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- Introduction Mathematics has an advantage over other subjects. Theorems are absolute. They are not subject to further discussion as to their correctness. No sane person can write a paper disputing the Pythagorean theorem and expect to be taken seriously. Once you have proved a theorem, it is forever true. On the other hand your theorem will not be accepted until it is proved. You cannot argue that you have checked 79 right triangles and have always found a^2 + b^2 = c^2 and thus have proved the Pythagorean theorem. Your study of examples is good work and leads you to conjecture that a^2 +b^2 = c^2 in any right triangle. But the result will only be a mathematical conjecture until it is proved. Some conjectures in mathematics have been checked for trillions of cases and mathematicians are 99.9999% certain they are valid but until a proof is supplied they are considered unsolved problems. A theorem is not established until it is proved. The statement in the theorem must be shown to always hold. A proof is a thoroughly convincing explanation, an explanation that leaves no shadow of doubt, as to why something is true. The amount of explanation and detail in this explanation depends upon the audience for the proof. The amount of explanation required between professors might be much less than that required between a professor and a student. Of course a purported theorem (a conjecture) can be shown to be false by the presentation of just one exception to the statement. Let’s suppose that your friend Emile says that he has discovered the following two part theorem.
Emile’s Theorem. A) The sum of any even integer and any odd integer is an odd integer. B) The product of any even integer and any odd integer is an odd integer.
Emile, like ALL OF US, is sometimes wrong so you decide the best way to check this out is to try to prove the theorem yourself. For A) you first try to understand the statement by considering some examples: 5 + 4 = 9, 11 + 20 = 31, 2 + 1 = 3. To prove the statement you need to know exactly what it says. This means knowing the definitions. What is an even integer? You could say “an even integer is any of the numbers 2, 4 , 6 , 8 ,.. ..” Is this correct? What about 0? Is 0 even? What about −2 or −10? Are they even. Yes, they are. Reading the statement of A) carefully we do not see the words “positive even integer” (2, 4 , 6 , 8 ,.. .) or “nonnegative even integer” (0, 2 , 4 ,.. .). Now, is this listing of all even integers the right description to use in our proof? We could make a similar listing of all odd integers (... , − 3 , − 1 , 1 , 3 , 5 ,.. .) but then to establish the result we need to show that if we add any number on the first list to any number on the second we have a number on
the second list. We cannot check this by hand or by computer (there are infinitely many cases). So this is not the best definition of even and odd to use. However when you check your notes you see the following definitions:
Definition. An integer m is odd if there exists an integer n so that m = 2n + 1. An integer m is even if there exists an integer n so that m = 2n.
Now let’s give your proof of A): Proof. Let m 1 be an even integer and let m 2 be an odd integer. Then there exist integers n 1 and n 2 so that m 1 = 2n 1 and m 2 = 2n 2 + 1. So m 1 + m 2 = 2n 1 + 2n 2 + 1 = 2(n 1 + n 2 ) + 1. So if n 3 = n 1 + n 2 then m 1 + m 2 = 2n 3 + 1. n 3 is an integer and so m 1 + m 2 is odd by the definition of odd.
Note that your proof required you to choose some notation. You could not let m 1 = 2n and m 2 = 2n + 1 as one might be tempted to do from the definition. Indeed this would mean that m 2 = m 1 + 1 which would be a restrictive case of the theorem and your proof would not be in full generality. Next you try to understand Theorem B). You first try is 2 · 3 = 6. You can stop now! You have shown that “Theorem B)” is false. The statement means that if you multiply any even integer by any odd integer you get an odd integer. But 2 · 3 = 6 shows this to be false. You investigate further: 4 · 7 = 28, 9 · 10 = 90 and you posit a revised
Theorem B)′. The product of an even integer and an odd integer is an even integer.
0.1. Prove Theorem B)′.
0.2. Formulate and prove an improvement of Theorem B)′.
In this course you will be proving many theorems. While they will often be more difficult than the examples above, the methodology you should use will be similar. First you should know all the definitions and previous results. This may seem impossible but if you have faithfully worked out the problems you will find that you DO know them. Maybe you need a peek now and then to refresh your memory but they are your results. You own them. They are friendly companions. The harder you work and the more problems you solve the easier the next problem will be. It may still be hard and challenging but you can do it. Read the theorem. Read it again. Make sure you understand precisely what it says. What is assumed. What must be proved. Try to construct examples that illustrate the theorem. How can you use the hypothesis to deduce the conclusion? When you have finished your proof, review it carefully. What was your argument? How did your use the
0.4. Prove or disprove each of the following.
a) If x is an integer then x^2 > 9 if and only if x > 3. b) If 5 > 3 then 4 is an even integer. c) If 5 < 3 then 4 is an odd integer. d) If 5 is an even integer then 4 > 3.
