## prove (n)⊇(m)⟺n∣m (n) \supseteq (m)\iff n\mid m\ (contains = divides for principal ideals)

For non-zero integers m and n, prove (m)⊂(n) iif n divides m, where (n) is the principal ideal. My attempt is following. For non-zero integers m and n, assume that (m)⊂(n). Then, mk∈(m) is also in (n). This means that ∃nh such that mk=nh. Then, we have m=nhk−1. Assume that n divides m for non-zero … Read more

## Find the greatest positive integer $x$ such that $23^{6+x}$ divides $2000!$

I’m currently reading Andreescu and Andrica’s Number Theory: Structures, examples and problems. Problem 1.1.7 states : Find the greatest positive integer $x$ such that $23^{6+x}$ divides $2000!$. The solution given is : The number $23$ is prime and divides every $23^{rd}$ number. In all, there are $\lfloor \frac{2000}{23} \rfloor$ = $86$ numbers from $1$ to … Read more

## $2(n-2)+1$ does not divide $(n-2)(n-3)/2$ for $n \ge 8$

For $n \ge 8$ the number $2(n-2)+1$ never divides $(n-2)(n-3)/2$. Any ideas how to prove this? I see that $(n-2)(n-3)/2 = 1 + 2 + \ldots + (n-3)$. If I suppose that $2(n-2)+1$ divides $(n-2)(n-3)/2$ then it should also divide their difference \begin{align*} (n-2)(n-3)/2 – (2(n-2) + 1) & = \sum_{k=1}^{n-3} k – (2n – … Read more

## Divisibility of 2n−n22^n-n^2 by 7

How many positive integers n<104 are there such that 2n−n2 is divisible by 7? Answer Just write out periods: 2^n:2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,|2… with period 3 n^2:1,4,2,2,4,1,0,1,4,2,2,4,1,0,1,4,2,2,4,1,0,|1… with period 7 So in every 21 numbers there are 6 numbers. Now just calculate 10^4\over 21=476 and since 10000 itself is not included there are 3 remainder left so totally … Read more

## Induction: Prove that 53n+72n−15^{3n} + 7^{2n-1} is divisible by 44

Prove that 53n+72n−1 is divisible by 4 for all n∈N. For n=1, ⇒53+71⇒132∣4 (which is divisible by 4) Let us assume given equation holds true for n=m, ⇒53m+72m−1|4 Now for n=m+1, 53m+3+72m+2−1 53m⋅53+72m−1⋅72 53m⋅125+72m−1⋅49 How do I go ahead from here? I am kind of stuck. Answer To complete your argument try to use the … Read more

## Dividing by something Undefined

I was thinking about trigonometry ratios, in particularly $\cot(\theta)$, which can be defined as $\cot(\theta) = \frac {1}{\tan(\theta)} = \frac {cos(\theta)}{sin(\theta)}$. Though $\tan(90)$ is not defined as you end up getting $\frac{1}{0}$. Though $\cot(90) = 0$. Though one could interpret that we had to divide it by something undefined as $\frac {1}{0}$, isn’t defined. Yet … Read more

## Are there integers a,b,ca,b,c such that aa divides bcbc, but aa does not divide bb and aa does not divide cc?

Are there integers a,b,c such that a divides bc, but a does not divide b and a does not divide c? I am not quite sure what to do with the given information. I know I could easily find an example. We know that a divides bc so, bc=aq for some integer q. And that a does … Read more

## What can we say about the prime factors of ​^{10}10+23​^{10}10+23?

In a video on ultrafinitism I saw a claim that the number ​^{10}10+23 does not have prime factorization. While I don’t accept the premise of ultrafinitism, I got curious, what can we say about the prime factors of this number? ​^{10}10 refers to the hyperoperation tetration. In other words, the number is equal to 10^{10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}}, … Read more

## Find the largest 4 digit positive integer n such that 10 divides n19+99nn^{19}+99^n

I’ve solved this problem and got the answer of 9991. I manually proved how some digits could or couldn’t be the ones digit of n, but I feel there is a faster way Answer n19+99n=n19+(100−1)n≡n19+(−1)nmod10 by Newton’s formula. We therefore need n19≡−1nmod10. Clearly both expressions cycle mod10. So you need only solve mod10, and the … Read more

## What is the largest perfect square that divides 20143−20133+20123−20113+…+23−132014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3

I’ve tried this but didn’t get the answer : Let S=20143−20133+20123−20113+…+23−13 Using n3−(n−1)3=3n2−3n+1, S=3(20142)−3(2014)+1+3(20122)−3(2012)+1+…+3(22)−3(2)+1=3(2014(2013)+2012(2011)+2010(2009)+…+2(1))+1(1007)=3(1007∑n=12n(2n−1))+1007=(1007∑n=14n2−1007∑n=12n)+1007=12(1007)(1008)(2015)6−2(1007)(1008)(3)2+1007 This is divisible by 1007 but not by 10072 which is the correct answer. Where have I gone wrong ? Answer 12(1007)(1008)(2015)6−2(1007)(1008)(3)2+1007=1007(12(1008)(2015)6−2(1008)(3)2+1)=1007(2(1008)(2015)−(1008)(3)+1)=1007(2(1007+1)(2015)−(1007+1)(3)+1)=1007(1007(2(2015)−3)+2(2015)−3+1)=1007(1007(2(2015)−3)+2(2015)−2)=1007(1007(2(2015)−3)+2(2014))=1007(1007(2(2015)−3)+4(1007))=10072(2(2015)−3+4)=10072(4031) AttributionSource : Link , Question Author : H G Sur , Answer Author : Siong Thye Goh