A Square and A Cube

Take 3 and multiply it by itself: 3 × 3 = 9. The number 9 is called the square of 3, and we write it as 3². In the same way 5² = 5 × 5 = 25 and 12² = 12 × 12 = 144.

The name comes from a shape, not from the multiplication. Put 9 dots on paper and you can arrange them in 3 rows of 3, a neat square. Put 16 dots and you get 4 rows of 4. Try the same with 10 dots and you cannot do it, because you will always be left with a row that is short. Numbers that can be arranged in a perfect square are called square numbers, or perfect squares.

Here are the first twenty squares. It is worth knowing them by heart, because almost everything else here becomes fast once you do.

n  :   1    2    3    4    5    6    7    8    9   10
n² :   1    4    9   16   25   36   49   64   81  100

n  :  11   12   13   14   15   16   17   18   19   20
n² : 121  144  169  196  225  256  289  324  361  400

The word perfect is doing real work in that phrase. 49 is a perfect square because 7 × 7 gives it exactly. 50 is not, because no whole number multiplied by itself lands on 50. Every whole number has a square, but only some whole numbers are squares.

The picture also runs backwards, and that is where square roots come from. If a square-shaped room has an area of 144 square metres, its side must be the number that gives 144 when multiplied by itself. That number is 12, so the side is 12 metres. We write √144 = 12 and read it as the square root of 144. Finding area asks you to square the side; finding the square root asks you to undo that.

One caution about the notation. 3² means 3 × 3, which is 9. It does not mean 3 × 2, which is 6. In the same way 10² is 100, not 20. It is easy to mix these up when you are rushing, so read the small raised 2 as an instruction to multiply the number by itself, and nothing else.

Squares grow quickly. From 1 to 10 they climb from 1 to 100; from 11 to 20 they climb from 121 to 400; by 30 the square is already 900, and 100² is 10000. That fast growth is why area shoots up when a plot of land gets wider. Doubling the side of a square plot from 20 m to 40 m does not double the area, it takes it from 400 square metres to 1600 square metres, which is four times as much. A plot priced at ₹2,000 per square metre would therefore cost four times as much, not twice as much.

1. The last digit is a filter. Square the digits 0 to 9 and look only at the unit digit of each answer:

0² = 0     1² = 1     2² = 4     3² = 9     4² = 16
5² = 25    6² = 36    7² = 49    8² = 64    9² = 81

The unit digits that turn up are 0, 1, 4, 5, 6 and 9. The digits 2, 3, 7 and 8 never turn up. The unit digit of any product depends only on the unit digits being multiplied, so this settles the matter for every number however large: a perfect square can never end in 2, 3, 7 or 8. That means 1057, 23453, 7928 and 222222 are all not perfect squares, and you know it without doing any real work.

Be careful with the reverse. Ending in 4 does not prove a number is a square, because 24 ends in 4 and is not one. The last digit can rule a number out, never rule it in.

2. Zeros come in pairs. 10² = 100, 20² = 400, 100² = 10000. A perfect square always ends in an even number of zeros. So 1000, which ends in three zeros, cannot be a perfect square.

3. Odd stays odd, even stays even. 7² = 49 is odd and 8² = 64 is even. Multiplying an odd number by itself keeps it odd, and multiplying an even number by itself keeps it even. So an odd square has an odd root and an even square has an even root, which is handy when you are narrowing down a guess.

4. The gap between consecutive squares is always odd. Take the list of squares and subtract each one from the next:

4 − 1 = 3       9 − 4 = 5       16 − 9 = 7
25 − 16 = 9    36 − 25 = 11    49 − 36 = 13

The differences are 3, 5, 7, 9, 11, 13, which are the odd numbers in order. Here is the reason. Think of (n + 1)² as (n + 1) rows with (n + 1) dots in each row. That is n rows of (n + 1) dots, plus one extra row of (n + 1) dots. The n rows give n × (n + 1) = n² + n dots, and the extra row gives n + 1 more. Adding them, (n + 1)² = n² + n + n + 1 = n² + 2n + 1. Subtract n² and what is left is 2n + 1. Any number of the form 2n + 1 is odd, so the gap is odd every single time.