- Division Algorithm, Euclidean Algorithm, Linear Diophantine Equations and Introduction to Modular Arithmetic Introduction and Goals. How can one natural number be expressed as the product of smaller natural numbers? This innocent sounding question leads to a vast field of intercon- nections among the natural numbers that mathematicians have been exploring for literally thousands of years. The adventure begins by recalling the arithmetic from our youth and looking at it afresh. In this section we will explore the Division Algorithm, greatest com- mon divisors, the Euclidean Algorithm, and some consequences of these to finding integer solutions to linear equations. We will develop skills in proving theorems by induction and other means. Many experiences in everyday life are cyclical—hours in the day, days in a week, motions of the planets. This concept of cyclicity gives rise to the idea of modular arithmetic, which develops the notion of numbers on a cycle. In this section, we will introduce the basic idea of modular arithmetic, which we will develop further in future sections.
Definitions.
- The natural numbers are the numbers { 1 , 2 , 3 , 4 ,.. .}.
- The integers are {... − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 ,.. .}.
- Suppose a and d are integers, then d divides a, denoted d|a, if and only if there is an integer k such that a = kd.
- Suppose that a, b, and n are integers, n > 0. We say that a and b are congruent modulo n if and only if n|(a − b). We denote this relationship as a ≡ b (mod n) and read these symbols as “a is congruent to b (mod n).” We begin now with the first set of questions. “Theorem” denotes a mathematical statement to be proved by you. For example,
Example Theorem. 3 | − 9. Then you would supply the proof, write it up clearly, and be prepared to present your proof at the board. Your write-up might look like this:
Example Theorem. 3 | − 9. Example Proof. By definition, 3| − 9 means that there exists an integer k such that −9 = 3k. So to prove that this definition is satisfied, we need to find an integer k such that −9 = 3k. Since −9 = 3 · (−3), the definition is satisfied, so 3| − 9 is true.
Note: If you are familiar with equivalence relations, you may note that the previous three theorems establish that congruence modulo n defines an equivalence relation on the set of integers. In the question before those theorems, 1.8, you described the equivalence classes modulo 3.
1.12. Theorem. Let a, b, c, d, and n be integers with n > 0. If a ≡ b (mod n) and c ≡ d (mod n), then a + c ≡ b + d (mod n).
1.13. Theorem. Let a, b, c, d, and n be integers with n > 0. If a ≡ b (mod n) and c ≡ d (mod n), then a − c ≡ b − d (mod n).
1.14. Theorem. Let a, b, c, d, and n be integers with n > 0. If a ≡ b (mod n) and c ≡ d (mod n), then ac ≡ bd (mod n).
1.15. Theorem. Let a, b, k, and n be integers with n > 0 and k > 0. If a ≡ b (mod n), then ak^ ≡ bk^ (mod n).
1.16. Question. Illustrate each of the above theorems with an example using actual numbers.
1.17. Theorem. Let a natural number n be expressed in base 10 as n = akak− 1 · · · a 1 a 0.
Then if m = ak + ak− 1 + · · · + a 1 + a 0 , n ≡ m (mod 3).
1.18. Theorem. A natural number that is expressed in base 10 is divisible by 3 if and only if the sum of its digits is divisible by 3.
1.19. Question. Devise and prove other divisibility criteria similar to the preceding one.
Well-Ordering Axiom for the Natural Numbers. Let S be any non-empty set of natural numbers, then S has a smallest element. We will just assume that the Well-Ordering Axiom for the Natural Numbers is true. So feel free to use it whenever you wish.
The Division Algorithm. Let n and m be natural numbers. Then (existence part) there exist integers q (for quotient) and r (for remainder) such that
m = nq + r and
0 ≤ r ≤ n − 1 (r is greater than or equal to 0 but less than or equal to n − 1). Moreover, (uniqueness part) if q, q′^ and r, r′^ are any integers that satisfy m = nq + r = nq′^ + r′^ with 0 ≤ r, r′^ ≤ n − 1, then q = q′^ and r = r′.
