Here's something you might find interesting. Pick two integers m and n, and find |m^2 - n^2|. What prime is most likely to divide this value? Subtle hint: (2a - 1)/(a^2)
Was anyone else thinking as he subtracted the fractions wow this is a long winded way to do that? n - 4 is obvious once you see that n(n-2)! = 8. I did my solutions a little differently as I tool the reciprocal of the fractions after subtracting the two terms.
You took the long way to find a common denominator. n! is a multiple of (n-1)!, so the common denominator is just n! so we need only to multiply the numerator and denominator of the 1/(n-1)! by n. 1/(n-1)! - 1/n! = n/n! - 1/n! = (n-1)/n! = (n-1)/(n*(n-1)*(n-2)!) = 1/(n*(n-2)!)
All up I liked this problem. Thankyou for it!
At the beginning, it would have been much quicker to say 1/(n-1)! = n/n!
Here's something you might find interesting. Pick two integers m and n, and find |m^2 - n^2|. What prime is most likely to divide this value?
Subtle hint: (2a - 1)/(a^2)
[1/(n-1)!][1-(1/n)]=⅛
If n=5 the LHS
Was anyone else thinking as he subtracted the fractions wow this is a long winded way to do that? n - 4 is obvious once you see that n(n-2)! = 8. I did my solutions a little differently as I tool the reciprocal of the fractions after subtracting the two terms.
1/(n-1)! - 1/n! = 1/8
n/n! - 1/n! = 1/8
(n-1)/n! = 1/8
n! = 8(n-1)
n! = 4(n-1)2
By inspection n=4
You took the long way to find a common denominator. n! is a multiple of (n-1)!, so the common denominator is just n! so we need only to multiply the numerator and denominator of the 1/(n-1)! by n.
1/(n-1)! - 1/n! = n/n! - 1/n! = (n-1)/n! = (n-1)/(n*(n-1)*(n-2)!) = 1/(n*(n-2)!)
❤❤
n/n! - 1/n! = 1/8
(n-1)/n! = 1/8
1/n(n-2)! = 1/8
n(n-2)! = 8
n(n-2)! = 2*4
n = 4
It turns out that there are an infinite number of solutions, including 2 solutions for n>0 😘