All Questions
Tagged with prime-factorization factorial
54
questions
26
votes
1
answer
531
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Prime factor wanted of the huge number $\sum_{j=1}^{10} j!^{j!}$
What is the smallest prime factor of $$\sum_{j=1}^{10} j!^{j!}$$ ?
Trial :
This number has $23\ 804\ 069$ digits , so if it were prime it would be a record prime. I do not think however that this ...
6
votes
1
answer
133
views
Conjecture: $\prod\limits_{k=0}^{n}\binom{2n}{k}$ is divisible by $\prod\limits_{k=0}^n\binom{2k}{k}$ only if $n=1,2,5$.
The diagram shows Pascal's triangle down to row $10$.
I noticed that the product of the blue numbers is divisible by the product of the orange numbers. That is (including the bottom centre number $...
1
vote
0
answers
62
views
Numbers $n$ such that $n!$ has all of its exponents odd
(First paragraph for motivation, second paragraph the actual problem)
I usually have to be on a stand as an embassador for my major (math) for highschoolers, and we usually have problems written on a ...
4
votes
0
answers
84
views
Finite many primes for every positive integer $b$?
Consider the function $$f(a,b):=\sum_{j=0}^a (bj)!=1+b!+(2b)!+\cdots +(ab)!$$
Given a positive integer $b$ , are there always only a finite number of positive integers $a$ such that $f(a,b)$ is prime ...
1
vote
1
answer
561
views
Let a rational number $\frac{a}{b}$ in its lowest form where $a$,$b$ are integers, with $0 < \frac{a}{b}< 1$, b > 1. How many of these have $ab = 15!$
Consider a rational number $\frac{a}{b}$ in its lowest form where $a$, $b$ are integers, with $0 < \frac{a}{b}< 1$, b > 1. How many of these have $ab = 15!$
Solution Given in Book:
$15!=2^{11}...
0
votes
1
answer
341
views
Another Doubt Regarding Brocard's Problem, Specifically about Where I Can Proceed after Prime Factorisation
I know this can draw downvotes as it's not a complete solution or sounds more like an opinion poll (or perhaps this must have appeared elsewhere, in which case you're free to let me know), but I would ...
0
votes
1
answer
63
views
If n is a positive integer that is four digits long and is relatively prime to 100!, why must n be prime?
Suppose there is some positive integer n that is four digits long and is relatively prime to 100! (meaning n and 100! have no common factors other than 1). n must be prime, but why?
100! is a ...
1
vote
1
answer
90
views
Median factor of factorials
What is the order of $m(N)$, the median prime in the prime factorization of $N!$, as $N\to\infty$? For example, $m(6)=2$ because $6!=2^4\times3^2\times5$ and the median of $\{2,2,2,2,3,3,5\}$ is $2$.
...
2
votes
1
answer
85
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Factorials and Place Value
I recently came across this question from a non-calculator exercise. The units and tens place value digits I can see as $0$ and $0$ since in $12!$ we have $10*5*2=100$ but is there a way to find $a$ ...
2
votes
1
answer
2k
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Find the power of a prime in the prime factorization of a large factorial
I have been trying to work through the following exercise:
Find the power of $5$ in the prime factorization of $2020!$.
So far I have worked out that the prime factorization of $2020$ is $2^2 \cdot 5^...
4
votes
1
answer
277
views
Proof that each prime power of ${2n \choose n}$ is $\leq \log_p 2n$
I'm trying to work through this proof of the prime number theorem.
Def: Let $P_p(x)$ be the prime power of $p$ in the prime factorization of $x$. I.e. for any natural number $x$, $x = \prod_{p\in \...
1
vote
1
answer
165
views
perfect factors from the prime factorization of a large number
This is probably an easy question, but I don't know how to do it:
In the prime factorization of $30!$, how many perfect factors occur?
This is from a timed competition, any answers that take more than ...
0
votes
3
answers
63
views
how (a!)/(b!) = (b + 1)×(b + 2)×⋯×(a − 1)×a [closed]
I was solving a problem in which i need to figure out the prime factorization of $\frac{a!}{b!}$ and i did that by computing (a!) and then (b!) by looping ((1 to a) & (1 to b)) and then derived n ...
6
votes
0
answers
99
views
Can we conclude $n=p-1$?
Let $\ n\ $ be a positive integer and $\ p\ $ a prime number such that $$\ p^2\mid (2n)! + n! + 1$$ The only pairs $\ (n,p)\ $ I found so far are
$(1,2)$ , $(2,3)$ , $(10,11)$ , $(106,107)$ , $(4930,...
0
votes
1
answer
67
views
Is the sequence infinite? Finite? Is there a general formula to determine n th term?
Sequence of numbers whose factorial on prime factorisation contains prime powers of prime numbers, whose power is greater than $1$ or contains multiplicity of one for all prime numbers less than equal ...