1.20. Illustrate the division algorithm for m = 22, n = 11; m = 33, n = 45; m = 277, n = 4.
1.21. Prove the existence part of the Division Algorithm. (Hint: Given n and m, how will you define q? Once you choose this q, then how is r chosen? Then show that 0 ≤ r ≤ n − 1.)
1.22. Prove the uniqueness part of the Division Algorithm. (Hint: If nq + r = nq′^ + r′, then nq − nq′^ = r′^ − r. Use what you know about r and r′^ as part of your argument that q = q′.)
1.23. Theorem. Let a, b, and n be integers with n > 0. Then a ≡ b (mod n) if and only if a and b have the same remainder when divided by n. Equivalently, a ≡ b (mod n) if and only if when a = nq 1 + r 1 (0 ≤ r 1 ≤ n − 1) and b = nq 2 + r 2 (0 ≤ r 2 ≤ n − 1), then r 1 = r 2.
Definitions.
- The greatest common divisor of two integers a and b is the largest integer d such that d|a and d|b. The greatest common divisor of two integers a and b is denoted gcd(a, b) or (a, b).
- If gcd(a, b) = 1, then a and b are said to be relatively prime. 1.24. Question. Find (36, 22), (45, −15), (296, −88), (0, 256), and (15, 28). 1.25. Theorem. Let a, n, b, r, and k be integers. If a = nb + r and k|a and k|b, then k|r.
1.26. Theorem. Let a, n 1 , b, r 1 , n 2 , and r 2 be integers. If a = n 1 b + r 1 , then (a, b) = (b, r 1 ). Similarly, if b = n 2 r 1 + r 2 , then (b, r 1 ) = (r 1 , r 2 ).
1.27. Question. As an illustration of the above theorem, note that 51 = 3 · 15 + 6 15 = 2 · 6 + 3 6 = 2 · 3 + 0.
Show that if a = 51 and b = 15, then (51, 15) = (6, 3) = 3.
other values, x = x 0 +h and y = y 0 +k, also satisfy ax+by = c? Formulate a conjecture that answers this question. Devise some numerical examples to ground your exploration. For example, 6(−2) + 15 · 2 = 18. Can you find other integers x and y such that 6x + 15y = 18? How many other pairs of integers x and y can you find? Can you find infinitely many other solutions?
1.38. Question (Euler). A farmer lays out the sum of 1,770 crowns in purchasing horses and oxen. He pays 31 crowns for each horse and 21 crowns for each ox. What are the possible numbers of horses and oxen that the farmer bought?
1.39. Theorem. Let a, b, c, x 0 , and y 0 be integers such that ax 0 + by 0 = c. Then the integers x = x 0 + (^) (a,bb) and y = y 0 − (^) (a,ba) also satisfy the linear Diophantine equation ax + by = c.
1.40. Question. If a, b, and c are integers and the linear Diophantine equation ax+by = c has at least one integer solution, find a general expression for all the integer solutions to that equation. Prove your conjecture.
1.41. Theorem. Let a, b, c, be integers. Then the equation ax + by = c has a solution if and only if (a, b)|c. If x 0 , y 0 is a solution, that is, ax 0 + by 0 = c, then for every integer k, the integers x = x 0 + (^) (a,bkb) and y = y 0 − (^) (a,bka) also satisfy the linear Diophantine equation ax + by = c. Moreover, every solution to the linear Diophantine equation ax + by = c is of this form.
1.42. Theorem. If a and b are integers and k is a natural number, then gcd(ka, kb) = k gcd(a, b).
1.43. Formulate a definition. For natural numbers a and b, give a suitable definition for the least common multiple of a and b, denoted lcm(a, b). Construct and compute some examples.
1.44. Theorem. If a and b are natural numbers, then gcd(a, b) lcm(a, b) = ab. 1.45. Corollary. If a and b are natural numbers, then lcm(a, b) = ab if and only if gcd(a, b) = 1.
1.46. Big Picture Question: How are the ideas of greatest common divisor and solutions to linear Diophantine equations related?
- Primes Primes.
- A natural number greater than 1 is prime if it cannot be expressed as the product of two smaller natural numbers.