The dot picture says the same thing more simply. To grow a 5 by 5 square of dots into a 6 by 6 square you add a column of 5 dots down one side, a row of 5 dots along the bottom, and 1 dot in the corner. That is 5 + 5 + 1 = 11 dots, which is exactly 36 − 25.

This makes some subtractions instant. 51² − 50² = 2 × 50 + 1 = 101, with no squaring at all.

5. Odd numbers add up to squares. Because each new square is the previous square plus the next odd number, adding the odd numbers starting from 1 keeps giving squares:

1                     =  1 = 1²
1 + 3                 =  4 = 2²
1 + 3 + 5             =  9 = 3²
1 + 3 + 5 + 7         = 16 = 4²
1 + 3 + 5 + 7 + 9     = 25 = 5²

The rule is that the sum of the first n odd numbers is n². So 1 + 3 + 5 + ... + 19 has ten terms and adds to 100, and you never have to do the addition.

6. How many non-squares sit between two squares. Between 4 and 9 lie 5, 6, 7 and 8, which is four numbers. Between 9 and 16 lie six numbers. Between 25 and 36 lie ten. The count is always 2n between n² and (n + 1)², which is just the gap 2n + 1 minus the one step you take to land on the next square. So between 81 and 100 there are 2 × 9 = 18 numbers that are not perfect squares.

Guessing works for small numbers, but nobody can guess √7056. Prime factorisation turns it into a mechanical job that always works.

The idea behind it is simple. If a number is a perfect square, it is something multiplied by itself, so every prime inside it must appear twice over, once in each copy. Take 36 = 6 × 6. Since 6 = 2 × 3, we get 36 = 2 × 3 × 2 × 3 = 2² × 3². The 2 appears twice and the 3 appears twice. That is the signature of every perfect square: each prime appears an even number of times.

So the method is: factorise the number, sort the primes into identical pairs, and take one prime out of each pair. Multiply together whatever you took out and that is the square root.

Worked example: find √7056. Divide by primes, smallest first:

2 | 7056
2 | 3528
2 | 1764
2 |  882
3 |  441
3 |  147
7 |   49
7 |    7
  |    1

So 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7, which is 2⁴ × 3² × 7². Now pair them: (2 × 2), (2 × 2), (3 × 3), (7 × 7). There are four complete pairs and nothing left over, so 7056 really is a perfect square. Take one number from each pair and multiply: 2 × 2 × 3 × 7 = 84. Therefore √7056 = 84. Checking, 84 × 84 = 84 × 80 + 84 × 4 = 6720 + 336 = 7056.

Worked example: find √5184. Dividing by 2 six times takes 5184 down to 81, and 81 = 3 × 3 × 3 × 3. So 5184 = 2⁶ × 3⁴. Pairing gives (2 × 2), (2 × 2), (2 × 2), (3 × 3), (3 × 3), so √5184 = 2 × 2 × 2 × 3 × 3 = 8 × 9 = 72.

Once you trust the method there is a shortcut: halve every exponent. In 2⁶ × 3⁴, half of 6 is 3 and half of 4 is 2, giving 2³ × 3² = 8 × 9 = 72. This works only because the exponents are even, which is just another way of saying the number is a perfect square.

When something is left unpaired. Factorise 2352 the same way and you get

2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7 = 2⁴ × 3 × 7²

The 2s form two complete pairs and the 7s form one pair, but the single 3 has no partner. That lone 3 is exactly what stops 2352 from being a perfect square, and it also tells you how to fix the number, in two different ways.

Multiply to complete the pair. Give the 3 a partner: 2352 × 3 = 7056 = 2⁴ × 3² × 7², and √7056 = 84. So the smallest number to multiply 2352 by is 3.

Divide to throw the odd one out. Remove the lone 3 instead: 2352 ÷ 3 = 784 = 2⁴ × 7², and √784 = 2 × 2 × 7 = 28. So the smallest number to divide 2352 by is also 3.

In general, collect every prime that is left unpaired and multiply those together. That product answers both the multiply question and the divide question. If a factorisation left one 3 and one 5 unpaired, the answer to both would be 3 × 5 = 15.

One more, to fix the pattern. 3645 = 3⁶ × 5, because dividing by 3 six times takes 3645 down to 5. The six 3s make three complete pairs, and the 5 stands alone. Multiply by 5 to get 18225 = 3⁶ × 5², whose square root is 3³ × 5 = 27 × 5 = 135.

