3+10^3$.

9. 9.What do varga, ghana and mula mean?Show answer

   Square, cube and root (of a plant), as used in ancient Indian mathematics.

10. 10.At which level do successive differences of cubes become constant?Show answer

    At the third level, where all the differences are 6.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain why only square-numbered lockers stay open?
• ☐Can you rule out a square from its units digit or its end zeros?
• ☐Can you find $(n+1{)}^2$ from $n^2$ by adding an odd number?
• ☐Can you find a square root and a cube root by prime factorisation?
• ☐Can you find the smallest multiplier or divisor that makes a perfect square or cube?
• ☐Can you find the smallest square divisible by three given numbers?
• ☐Can you place $\sqrt{300}$ between two whole numbers and say which it is nearer?
• ☐Can you guess the cube root of a 5- or 6-digit perfect cube?
• ☐Can you write $n^3$ as a sum of n consecutive odd numbers?
• ☐Can you state why 1729 is called a taxicab number?

A Square and A Cube: frequently asked questions