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The 4 by 4 "multiplication table" below is completely familiar. \(\begin{array}{rrrr}1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \\ 4 & 8 & 12 & 16\end{array}\) What is the total of all the numbers in the 4 by 4 square? How should one write this answer in a way that makes the total obvious?

Short Answer

Expert verified
The total is 100, and it is written as \( (1 + 2 + 3 + 4)^2 = 100 \).

Step by step solution

01

Identify the Table Elements

The 4 by 4 table is generated from the multiplication of numbers from 1 to 4. This means each entry in the table is given by the product of the row number and the column number.
02

Calculate the Total

Each row in this 4 by 4 table consists of multiples of the row number being added together. Therefore, each row sum is equal to the row number times the sum of numbers from 1 to 4. The sum of numbers from 1 to 4 is: \(1 + 2 + 3 + 4 = 10\).
03

Compute Row Sums

Using the row number multiplied by 10 (the sum of 1 to 4), calculate each row: - First row: \(1 \times 10 = 10\)- Second row: \(2 \times 10 = 20\)- Third row: \(3 \times 10 = 30\)- Fourth row: \(4 \times 10 = 40\)
04

Sum All Row Totals

Add all the row totals together to get the total:\[10 + 20 + 30 + 40 = 100\]
05

Write the Total in an Obvious Way

The total of all numbers in the table is equal to \((1 + 2 + 3 + 4)^2\), because the table is effectively adding these series up in both rows and columns: \( (1 + 2 + 3 + 4) \times (1 + 2 + 3 + 4) = 10 \times 10 = 100\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Operations
Matrix operations involve manipulating matrices using certain rules. In the context of the 4 by 4 multiplication table, each element in the matrix is the product of its row index and column index. Such operations are foundational in linear algebra. Here, every number in the matrix corresponds to an intersection of these indices, following a multiplication pattern.

When you perform matrix operations, you often focus on combining elements, either by addition, subtraction, or other transformations. In solving the problem of finding the sum of the entire table, we engage in additive operations by summing across rows and columns. This involves systematic calculations where each component is part of a larger structured grid.
  • Each matrix element is a reflection of basic multiplication principles.
  • It showcases structured relationships and symmetries within numbers.
  • Matrix operations help streamline calculations that might otherwise seem complex if considered individually.
As you can see, understanding matrix operations allows for powerful and efficient ways of problem-solving.
Elementary Arithmetic
Elementary arithmetic is the foundation for most mathematical reasoning and problem solving. In this exercise, we work through a simple multiplication table, a fundamental concept in arithmetic.

Multiplication is a form of repeated addition. Here, each row in the 4 by 4 table is achieved by consistently adding the basic number being multiplied. When learning multiplication, it's crucial to grasp that it simplifies and shortens what could be long sequences of additions.
  • For the first row, each number is essentially added 'x' times, where 'x' is the multiplier.
  • Multiplicands form the basic units of larger products, as seen in the repeated multiplication along rows.
These arithmetic operations provide the tools to approach more complex mathematical concepts. Understanding multiplication as repeated addition can ease the move to grasping larger patterns, such as those found in sequences or matrices.
Sum of Sequences
The sum of sequences is a concept where numbers in a sequence are added together to produce a total. In solving our multiplication table problem, the sum of sequences was pivotal.

The numbers in each row of the table are a series of multiples, and when added together, they give the row sum. Furthermore, the total of the table is found through adding these row sums. The series of numbers from 1 to 4, whose sum is 10, acts as a repeated sequence in our operation. Thus, calculating the sum of a sequence efficiently reduces what could otherwise be a cumbersome task.
  • The sequence 1, 2, 3, 4 repeats across rows and columns.
  • The sum, when squared, forms the total result due to consistent repetition.
Understanding the sum of sequences concept allows for more elegant mathematical formulations, such as transforming a repeated addition into a concise multiplication.

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Most popular questions from this chapter

(a)(i) Expand \((a+b)^{2}\) and \((a+b)^{3}\). (ii) Without doing any more work, write out the expanded forms of \((a-b)^{2}\) and \((a-b)^{3}\). (b) Factorise (i) \(x^{2}+2 x+1\) (ii) \(x^{4}-2 x^{2}+1\) (iii) \(x^{6}-3 x^{4}+3 x^{2}-1\). (c)(i) Expand \((a-b)(a+b)\). (ii) Use (c)(i) and (a)(i) to write down (with no extra work) the expanded form of $$ (a-b-c)(a+b+c) $$ and of $$ (a-b+c)(a+b-c) $$ (d) Factorise \(3 x^{2}+2 x-1\).

Let \(\Delta=\operatorname{area}(\triangle A B C)\) (a) Prove that $$ \Delta=\frac{1}{2} \cdot a b \cdot \sin C . $$ (b) Prove that \(4 R \Delta=a b c\).

Problem 27 (Overlapping squares) A square \(A B C D\) of side 2 sits on top of a square \(P Q R S\) of side \(1,\) with vertex \(A\) at the centre \(O\) of the small square, side \(A B\) cutting the side \(P Q\) at the point \(X,\) and \(\angle A X Q=\theta\) (a) Calculate the area of the overlapping region. (b) Replace the two squares in part (a) with two equilateral triangles. Can you find the area of overlap in that case? What if we replace the squares (i.e. regular 4-gons) in part (a) with regular \(2 n\) -gons? \(\Delta\)

Reveals the triple (3,4,5) as the first instance \((m=1)\) of a one-parameter infinite family of triples, which continues $$ (5,12,13)(m=2),(7,24,25)(m=3),(9,40,41)(m=4), \ldots $$ whose general term is $$ (2 m+1,2 m(m+1), 2 m(m+1)+1) $$ The triple (3,4,5) is also the first member of a quite different "one- parameter infinite family" of triples, which continues $$ (6,8,10),(9,12,15), \ldots $$ Here the triples are scaled-up versions of the first triple (3,4,5) . In general, common factors simply get in the way: If \(a^{2}+b^{2}=c^{2}\) and \(H C F(a, b)=s,\) then \(s^{2}\) divides \(a^{2}+b^{2},\) and \(a^{2}+b^{2}=c^{2} ;\) so \(s\) divides \(c .\) And if \(a^{2}+b^{2}=c^{2}\) and \(H C F(b, c)=s,\) then \(s^{2}\) divides \(c^{2}-b^{2}=a^{2},\) so \(s\) divides \(a .\) Hence a typical Pythagorean triple has the form \((s a, s b, s c)\) for some scale factor \(s,\) where \((a, b, c)\) is a triple of integers, no two of which have a common factor: any such triple is said to be primitive (that is, basic - like prime numbers). Every Pythagorean triple is an integer multiple of some primitive Pythagorean triple. The next problem invites you to find a simple formula for all primitive Pythagorean triples.

Which is bigger: \(17 \%\) of nineteen million, or \(19 \%\) of seventeen million?

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