Prime factorisation is exact, but it only finishes the job on perfect squares. Factorise 50 and you get 2 × 5²; the lone 2 will not pair with anything, so 50 has no whole-number square root. The square root of 50 still exists, it is simply not a whole number, and for most practical work a good estimate is all anyone needs.

Step 1: trap the number between two squares. Run through the squares you know until you find the two that sandwich your number. For 50, the squares on either side are 49, which is 7², and 64, which is 8². Since 49 is smaller than 50 and 50 is smaller than 64, √50 lies somewhere between 7 and 8. Already you know the answer starts with 7.

Step 2: judge how far along the gap it sits. The number 50 is just one step above 49, while the whole gap from 49 to 64 is fifteen steps. So 50 is barely into that gap and √50 must be only a little above 7. A sensible estimate is 7.1, and the true value is about 7.07.

Step 2 can be sharpened with a small piece of arithmetic. The gap between n² and (n + 1)² is 2n + 1, which you met earlier. So the fraction of the way along is

(N − n²) ÷ (2n + 1)

and adding that fraction to n gives the estimate. Because the fraction treats the climb from one square to the next as even, when it is really steeper at the start, this estimate always comes out a shade below the true root, never above it. It is never more than about a tenth short, so the first decimal place is either exactly right or one tenth low.

Worked example: estimate √300. The squares around it are 17² = 289 and 18² = 324, so √300 lies between 17 and 18. The gap is 2 × 17 + 1 = 35, and the number is 300 − 289 = 11 above the lower square. The fraction is 11 ÷ 35, which is about 0.31. The estimate is 17 + 0.31 = 17.31, which is 17.3 to one decimal place. The true value is about 17.32, so one decimal place has come out right.

Worked example: estimate √1000. Here 31² = 961 and 32² = 1024, so the answer is between 31 and 32. The gap is 2 × 31 + 1 = 63, and 1000 − 961 = 39. The fraction 39 ÷ 63 is about 0.62, giving 31.6. The true value is about 31.62.

Worked example: estimate √80. Since 8² = 64 and 9² = 81, and 80 is only one below 81, √80 must be only just below 9. The estimate is 8.9, and the true value is about 8.94. Notice how much work the trapping step did on its own here.

If a question only asks between which two whole numbers a square root lies, step 1 by itself is the complete answer. For example √70 lies between 8 and 9, because 64 is smaller than 70 and 70 is smaller than 81. Do not be tempted to halve the number instead: half of 70 is 35, which is nowhere near the answer.

The same trapping idea settles questions about arrangements. Suppose a gardener has 1000 saplings and wants to plant them with the number of rows equal to the number of plants in each row. That needs a perfect square, and 1000 is not one. The next perfect square above 1000 is 1024, so he needs 1024 − 1000 = 24 more saplings and will end up with 32 rows of 32. If instead he is willing to leave some out, he drops to 961 and removes 1000 − 961 = 39 saplings, giving 31 rows of 31.

The cube of a number is that number used as a factor three times: 4³ = 4 × 4 × 4 = 64. Once again the name comes from a shape. Build a solid cube out of small unit blocks with 4 blocks along each edge and you will use 64 blocks altogether. A cube of edge 2 uses 8 blocks and a cube of edge 3 uses 27. Numbers you can build a solid cube from are called cube numbers or perfect cubes.

Here are the first twenty cubes:

n  :    1     2     3     4     5     6     7     8     9    10
n³ :    1     8    27    64   125   216   343   512   729  1000

n  :   11    12    13    14    15    16    17    18    19    20
n³ : 1331  1728  2197  2744  3375  4096  4913  5832  6859  8000

Cubes grow much faster than squares. 10² is 100 but 10³ is 1000; 20² is 400 but 20³ is 8000. This is why a water tank holds so much more when you widen it. A cubical tank of edge 1 metre holds 1 cubic metre of water, but doubling the edge to 2 metres makes it hold 8 cubic metres, not 2. If water were billed at ₹20 per cubic metre, filling the bigger tank would cost ₹160 against ₹20 for the smaller one.

Perfect cubes thin out very quickly. From 1 to 100 there are only four of them: 1, 8, 27 and 64. From 1 to 1000 there are just ten. Compare that with perfect squares, of which there are 31 from 1 to 1000, since 31² = 961.