- A natural number greater than 1 that is not prime, and therefore is the product of two smaller natural numbers, is composite. The study of primes is one of the main focuses of number theory. As we shall prove, every natural number greater than 1 is either prime or it can be expressed as a product of primes. The prime numbers give us a world of questions to explore. We will prove that there are infinitely many primes, but how are they distributed among the natural numbers? How many primes are there less than a natural number n? How can we find them? How can we use them? These questions and others have been among the driving questions of number theory for centuries and have led to an incredible amount of beautiful mathematics. Some of the basic techniques of analysis that arise in this study include: doing an example and then generalizing the method; proving a statement by contradiction, that is, assuming the statement is not true and then logically deducing a contradiction from that (false) assumption, thereby proving that the theorem must be true.
2.1. Theorem. If a natural number n > 1 is not prime, then there exists a prime p such that p|n.
2.2. Exercise (Sieve of Eratosthenes). Write down all the natural numbers from 1 to 100, perhaps on a 10 × 10 array. Circle the number 2, the smallest prime. Cross off all numbers divisible by 2. Circle 3, the next number that is not crossed out. Cross off all larger numbers that are divisible by 3. Continue to circle the smallest uncrossed out number and cross out its multiples. (a) Why are the circled numbers all the primes less than 100? (b) For each natural number n, π(n) denotes the number of primes less than or equal to n. Graph π(n) for n = 1, 2 ,... , 100. (c) Make a guess about approximately how large π(n) is relative to n. In particular, do you suspect that π( nn )is generally an increasing function or a decreasing function, and do you suspect that it approaches some specific number as a limit as n goes to infinity?
2.3. Theorem. A natural number n is prime if and only if for all primes p ≤ √n, p does not divide n.
2.4. Exercise. Use the preceding theorem to verify that 101 is prime.
2.13. Question. Find lcm(3^14722115173 , 52114138 17). 2.14. Conjecture. Make a conjecture that generalizes the ideas you used to answer the two previous questions.
2.15. Question. Do you think this method is always a better or always a worse method than using the Euclidean Algorithm to find (a, b)? Why?
2.16. Theorem. Let a, b, and n be integers. If a|n, b|n, and (a, b) = 1, then ab|n. 2.17. Theorem. Let p be a prime and a be an integer. If p does not divide a, then (a, p) = 1.
2.18. Theorem. Let p be a prime and a and b be integers. If p|ab, then p|a or p|b. 2.19. Theorem. Let a, b, and c be integers. If (b, c) = 1, then (a, bc) = (a, b)(a, c). 2.20. Theorem. Let a, b, and c be integers. If (a, b) = 1 and (a, c) = 1, then (a, bc) = 1. 2.21. Theorem. Let a, b be integers. If (a, b) = d, then ( (^) a d ,^ db
2.22. Theorem. Let a, b, u, and v be integers. If (a, b) = 1, u|a, and v|b, then (u, v) = 1. 2.23. Theorem. For all natural numbers n, (n, n + 1) = 1. 2.24. Theorem. Let k be a natural number. Then there exists a natural number n (which will be larger than k) such that no natural number less than k and greater than 1 divides n.
2.25. Theorem (Infinitude of Primes Theorem). There are infinitely many prime numbers.
2.26. Question. After you have devised a proof or proofs for 2.25, what were the most clever or most difficult parts of your arguments?
2.27. Question. If you are given n+1 natural numbers less than or equal to the natural number 2n, is some pair necessarily relatively prime?
2.28. Theorem. There exist arbitrarily long sequences of composite numbers. That is, for any natural number n, there exists a run of n consecutive composite (that is, non-prime) natural numbers.
Definitions. A rational number is a number that can be written as a b where a and b are integers (b = 0). A real number that is not rational is irrational.
The Fundamental Theorem of Arithmetic can be used to prove that certain equations do not have integer solutions.
2.29. Theorem. There do not exist natural numbers m and n such that 7m^2 = n^2.
2.30. Theorem.
7 is irrational, that is, there do not exist natural numbers n and m such that
7 = (^) mn.
2.31. Question. For what other numbers can you prove their irrationality? Make and prove the most general conjecture you can.
2.32. Theorem. If r 1 , r 2 ,... , rm are natural number and each one is congruent to 1 (mod 4), then r 1 r 2... rm is also congruent to 1 (mod 4).
2.33. Theorem (Infinitude of 4k +3 Primes Theorem). There are infinitely many prime numbers that are congruent to 3 (mod 4).
In fact, the following much more general theorem is true, however, its proof is quite difficult and we will not attempt it in this course: Infinitude of ak + b Primes Theorem. If a and b are relatively prime, then there are infinitely many integers k for which ak + b is prime. Mersenne primes. A Mersenne prime is a prime of the form 2p^ − 1 where p is a prime.
