A special case of what you have discovered in the course of investigating the nature of solutions to the congruence equation

axb(modn) is the following:If

pis prime andais not a multiple ofp, then there exists an integerbsuch thatab1 (modp).Now fix

pas a prime number, and suppose thatais an integer with 0 <a<p. Then there is an integerb, also between 0 andp, such thatab1 (modp). Sinceab1 (modp) implies thatba1 (modp), we can try to pair off the integers between 1 andp– 1 so that the product of each pair is 1. The applet below takes a prime number as input, and displays the inverse for each value ofabetween 1 andp– 1. Here it is in action withp= 11:You may find this applet useful when working on Research Question 6.

## 5.4.1(a) Factorials

Recall the definition of

nfactorial:n! = 1 · 2 · 3 ···n. The on-line calculator will compute factorials using the usual notation. Try it:In Research Question 6, you are asked to investigate the behavior of factorials modulo

n.

## Research Question 6

(a) Find a formula for

n! %n.

(b) Find a formula for (n– 2)! %n.

(c) Find a formula for (n– 1)! %n.

## 5.4.1(b) Hint:

If you have a conjecture that you believe to be correct, but are having trouble finding a proof, it may be helpful to review the proof of the formula for the sum

(1 + 2 + 3 + ··· + n) %n.

Section 5.1 | Section 5.2 | Section 5.3 | Section 5.4 | Section 5.5 | Section 5.6

Copyright © 2001 by W. H. Freeman and Company