Pattern 1: the unit digit. Cube the digits 0 to 9 and look at the last digit of each answer:

0³ = 0      1³ = 1      2³ = 8      3³ = 27     4³ = 64
5³ = 125    6³ = 216    7³ = 343    8³ = 512    9³ = 729

The unit digits produced are 0, 1, 8, 7, 4, 5, 6, 3, 2 and 9. Every digit from 0 to 9 appears exactly once. So, unlike squares, a cube can end in any digit at all, and the last digit can never rule a number out. What it can do instead is hand you the last digit of the cube root, because the matching is one to one. The digits 0, 1, 4, 5, 6 and 9 each pair with themselves, while 2 and 8 swap places and 3 and 7 swap places. So a cube ending in 3 must come from a root ending in 7, and a cube ending in 2 must come from a root ending in 8. Check it against the table: 4913 ends in 3 and its cube root is 17; 512 ends in 2 and its cube root is 8.

Pattern 2: odd and even. The cube of an even number is even and the cube of an odd number is odd, exactly as with squares. 6³ = 216 is even and 7³ = 343 is odd.

Pattern 3: the odd numbers turn up again. Break the odd numbers into groups of one, then two, then three, then four, and each group adds up to a cube:

1                      =   1 = 1³
3 + 5                  =   8 = 2³
7 + 9 + 11             =  27 = 3³
13 + 15 + 17 + 19      =  64 = 4³
21 + 23 + 25 + 27 + 29 = 125 = 5³

Each new group simply carries on from where the previous group stopped, and the number of terms in a group matches the number being cubed. You can check the fourth line by hand: 13 + 15 = 28, then 28 + 17 = 45, then 45 + 19 = 64.

Pattern 4: cubes add up to squares. Add the cubes starting from 1 and every total is a perfect square:

1³ + 2³            =   9 =  3² = (1 + 2)²
1³ + 2³ + 3³       =  36 =  6² = (1 + 2 + 3)²
1³ + 2³ + 3³ + 4³  = 100 = 10² = (1 + 2 + 3 + 4)²

The total is always the square of the sum of the numbers being cubed. Adding the cubes from 1³ to 5³ gives 1 + 8 + 27 + 64 + 125 = 225, which is 15², and 15 is exactly 1 + 2 + 3 + 4 + 5.

The cube root of a number is the number that was cubed to make it. Since 5 × 5 × 5 = 125, the cube root of 125 is 5, and we write ∛125 = 5. Volume is the natural place for it: if a cubical box holds 125 cubic cm, its edge must be 5 cm.

The prime factorisation method is the square-root method with one change. A perfect cube is something used as a factor three times, so every prime inside it must appear three times over, once for each copy. Take 216 = 6 × 6 × 6. Since 6 = 2 × 3, we get 216 = 2 × 3 × 2 × 3 × 2 × 3 = 2³ × 3³, and each prime appears exactly three times. That is the signature: in a perfect cube, every prime appears a number of times that is a multiple of 3.

So the method is: factorise, sort the primes into identical triples, and take one prime out of each triple.

Worked example: find ∛13824. Keep dividing by 2 for as long as you can:

13824 ÷ 2 = 6912      6912 ÷ 2 = 3456
 3456 ÷ 2 = 1728      1728 ÷ 2 =  864
  864 ÷ 2 =  432       432 ÷ 2 =  216
  216 ÷ 2 =  108       108 ÷ 2 =   54
   54 ÷ 2 =   27        27 = 3 × 3 × 3

Count the divisions by 2 and there are nine of them, then three 3s at the end. So 13824 = 2⁹ × 3³. Group the primes in threes: (2 × 2 × 2), (2 × 2 × 2), (2 × 2 × 2), (3 × 3 × 3). Every prime sits in a complete triple, so 13824 is a perfect cube. Take one number from each triple: ∛13824 = 2 × 2 × 2 × 3 = 24. Checking, 24 × 24 = 576 and 576 × 24 = 13824.

The exponent shortcut works here too, but with 3 instead of 2: divide every exponent by 3. In 2⁹ × 3³, 9 divided by 3 is 3 and 3 divided by 3 is 1, giving 2³ × 3 = 8 × 3 = 24.

