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Chapter 1: Problem 4

(a) Factorise 12345 as a product of primes. (b) Using only mental arithmetic, make a list of all prime numbers up to 100 . (c)(i) Find a prime number which is one less than a square. (ii) Find another such prime. There are 4 prime numbers less than \(10 ; 25\) prime numbers less than \(100 ;\) and 168 prime numbers less than 1000 . Problem \(\mathbf{4}(\mathrm{c})\) is included to emphasise a frequently neglected message: Words and images are part of the way we communicate. But most of us cannot calculate with words and images. To make use of mathematics, we must routinely translate words into symbols. For example, unknown numbers need to be represented by symbols, and points in a geometric diagram need to be properly labelled, before we can begin to calculate, and to reason, effectively.

Short Answer

Expert verified
(a) 12345 = 3 脳 5 脳 823. (b) Primes up to 100: 2, 3, 5, 7, ... , 97. (c) 3 and 47.

Step by step solution

01

Understanding Prime Factorization

Prime factorization is the process of expressing a number as a product of its prime factors. We'll be able to say a number is fully factorised into primes when it's expressed only in terms of prime numbers.
02

Finding Prime Factors of 12345

To factorize 12345, we start by checking divisibility with the smallest primes (2, 3, 5, 7, etc.). 12345 is not divisible by 2 (it's odd), but is divisible by 3 (sum of digits 15 is divisible by 3). Dividing, 12345 梅 3 = 4115. Further checking, 4115 is divisible by 5 (last digit is 5), giving 4115 梅 5 = 823. We check 823, which does not divide further by primes up to its square root, confirming it is a prime. Thus, 12345 = 3 脳 5 脳 823.
03

Listing Primes up to 100

Prime numbers are divisible only by 1 and themselves. Using mental arithmetic, list all prime numbers below 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. Count them to confirm there are 25 primes, as expected.
04

Finding a Prime Number Less Than a Square

To find a prime that is one less than a square, we try specific squares to see if subtracting one gives us a prime. Testing, 1. Check 4 - 1 = 3 (prime) 2. Check 9 - 1 = 8 (not prime) 3. Check 25 - 1 = 24 (not prime) 4. Check 49 - 1 = 48 (not prime) Only 3 was prime.
05

Finding Another Prime Number Less Than a Square

Continuing from Step 4, 1. Check 16 - 1 = 15 (not prime) 2. Check 36 - 1 = 35 (not prime) 3. Check 64 - 1 = 63 (not prime) 4. Check 81 - 1 = 80 (not prime) These were not prime solutions. Continuing: 5. Check 121 - 1 = 120 (not prime) 6. Check 49 - 1 = 47 (prime) We find 47 is another solution.

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Key Concepts

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

Prime Numbers
Prime numbers are the building blocks of mathematics. They are numbers greater than one with no divisors other than 1 and themselves. Understanding prime numbers is crucial because every number can be uniquely expressed as a product of primes. This is known as the Fundamental Theorem of Arithmetic.

When we talk about prime numbers, a few properties stand out:
  • The only even prime number is 2. All other even numbers can be divided by 2, so they cannot be prime.
  • Prime numbers are infinitely abundant. No matter how large a prime you find, there is always another one lurking out there.
  • They become less frequent as numbers get larger, but they never completely disappear.
The smallest prime numbers are 2, 3, 5, 7, 11, and so on. Knowing these small primes helps in tasks like prime factorization, where larger numbers are broken down into these basic elements.
Mental Arithmetic
Mental arithmetic is a wonderful skill that involves performing calculations in your head without the use of calculators or paper. When it comes to identifying prime numbers and factorizing them, mental arithmetic becomes an invaluable tool.

Here鈥檚 how you can practice mental arithmetic with primes:
  • To check if a smaller number is prime, quickly test if it is divisible by any primes up to its square root. For instance, check if 29 is divisible by 2, 3, or 5 to see if it's a prime.
  • Use basic division and multiplication facts to evaluate divisibility quickly. For example, if a number ends in 0 or 5, it's divisible by 5.
  • Practice listing primes using simple divisibility rules and memorization techniques, like remembering the first few primes up to 100.
By developing mental arithmetic skills, you enhance your number sense and gain a deeper appreciation of the elegance of prime numbers.
Understanding Primes
Understanding primes means recognizing their unique role in mathematics. They are not just numbers. They serve as the foundation for various mathematical concepts, including cryptography and number theory.

Here鈥檚 what to grasp about primes:
  • Recognize that every integer greater than one is either a prime itself or can be represented as a product of prime numbers. This decomposition is always the same, which exemplifies the unique role of primes in mathematics.
  • Explore the concept of prime factorization. This involves expressing any number as a product of primes, offering a clearer view of its structure and divisors.
  • Discover the applications of prime numbers in real-world contexts, such as secure communications where large primes are used in encryption algorithms like RSA.
Understanding primes is not only about knowing them by definition, but also appreciating their universality and applications in mathematics and beyond.

