Cube patterns in mathematics: Patterns in Cube of Number: Definition, Properties

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3} = 4 \times 4 \times 4 = 64\) shown by the figure below, a cube with length, breadth, and height equal to \(4\) units.

Have you seen a Rubik’s cube? Look at the picture of Rubik’s cube given below.

It has \(3 \times 3 \times 3\) i.e., \(27\) cubes.

Thus, \(27\) is a cube number.

Thus, the product obtained by multiplying a number three times is called the cube of that number.

? Learn about Cubes and Cube Roots here

How to Find a Perfect Cube?

Now, observe the following table.

Number Cube Number Cube
\(1\) \(1 \times 1 \times 1 = 1\) \(6\) \(6 \times 6 \times 6 = 216\)
\(2\) \(2 \times 2 \times 2 = 8\) \(7\) \(7 \times 7 \times 7 = 343\)
\(3\) \(3 \times 3 \times 3 = 27\) \(8\) \(8 \times 8 \times 8 = 512\)
\(4\) \(4 \times 4 \times 4 = 64\) \(9\) \(9 \times 9 \times 9 = 729\)
\(5\) \(5 \times 5 \times 5 = 125\) \(10\) \(10 \times 10 \times 10 = 1000\)

The cubes obtained in the table are perfect. 3} = 1 + 5 \times 4 \times 3\)

Properties of Cubes

The cube of natural numbers have the following interesting properties:

1. Cubes of all even natural numbers are even.

2. Cubes of all odd natural numbers are odd.

3. The sum of the cubes of first \(n\) natural numbers is equal to the square of their sum.

4. Cubes of the numbers ending with \(4, 5, 6\) and \(9\) are the numbers ending in the same digit. Cubes of numbers ending in digit \(2\) ending in digit \(8\), and the cubes are ending in digit \(8\) ending in digit \(2\). The cubes of the numbers ending in digits \(3\) and \(7\) ends in digits \(7\) and \(3\) respectively.

Computation of Cube Root through a Pattern

In the cube formula, we multiply a number three times to get its cube, so to find the cube root of a number, break down the number to be expressed as a product of three equal numbers, and thus, we get the cube root. Cube root is the number that needs to be multiplied three times to get the original number. 3}\). In other words, the cube root of a number \(n\) is that number \(m\) whose cube gives \(n\).

We can use the following method to compute the cube root of small numbers, which are perfect cubes of natural numbers.

Let us look at the procedure first.

Step 1: Obtain the natural number.

Step 2: Subtract \(1\) from it. If you get zero as a result, then the cube root of the number is \(1\), else go to the next step.

Step 3: Subtract \(7\left( { = 1 + \frac{{2 \times 1}}{2} \times 6} \right)\) from the resulting number obtained in step 2. If the result is zero, the cube root of the given number is \(2\), else go to the next step.

Step 4: Subtract \(19\left( { = 1 + \frac{{3 \times 2}}{2} \times 6} \right)\) from the resulting number obtained in step 3. If the result is zero, the cube root of the given number is \(3\), else go to the next step.

Step 5: Subtract \(37\left( { = 1 + \frac{{4 \times 3}}{2} \times 6} \right)\) from the resulting number obtained in step 4. 3} = 1 + 51 \times 50 \times 3 = 7651\)

Q.4. Is \(216\) a perfect cube? What is that number whose cube is \(216\)?
Factors of \(216\) are,
\(216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\)
Now, grouping the factors in triples of equal factors, we get,
\(216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3)\)
We find that the prime factors of \(216\) can be grouped into triples of equal factors, and no factor is left over.
Therefore, \(216\) is a perfect cube.
Thus, taking one factor from each triple, we get,
\(2 \times 3 = 6\)
Hence, \(216\) s the cube of \(6\).

Q.5. Write cubes of the first four positive integers, which can be written in the form \((3\,n + 1)\). For example, \(4, 7, 10\),… and examine the following: “Cube of positive integer which can be written in the form \((3\,n + 1)\) can also be written in this form.”
Four numbers can be written in form \(3\,n + 1\), when \(n = {\rm{1,2,3,4}}\) are \(4, 7, 10, 13\). 3}\).

Q.2. What is a number pattern? Explain.
Numbers are fascinating and astounding because they contain many beautiful patterns and sequences that are incredibly fascinating. A list of numbers with a common trait is called a pattern of numbers. In math, solving problems with number patterns increases a student’s logical thinking and mathematical reasoning ability. To answer any inquiry about whole number patterns in a sequence, the rule used to create the pattern is to be initially understood.

Q.3. What are examples of cube numbers?
Examples of cube numbers are \(1331,3375,8000,15625\) etc.