Worked example: find ∛9261. The number is odd, so there is no point trying 2. Start at 3: 9261 ÷ 3 = 3087, then 3087 ÷ 3 = 1029, then 1029 ÷ 3 = 343, and 343 = 7 × 7 × 7. So 9261 = 3³ × 7³ and ∛9261 = 3 × 7 = 21.

When a triple is incomplete. Factorise 1323. It is odd, and 1323 ÷ 3 = 441, then 441 ÷ 3 = 147, then 147 ÷ 3 = 49, and 49 = 7 × 7. So 1323 = 3³ × 7². The three 3s form a complete triple, but the 7s are one short of a triple.

Multiply to finish the triple. The 7s need one more 7, so multiply: 1323 × 7 = 9261 = 3³ × 7³, whose cube root is 21. The smallest multiplier is 7. You must supply exactly what is missing, one 7 and no more, or the answer will not be the smallest.

Divide to clear the incomplete group. Now factorise 704. Dividing by 2 gives 352, then 176, then 88, then 44, then 22, then 11, which is six divisions. So 704 = 2⁶ × 11. The six 2s make two complete triples, but the single 11 is stranded and can never be part of one. Divide it away: 704 ÷ 11 = 64 = 2⁶, and ∛64 = 2 × 2 = 4. The smallest divisor is 11.

The rule has the same shape as it did for squares. Look at every prime whose count is not a multiple of 3. To make a cube by multiplying, supply whatever is needed to push each of those counts up to the next multiple of 3. To make a cube by dividing, remove the incomplete groups altogether.

Prime factorisation is reliable but slow on large numbers. If you already know that a number is a perfect cube, there is a much faster route that uses the digit patterns from earlier.

The grouping method, step by step. Take ∛17576.

Step 1. Starting from the right, mark off the digits in groups of three: 17 and 576. The left-hand group may end up with one, two or three digits, and that is fine.

Step 2. Look at the right-hand group and only at its last digit, which here is 6. From the cube pattern, a cube ending in 6 comes from a root ending in 6. So the answer ends in 6.

Step 3. Look at the left-hand group, 17, and find the largest cube that does not exceed it. The cubes run 1, 8, 27, and 8 is the largest one that is not more than 17. Since 8 = 2³, the tens digit of the answer is 2.

Step 4. Put the two digits together: ∛17576 = 26. A quick check: 26 × 26 = 676, and 676 × 26 = 676 × 25 + 676 = 16900 + 676 = 17576.

Another one: ∛110592. Group the digits as 110 and 592. The right-hand group ends in 2, and a cube ending in 2 comes from a root ending in 8, so the answer ends in 8. The left-hand group is 110; the cubes run 1, 8, 27, 64, 125, and the largest one not exceeding 110 is 64 = 4³, so the tens digit is 4. Therefore ∛110592 = 48.

And one more: ∛91125. Group as 91 and 125. The right-hand group ends in 5, and 5 comes only from 5, so the answer ends in 5. The left-hand group is 91, and the largest cube not exceeding 91 is 64, so the tens digit is 4. Therefore ∛91125 = 45.

This method assumes you have been told that the number is a perfect cube. If it is not one, the digits will still produce an answer and that answer will simply be wrong, so use the method only when a perfect cube is promised, or check by multiplying your answer back out.

Estimating a cube root that is not exact. When the number is not a perfect cube, do exactly what you did with square roots and trap it. For ∛100 the cubes on either side are 4³ = 64 and 5³ = 125, so the answer lies between 4 and 5. The number 100 sits 36 above 64 and 25 below 125, so it is nearer the top of the range, and a reasonable estimate is 4.6. The true value is about 4.64.

For ∛500 the surrounding cubes are 7³ = 343 and 8³ = 512. Since 500 is only 12 below 512 but 157 above 343, it sits very close to 8, so 7.9 is a good estimate. The true value is about 7.94.

For ∛2000 the cubes are 12³ = 1728 and 13³ = 2197. The number is 272 above 1728 and 197 below 2197, which puts it a little past the middle, so about 12.6. The true value is about 12.6 as well.

Cube roots move slowly, and that is what makes estimating them forgiving. The whole range from 12 to 13 has to cover every number from 1728 all the way to 2197. So getting the whole-number part right already puts you close to the answer, and one decimal place is normally as far as you need to go.