From e3490f30d615f42d58f3bc68ffb52bd91b2d4bc3 Mon Sep 17 00:00:00 2001 From: Pepsi Date: Wed, 28 Jun 2023 21:38:27 +1000 Subject: [PATCH] correct formatting in q3 part 1 --- q3/part1.md | 16 ++++++++-------- 1 file changed, 8 insertions(+), 8 deletions(-) diff --git a/q3/part1.md b/q3/part1.md index 194dbe4..443f349 100644 --- a/q3/part1.md +++ b/q3/part1.md @@ -2,18 +2,18 @@ ### Part 1 -Let d be an integer. The three requirements for d are that +Let $d$ be an integer. The three requirements for $d$ are that -d|n+5 &(1) \\d|n^2+2 &(2) \\d|27&(3), +$$d|n+5 &(1) \\d|n^2+2 &(2) \\d|27&(3),$$ -Where n is any integer. +Where $n$ is any integer. - Assume that (1) and (2) are true. The consequence of this is that there must be 2 integers, (let them be a and b), such that + Assume that $(1)$ and $(2)$ are true. The consequence of this is that there must be 2 integers, (let them be $a$ and $b$), such that -n+5&=ad &(3)\text{, derived from }(1)\\n^{2}+2&=bd &(4)\text{, derived from }(2). +$$n+5&=ad &(3)\text{, derived from }(1)\\n^{2}+2&=bd &(4)\text{, derived from }(2).$$ -Equation (3) can be rearranged such that n=ad-5. Substituting this into (4), +Equation $(3)$ can be rearranged such that $n=ad-5$. Substituting this into $(4)$, -\begin{aligned}bd&=(ad+5)^{2}+2\\&=a^{2}d^{2}-10ad+25+2\\27&=bd-a^{2}d^{2}+10ad \\27&=d\left(b-a^{2}d+10a\right). &(5)\end{aligned} +$$\begin{aligned}bd&=(ad+5)^{2}+2\\&=a^{2}d^{2}-10ad+25+2\\27&=bd-a^{2}d^{2}+10ad \\27&=d\left(b-a^{2}d+10a\right). &(5)\end{aligned}$$ -Since a,b, and d are all defined as integers, the right hand side of (5) is an integer which is divisible by d. Therefore, 27 must also be divisible by d, satisfying (3). As such, we can conclude that for all integers n, if d is an integer such that d|n+5 and d|n^2+2, then d|27. \ No newline at end of file +Since $a$,$b$, and $d$ are all defined as integers, the right hand side of $(5)$ is an integer which is divisible by $d$. Therefore, 27 must also be divisible by $d$, satisfying $(3)$. As such, we can conclude that for all integers $n$, if $d$ is an integer such that $d|n+5$ and $d|n^2+2$, then $d|27$. \ No newline at end of file