Q.4. What are the cube numbers in order?
The first \(10\) cube number in order are, \({\rm{1,8,27,64,125,216,343,512,729}}\) and \(1000\).

Q.5. Can a perfect cube end with two zeros?
No, a perfect cube cannot end with two zeros; instead, a perfect cube end with three zeros.

Learn about Volume of Cube here

We hope you find this article on ‘Patterns in Cube of Numbers helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them. 

Counting Cubes | NZ Maths


This is a level 4 algebra strand activity from the Figure It Out series.

A PDF of the student activity is included.

Achievement Objectives

NA4-9: Use graphs, tables, and rules to describe linear relationships found in number and spatial patterns.

AO elaboration and other teaching resources

Student Activity

Click on the image to enlarge it. Click again to close. Download PDF (151 KB)

Specific Learning Outcomes

write a rule to describe a pattern

use a rule to make predictions

Required Resource Materials

FIO, Level 4+, Algebra, Book Four, Counting Cubes, page 15

multilink cubes (optional)


In this activity, students will need to use their spatial skills to visualise the growing patterns of cubes made by Vyshan and Hema. Each of Vyshan’s buildings has an inner solid core in the shape of a cube. The core cannot be seen because a layer of cubes covers each of its 6 faces. The following diagram shows the 6 layers for Vyshan’s pattern 2.

In the pattern, there are 2 x 2 x 2 = 8 cubes making up the central core. Each of the 6 faces of the core has a single layer of 2 x 2 = 4 cubes attached to it. So pattern 2 has 23 + 6 x 22 = 32 cubes altogether.

The following table shows the number of cubes making up each of Vyshan’s patterns and how the patterns in each column lead to algebraic rules.

There are various ways of looking at Hema’s pattern, but the simplest way to express a rule is explained in the Answers

The following table shows how this rule works:

Note that the number of cubes for any pattern is 12 more than in the previous pattern because 1 cube is added to each of the 12 edges of a pattern to make the next building in Hema’s pattern.

In answer to question 3a, students with a good spatial sense may notice that there is a simple connection between Vyshan’s and Hema’s buildings. For example, Vyshan’s first pattern fits into Hema’s first pattern to make a 3 x 3 x 3 = 27 solid cube. Their second patterns fit together to make a 4 x 4 x 4 = 64 solid cube, and so on. Geometrically, we say that the patterns complement one another.

The following table shows how this relationship can be used. Note that the number of cubes on each edge is 2 more than the pattern number. (So in pattern 1, there are 1 + 2 = 3 cubes on each edge.)

If we know any two of the column entries for a given pattern, we can easily calculate the third. For example, the number of cubes in each of Vyshan’s patterns can be found by subtracting the numbers of cubes in Hema’s corresponding pattern from the total number of cubes in the joined patterns.

Answers to Activity
  1. a. Vyshan’s pattern:
    Pattern 1 has 7 cubes.
    Pattern 2 has 32 cubes.
    Pattern 3 has 81 cubes.

    Hema’s pattern:
    Pattern 1 has 20 cubes.
    Pattern 2 has 32 cubes.
    Pattern 3 has 44 cubes.

    b. Each building has 12 edges surrounding the hollow space. The cubes are increasing by 12 because 1 cube is added to each of the 12 edges to make the next building. A possible rule is: the number of cubes in the previous model plus 12.

    Each pattern has 8 corner cubes. The number of cubes that link each corner grows with each pattern and is the same number as the pattern number. So a rule for the number of cubes in each pattern can be expressed as “12 times the pattern number + 8”. 2.

  2. 216 cubes. (Hema will need 56 cubes, and Vyshan will need 160 cubes (64 + 96).)
  3. a. Vyshan’s first pattern fits into Hema’s first pattern to make a 3 by 3 by 3 solid cube, his second pattern fits into Hema’s second pattern to make a 4 by 4 by 4 solid cube, and so on.

    b. Hema’s next building will have 6 cubes along each edge. Vyshan’s next building fits inside Hema’s building to make a solid cube, so they will need a total of 6 x 6 x 6 (or 63) = 216 cubes.


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Level Four

Rubik’s cube in mathematics

Studying the sources of literature and solving problems existing for the cube, I found an interesting problem from the Mathematical Olympiad for schoolchildren on the estate of G.P. sawed it into 8 cubes and put the cubes back in the form of a cube, the entire surface of which is painted. Hurvinek looks at the cube and sees, of course, not all the faces, but only three, turned towards him. But he claims to know which cube lies in the corner farthest from him. Which?

Dunno dismantled the Rubik’s Cube in an arbitrary direction. Znayka looks at the cube and sees, of course, not all the faces, but only three, turned towards it. But he claims that if Dunno shows him the bottom invisible face, then he will be able to determine which cubes lie on the tops of the far back invisible face. What answer did Znayka give?

