Try this beautiful problem from the PRMO, 2017 based on GCD and Primes.
For each positive integer n, consider the highest common factor \(h_n\) of the two numbers n!+1 and (n+1)! for n<100, find the largest value of \(h_n\).
GCD
Primes
Inequalities
But try the problem first...
Answer: is 97.
PRMO, 2017, Question 29
Elementary Number Theory by David Burton
First hint
n! +1 is not divisible by 1,2,.....,n (n+1)! divisible by 1,2,....,n then \(hcf \geq (n+1)\) and (n+1)! not divisible by n+2, n+3,...... then hcf= (n+1)
Second Hint
let n=99, 99! +1 and (100)! hcf=100 not possible as 100 |99! and 100 is non prime
Final Step
let n=97 96! + 1 and 97! both divisible by 97 then hcf=97.
Try this beautiful problem from the PRMO, 2017 based on GCD and Primes.
For each positive integer n, consider the highest common factor \(h_n\) of the two numbers n!+1 and (n+1)! for n<100, find the largest value of \(h_n\).
GCD
Primes
Inequalities
But try the problem first...
Answer: is 97.
PRMO, 2017, Question 29
Elementary Number Theory by David Burton
First hint
n! +1 is not divisible by 1,2,.....,n (n+1)! divisible by 1,2,....,n then \(hcf \geq (n+1)\) and (n+1)! not divisible by n+2, n+3,...... then hcf= (n+1)
Second Hint
let n=99, 99! +1 and (100)! hcf=100 not possible as 100 |99! and 100 is non prime
Final Step
let n=97 96! + 1 and 97! both divisible by 97 then hcf=97.
