The Riemann Zeta Club is an enrichment maths club for Lower Sixth Formers, aimed as preparation for the Maths Challenge, Maths Olympiad and Oxbridge entry. This chapter covers divisibility, modular arithmetic, digit problems, Diophantine Equations and general theory concerning the integers.

- Download all files (zip)
- RZC-Chp2-NumberTheory-Slides.pptx
*(Slides)* - RZC-NumberTheory-Worksheet1.docx
*(Worksheet)* - RZC-NumberTheory-Worksheet2.docx
*(Worksheet)* - RZC-NumberTheory-Worksheet3.docx
*(Worksheet)* - RZC-NumberTheory-Worksheet4.docx
*(Worksheet)* - CoprimeDemo.xlsx
*(Spreadsheet)*

## Mr D Knowles

### 12th Oct 2020 Flag Comment

I may be way off here but for Worksheet 1 Q3 - can't we also use the fact that either n-1 or n+1 is divisible by 4? Therefore grab an extra factor of 2 making the answer 48 instead of 24?

## F Ding

### 27th Aug 2020 Flag Comment

Thank you so much! I did some of these slides over lockdown in preparation for SMC! Thank you!!!!

## B Solomons

### 26th Nov 2018 Flag Comment

I'm loving this - super helpful! Lil error on number theory worksheet 1 though, for the optional question 2. When you say "if the prime factor p is greater than fifty, it will appear twice in the denominator" - but that assumes that it's less than 100, else it won't appear in the denominator at all! 199 is prime, will show up in the numerator once, and won't show up in the denominator, therefore being the largest possible prime factor of the number. Really loving the resources though!

## Shrimat

### 25th Aug 2017 Flag Comment

Hi Dr Frost, I think there is an error in the solution for Q5b because I believe that you have missed out the prime factor 19 and thus part c is incorrect as well, as you have not multiplied the solution by 2 once more for the extra prime factor. Good stuff!

## Thomas Bolton

### 2nd Feb 2017 Flag Comment

Hi sir. I believe there's a mistake on Number Theory Worksheet 3, Q6. We need to find all p such that p itself and p^2 + 14 are prime. The solutions say that there are no solutions with possible residues of 1,2 mod 3. It discounts, understandably, the instance of p= 0 mod 3 as then p would no longer be prime, except for when p = 3. In this case p^2 +14 = 23 which is a prime.

## ben horspool

### 1st Oct 2016 Flag Comment

2*5+3*5=25, therefore 2n add 3n can equal a perfect square, you made an error on slide 77. other than that very helpful stuff

## D Person

## 22nd Oct 2021 Flag Comment