Obviously, after sawing, exactly three faces of 8 small cubes are painted, because these cubes are on the tops, and there are 8 tops, which is clearly visible on the scans. Let us write out combinations of colorings of the faces of all 8 corner cubes, according to the development: ZHZK, ZHZO, ZhSO, ZHKS, BOZ, BSO, BSK, BZK.

(Y-yellow, G-green, R-red, B-white, O-orange, S-blue).

We see a cube in which all sides of the UPC are visible, for clarity, we will color it in gray on the presented possible combinations of corner cubes. Let’s look at the bottom layer. On the front side there is a LDL cube, and the yellow one lies on the invisible side. Let’s color it grey.

To the left lies the HP cube, and the yellow one lies on an invisible side. On the presented possible combinations of LS, the colors can belong to the combinations of LS and LS, therefore, for clarity, we paint the LS sides on these combinations in gray.

On the right side of the bottom layer is a CC cube, with the yellow one lying on an invisible face. On the presented possible combinations of CS, the colors can belong to combinations of HSC and BSC, therefore, for clarity, let’s paint the CS sides on these combinations in gray.

It is clearly seen that LC colors were painted over in combinations of LSO and JCL, CL colors were painted over in combinations of HSC and BSC.

To the right is the HP cube. On the presented all possible combinations of corner cubes, ZhS colors can belong to combinations of ZhSO and ZhKS, the colors of ZhS in combinations of ZhSO and ZhKS are already painted in gray.

On the left is a CO cube. On the presented, possible combinations of corner cubes, CO colors can belong to combinations of BSO and ZSO, therefore, for clarity, let’s paint the OS sides on these combinations in gray.

It is clearly seen that the LS colors are painted in gray in combinations of LSO and ZHSS, the color SMs are painted over in combinations of BSO and ZhSO. As a result of the exclusion (coloring), we still have the combinations of the LZK and BZ intact, which means they belong to the invisible far edge.

Answer: On the far invisible side lie the HP and BOZ cubes. Moreover, the BOZ cube lies in the upper layer (because we see the white side of the far cube in the upper layer), and the ZhZK cube lies in the lower one.

I hope this problem will be used in logical problems for grades 8-9.

If you have any questions about solving the problem, be sure to write in the comments, I will be happy to answer!

Non-transitive cubes • Nikolay Avilov • Popular science problems on «Elements» • Mathematics

There are many paradoxes in probability theory, which, at first glance, contradict logic. In this problem, we considered a variation on the theme of one of these paradoxical cases. The first example of non-transitive cubes was given by the American mathematician Bradley Efron. Martin Gardner told the world about them in his column in the magazine Scientific American and in the wonderful book Tic-Tac-Toe. This set contains four cubes A, B, C and D, on the faces of which numbers are placed, and the relation A > B > C > D > A is satisfied.

There are several implementations of Efron’s cubes. In one of them: the numbers 1, 2, 3, 9, 10, 11 are placed on the die A, the numbers 0, 1, 7, 8, 8, 9 are placed on the die B, the numbers 5, 5, 6 are placed on the die C, 6, 7, 7, and the D die has numbers 3, 4, 4, 5, 11, 12.

also throws it away. The one with the most points wins. It is clear that none of the players can secure a win on every roll, but one of them can increase his chances of winning. Who? Let’s look into this issue with the help of tables (Fig. 4).

In these tables, as in the solution, the cells are shaded, which correspond to the case of a larger number on the die corresponding to the vertical row of numbers. Counting the filled cells in the tables, it is easy to check that die A is stronger than die B, die B is stronger than die C, and die C is stronger than die D — in all three cases, the stronger die wins in 22 out of 36 possible outcomes of the roll. But it is easy to see that the «weakest» die D is stronger than die A, and the probability of its victory is also 11/18. There is no seeming paradox here, as in the set of cubes from our problem, since the comparison of probabilities does not have the property of transitivity.

It turns out that the cubes can be ranked according to the closed cycle A > B > C > D > A, which means that the second player has more chances to win, because no matter which cube the first player chooses, he will find a stronger cube — it and take the second.

Efron invented non-transitive cubes more than 50 years ago. Since then, as already mentioned, many different variations on this theme have been invented. For example, Vadim Kryakvin proposed a set of four dice (like Efron’s), which uses the numbers from 1 to 24 once (Fig. 5).

Moreover, this set is obtained from Efron’s cubes as a result of a very simple operation: you need to write out all 24 numbers from Efron’s cubes in non-decreasing order and number them with numbers from 1 to 24. After that, it remains to replace each number on Efron’s cubes with its number and get Kryakvin’s cubes . Interestingly, as a result, the probability of winning the “strong” die in each pair changes: now it is equal to 2/3.