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Most popular questions from this chapter

(a) Using only mental arithmetic: (i) Determine which is bigger: $$ \frac{1}{2}+\frac{1}{5} \quad \text { or } \quad \frac{1}{3}+\frac{1}{4} ? $$ (ii) How is this question related to the observation that \(10<12 ?\) (b) [This part will require some written calculation and analysis.] (i) For positive real numbers \(x\), compare $$ \frac{1}{x+2}+\frac{1}{x+5} \quad \text { and } \quad \frac{1}{x+3}+\frac{1}{x+4} $$ (ii) What happens in part (i) if \(x\) is negative?

(a)(i) Explain why any integer that is a factor (or a divisor) of both \(m\) and \(n\) must also be a factor of their difference \(m-n,\) and of their sum \(m+n\). (ii) Prove that $$ H C F(m, n)=H C F(m-n, n) $$ (iii) Use this to calculate in your head \(H C F(1001,91)\) without factorising either number. (b)(i) Prove that: \(H C F(m, m+1)=1\). (ii) Find \(H C F(m, 2 m+1)\). (iii) Find \(H C F\left(m^{2}+1, m-1\right)\).

(Pages of a newspaper) I found a (double) sheet from an old newspaper, with pages 14 and 27 next to each other. How many pages were there in the original newspaper?

(a) Which of the prime numbers \(<100\) can be written as the sum of two squares? (b) Find an easy way to immediately write \(\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\) in the form \(\left(x^{2}+y^{2}\right)\). (This shows that the set of integers which can be written as the sum of two squares is "closed" under multiplication.) (c) Prove that no integer (and hence no prime number) of the form \(4 k+3\) can be written as the sum of two squares. (d) The only even prime number can clearly be written as a sum of two squares: \(2=1^{2}+1^{2}\). Euler \((1707-1783)\) proved that every odd prime number of the form \(4 k+1\) can be written as the sum of two squares in exactly one way. Find all integers \(<100\) that can be written as a sum of two squares. (e) For which integers \(N<100\) is it possible to construct a square of area \(N\), with vertices having integer coordinates? In Problem 25 parts (a) and (d) you had to decide which integers \(<100\) can be written as a sum of two squares as an exercise in mental arithmetic. In part (b) the fact that this set of integers is closed under multiplication turned out to be an application of the arithmetic of norms for complex numbers. Part (e) then interpreted sums of two squares geometrically by using Pythagoras' Theorem on the square lattice. These exercises are worth engaging in for their own sake. But it may also be of interest to know that writing an integer as a sum of two squares is a serious mathematical question \- and in more than one sense. Gauss \((1777-1855),\) in his book Disquisitiones arithmeticae (1801) gave a complete analysis of when an integer can be represented by a 'quadratic form', such as \(x^{2}+y^{2}\) (as in Problem 25) or \(x^{2}-2 y^{2}\) (as in Problem \(\mathbf{5 4}(\mathrm{c})\) in Chapter 2 ). A completely separate question (often attributed to Edward Waring \((1736-1798))\) concerns which integers can be expressed as a \(k^{\text {th }}\) power, or as a sum of \(n\) such powers. If we restrict to the case \(k=2\) (i.e. squares), then: \- When \(n=2,\) Euler \((1707-1783)\) proved that the integers that can be written as a sum of two squares are precisely those of the form $$ m^{2} \times p_{0} \times p_{1} \times p_{2} \times \cdots \times p_{s} $$ where \(p_{0}=1\) or \(2,\) and \(p_{1}

In how many different ways can the missing digits in this short multiplication be completed? $$ \begin{array}{r} \quad \square 6 \\ \times \quad \square \\ \hline \square 28 \\ \hline \end{array} $$ One would like students not only to master the direct operation of multiplying digits effectively, but also to notice that the inverse procedure of "identifying the multiples of a given integer that give rise to a specified output" depends on the HCF of the multiplier and the base (10) of the numeral system. \- Multiplying by \(1,3,7,\) or 9 induces a one-to-one mapping on the set of ten digits \(0-9\); so an inverse problem such as \(" 7 \times \square\) ends in 6 " has just one digit-solution. \- Multiplying by \(2,4,6,\) or 8 induces a two-to-one mapping onto the set of even digits (multiples of 2); so an inverse problem such as " \(6 \times \square\) ends in 4" has two digit-solutions, and an inverse problem such as " \(6 \times \square\) ends in 3 " has no digit-solutions. \- Multiplying by 5 induces a five-to-one mapping onto the multiples (0 and 5 ) of 5 , so an inverse problem such as " \(5 \times \square\) ends in 0 " has five digit-solutions and an inverse problem such as " \(5 \times \square\) ends in 3 " has no digit-solutions at all. \- Multiplying by 0 induces a ten-to-one mapping onto the multiples of 0 (namely 0 ); so an inverse problem such as " \(0 \times \square\) ends in 0 " has ten digit-solutions and an inverse problem such as " \(0 \times \square\) ends in 3 (or any digit other than 0 )" has no digit-solutions at all. The next problem shows - in a very simple setting - how elusive inverse problems can be. Here, instead of being asked to perform a direct calculation, the rules and the answer are given, and we are simply asked to invent a calculation that gives the specified output.
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