A Nice Factorial Equation

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[@MsAcpaul]

All up I liked this problem. Thankyou for it!

[@selahattinkara-o5h]

👏👏👏👏👏👏

[@martinepstein9826]

Multiply both sides by n!. n - 1 = n!/8 n =/= 0 or 1. So we can write n! as n(n - 1)(n - 2)!. n - 1 = n(n - 1)(n - 2)!/8 Multiply both sides by 8/(n - 1) 8 = n(n - 2)! n(n - 2)! is strictly increasing so there is at most 1 solution, and it's easy to guess and check. n = 4

[@nasrullahhusnan2289]

[1/(n-1)!][1-(1/n)]=⅛ If n=5 the LHS

[@اسماعیلخسروی-خ6ظ]

❤❤

[@RAG981]

At the beginning, it would have been much quicker to say 1/(n-1)! = n/n!

[@Qermaq]

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)

[@MsAcpaul]

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.

[@brianwade4179]

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

[@pedrojose392]

(n!-(n-1)!)/(n!*(n-1)!)=1/8 (n-1)!*(n-1)/(n!*(n-1)!)=1/8 (n-1)/n!=1/8 The exponent of 2 must be greater than or equal to 3 in n! So n>=4 n=5 does not work as 5 does not divide 8 n>=6, for sure, (n-1)/n!

[@bsmith6276]

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)!)

[@rob876]

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

[@alexandermorozov2248]

It turns out that there are an infinite number of solutions, including 2 solutions for n>0 😘