2.34. Question. What is x xm−− 11 if x = 1?
2.35. Theorem. If n is a natural number and 2n^ − 1 is prime, then n must be a prime.
2.36. Question. Find the first few Mersenne primes.
2.37. For extra credit, an A in the class, and a Ph.D. in mathematics, prove that there are infinitely many Mersenne primes or prove that there aren’t.
Distribution of primes. How are the primes distributed among the natural numbers? Is there some pattern to their distribution? There are infinitely many primes, but how rare are they among the numbers? What proportion of the natural numbers are prime numbers? To explore these questions, the best way to start is to look at the natural numbers and the primes among them. Here then are the first few natural numbers with the primes printed in bold:
1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 ,...
n π(n) Prop. of primes: 1 / Ln(n) π(n) ln( n n) π(n))/n 10 4. 4. 43429 .... 92104 ... 100 25. 25. 21714 ... 1. 15133 ... 1000 168. 168. 14476 ... 1. 16054 ... 10000 1229. 1229. 10857 ... 1. 13199 ... 100000 9592. 09592. 08685 ... 1. 10443 ... 1000000 78498. 078498. 07238 ... 1. 08453 ... 10000000 664579. 0664579. 06204 ... 1. 07121 ... 100000000 5761455. 05761455. 05428 ... 1. 06143 ... 1000000000 50847534. 050847534. 04825 ... 1. 05383 ...
Notice how the last column seems to be getting closer and closer to zero—that is, the proportion of primes in the first n natural numbers is approximately 1/ Ln(n) and the fraction π(n)/n is becoming increasing closer to 1/ Ln(n) as n grows without bound.
The Prime Number Theorem. As n approaches infinity, the number of prime num- bers less than or equal to n approaches n/ Ln(n). Specifically, limn→∞ π(n) ln( n n)= 1.
The proofs of this theorem are difficult. If you prove it independently, you don’t have to take the final. Here are two famous open questions about prime numbers. If you solve them, you don’t have to take the final.
The Twin Prime Question. Are there infinitely many pairs of prime numbers that differ from one another by two? (11 and 13, 29 and 31, 41 and 43 are examples of some such pairs.)
The Goldbach Question. Can every positive, even number greater than 2 be written as the sum of two primes? (Pick some even numbers at random and see whether you can write them each as a sum of two primes.)
The Goldbach Conjecture has been verified by computer for all even numbers up to 400,000,000,000,000. As the even numbers get larger, there seem to be more ways to write them as a sum of two primes. For example, the number 100,000,000 can be written as the sum of two primes in 219,400 different ways.
- Modular arithmetic Introduction and goals. In the first section, we established the basics of modular arith- metic. Now we proceed to see how modular arithmetic can be used to deal with other familiar algebraic constructions such as functions and equations, and how it can help us to better understand primes and composite numbers. Modular arithmetic is interesting as an abstract topic in number theory, but it also plays a daily role in real life. It is the basis for public key cryptography and check digits associated with error detection. Here we develop the idea of modular arithmetic and later explore some of its applications outside mathematics. Recall the definition. Suppose that a, b, and n are integers, n > 0. We say that a and b are congruent modulo n if and only if n|(a − b). We denote this relationship as
a ≡ b (mod n)
and read these symbols as “a is congruent to b (mod n).”
3.1. Question. Show that 41|(2^20 − 1) by following these steps. Explain why each step is true.
i) 2^5 ≡ −9 (mod 41) ii) (2^5 )^4 ≡ (−9)^4 (mod 41) iii) 2^20 ≡ 812 (mod 41) ≡ (−1)^2 (mod 41) iv) 2^20 − 1 ≡ 0 (mod 41). 3.2. Question. In your head, find k (0 ≤ k ≤ 11) such that k ≡ 37453 (mod 12). (Hint: Don’t try to multiply it out and then divide by 12.)
3.3. Question. In your head or using paper and pencil but no calculator, find k, 0 ≤ k ≤ 6, such that 2^50 ≡ k (mod 7).
3.4. Question. Using paper and pencil, but no calculator, find k, 0 ≤ k ≤ 11, such that 39^453 ≡ k (mod 12). (Hint: First replace 39 by an integer between 0 and 11. Express 453 in base 2. For example, 23 = 1 · 24 + 1 · 22 + 1 · 21 + 1 · 20. Compute 3(2n)^ (mod 12) for all values of n up to the largest needed to write 453 in base 2. Note that 3(2n+1^ )(mod 12) = 3 (2n)^ · 3 (2n^ )(mod 12). With this hint, you can do this problem without ever multiplying numbers larger than 12. You do not have to do more than 10 steps of multiplying two numbers less than 12 and reducing the answer mod 12.)