Our problem, in fact, turned out as a natural development of the idea of ​​V. Kryakvin: we wanted to obtain non-transitive sets with b about with a greater number of cubes, on the faces of which the initial numbers of the natural series would be written once. I came up with a set of five cubes, in which each next cube is stronger than the previous one in a cycle (Fig. 6). Each die in this set is also twice as likely to win against the next one.

The problem with five cubes was proposed on the «problem» site Readers have found many solutions to this problem, but most importantly, new methods for finding sets of non-transitive cubes have been proposed and constructions have been invented for a general problem in which a set of n cubes \(A_i\) \((i=1,\ldots,n)\) such that \(A_1>A_2>\ldots>A_n>A_1\). Further, it will be convenient for us to represent a set of cubes in the form of a list (or, if you like, a table) of rows, in which each row is the numbers from the faces of the next cube, written in ascending order.

Alexander Domashenko came up with a simple and understandable way to construct a set containing n non-transitive cubes. This method works for \(n\ge4\).

Let’s start filling out our list with numbers from 1 to 9(they are marked in red below). The remaining numbers from 10 to 6 n are divided into sixes — in order, starting with the number 10, and these sixes are entered line by line in the list from bottom to top (from the last line). The recording of sixes continues until the fourth line of the list is filled. At this point, 9 numbers from \(6n-8\) to \(6n\) will remain unused. We will add them to the first three lines — also moving from bottom to top. The list will look like this:

\(A_1\): 6, 7, 8, 9, \(6n-1\), \(6n\)
\(A_2\): 3, 4, 5, \(6n-4\), \(6n-3\), \(6n-2\)
\(A_3\): 1, 2, \(6n -8\), \(6n-7\), \(6n-6\), \(6n-5\)
\(A_4\): \(10+6(n-4)\), \( 11+6(n-4)\), \(12+6(n-4)\), \(13+6(n-4)\), \(14+6(n-4)\), \(15+6(n-4)\)
\(A_5\): \(10+6(n-5)\), \(11+6(n-5)\), \(12+6 (n-5)\), \(13+6(n-5)\), \(14+6(n-5)\), \(15+6(n-5)\)
. ..
\(A_{n-1}\): 16, 17, 18, 19, 20, 21
\(A_n\): 10, 11, 12, 13, 14, 15

non-transitive, it is easy to check. Indeed, comparing the cubes \(A_1\) and \(A_2\), we note that each of the numbers 6, 7, 8 and 9cube \(A_1\) is greater than the numbers 3, 4 and 5 of the cube \(A_2\), and the numbers \(6n-1\) and \(6n\) of the cube \(A_1\) are greater than any number of the cube \(A_2\) , so the die \(A_1\) is stronger than the die \(A_2\) in \(4\cdot3+2\cdot6=24\) cases out of 36 possible, so the probability of winning is \(P(A_1>A_2)=\frac{ 24}{36}=\frac23\). Similarly, we check that \(P(A_2>A_3)=P(A_3>A_4)=P(A_n>A_1)=\frac23\). In the remaining \(n-4\) pairs of cubes in a cycle, each of the six numbers of the upper cube is greater than any number of the lower one, so the probabilities are equal to 1.

Mikhail Vatnik came up with another method for constructing sets of non-transitive cubes, which can be called recurrent: having a non-transitive set of n cubes, he allows you to get a non-transitive set of n +1 cubes.

This algorithm consists of three steps:
1) Double all numbers on all faces of all cubes.
2) Add another die, in which each number is 1 more than the corresponding number of the first die (if the numbers \(2a\), \(2b\), \(2c\), \(2d) are written on the first die after doubling \), \(2e\), \(2f\), then a cube with numbers \(2a+1\), \(2b+1\), \(2c+1\), \(2d+1\ ), \(2e+1\), \(2f+1\) The order of the cubes, of course, cannot be changed!0071 3) Renumber all numbers on all cubes in ascending order (that is, do what Kryakvin did with Efron’s cubes).

It is easy to verify that the constructed set of cubes has the property of non-transitivity. Moreover, in all pairs, except for the last one, each cube is stronger than its counterpart with the same probability as pairs of cubes in a set with \(n\) cubes. For the last pair: \(P(A_{n+1}>A_1)=\frac{1+2+3+4+5+6}{36}=\frac{7}{12}>\frac{1 }{2}\). Hence, the constructed set of \(n+1\) cubes is also non-transitive.

Let’s demonstrate this technique by constructing a set of seven cubes based on the set of six cubes that we considered in the solution.

By alexxlab

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