3.5. Question. Show that 39|(53^103 + 103^53 ).
3.16. Note: This theorem implies that we cannot find a magical polynomial that produces only prime values for every integer input. Nevertheless, some polynomials do pretty well. The polynomial f (x) = x^2 + x + 41 has a prime value, that is, f (n) is prime, for 80 consecutive integer inputs, n = − 40 , − 39 ,... , 38 , 39. Try a few values to test this assertion.
3.17. Theorem. Given any integer a and natural number M , there exists a unique integer t in { 0 , 1 , 2 ,... , M − 1 } such that a ≡ t (mod M ).
Definitions.
- The set { 0 , 1 , 2 ,... M − 1 } is the canonical complete residue system mod M.
- Any set of integers {a 1 , a 2 ,... aM } for which every integer is congruent to exactly one element of the set (mod M ) is a called a complete residue system modulo M.
3.18. Question. Give three complete residue systems modulo 4: the canonical complete residue system, one containing negative numbers, and one containing no two consecutive numbers.
Linear congruences. Our next goal is to determine when there are solutions to the general linear congruence ax ≡ b (mod m) (that is, when there are values for x that make the congruence true) and how to find all the solutions. Let’s start with some examples.
3.19. Question. Find all solutions in the appropriate canonical complete residue system mod n that satisfy the following linear congruences:
i) 26x ≡ 14 (mod 3). ii) 2x ≡ 3 (mod 5). iii) 4x ≡ 7 (mod 8). iv) 24x ≡ 123 (mod 213). This one would be tedious to do by trial and error, so perhaps we should defer work on it for now and instead try to develop some techniques that might help. 3.20. Theorem. Let a, b, and m be integers with m > 0. Show that ax ≡ b (mod m) has a solution if and only if there exist integers x and y such that ax + my = b.
3.21. Theorem. Let a, b, and m be integers with m > 0. The equation ax ≡ b (mod m) has a solution if and only if (a, m)|b.
3.22. Question. What does the preceding theorem tell us about the congruence (iv) above?
3.23. Technique. (a) Use the Euclidean Algorithm to find a member x of the canonical complete residue system (mod 213) that satisfies 24x ≡ 123 (mod 213). (b) Find all members x of the canonical complete residue system (mod 213) that satisfy 24 x ≡ 123 (mod 213).
3.24. Question. Let a, b, and m be integers with m > 0. How many solutions are there to the linear congruence ax ≡ b (mod m) in the canonical complete residue system mod m?
3.25. Theorem. Let a, b, and m be integers with m > 0. Then (i) the congruence ax ≡ b (mod m) is solvable in integers if and only if (a, m)|b. (ii) If ax ≡ b (mod m) has a solution, then there are exactly (a, m) solutions in the canonical complete residue system modulo m. (iii) If x 0 is a solution to the congruence ax ≡ b (mod m), then x is a solution if and only if x ≡ x 0 + (^) (a,mm) n (mod m) for some n ∈ { 0 , 1 , 2 ,... , [(a, m) − 1]}. Sometimes in real life, we are confronted with problems involving simultaneous linear congruences. For example,
3.26. Puzzle. A band of 17 M328K students stole a pack of 1-hryvnia bills from their professor. When they tried to divide the fortune into equal portions, 3 bills remained. In the ensuing brawl over who should get the extra bills, Rob was killed. The bills were redistributed, but this time an equal division left 10 bills. Again they fought about who should get the remaining bills and Ben was killed. Now, fortunately, the bills could be divided evenly among the surviving 15 students. What was the fewest number of bills that could have been in the pack?
3.27. Puzzle (Brahmagupta, 7th century A.D.) When eggs in a basket are removed 2, 3, 4, 5, 6 at a time, there remain, respectively, 1, 2, 3, 4, 5 eggs. When they are taken out 7 at a time, none are left over. Find the smallest number of eggs that could have been contained in the basket.
3.28. Theorem (Chinese Remainder Theorem). Suppose m 1 , m 2 ,... , mL are positive integers that are pairwise relatively prime. Then the system of L congruences x ≡ a 1 (mod m 1 ) x ≡ a 2 (mod m 2 ) · · · x ≡ aL (mod mL)
