A Quadratic Factorial Equation

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[@JossWainwright]

Because n! dominates n^2 we can reduce to a tiny handful of candidates. It's not necessary to even write anything on the paper. My first guess was n=5 because 11*n and n^2 will both end in a five, and add to ending in a zero. And 5! (and all higher factorials) end in zero. So the the expression tells you it's going to be five (since it also has 40 as a term, which ends in a zero). Anything much higher than five and the LHS shoots through the roof.

[@experimentingalgorithm1546]

I guess most of us guessed 5 right away and answered correctly 😂

[@mcumer]

Obviusly n! > 40 and therefore n>4 . If we put n=5, we obtain 5!=5^2+11*5+40, id est 120=25+55+40 which is 120=120.. it's easy to show that n=5 is the only one solution.. indeed, consider the 2 sequences an= n! and bn= n^2+11n+40: they are both increasing n>0.. furthermore for n=6 an>bn.. therefore an>bn for n>5 and n=5 is the only one solution

[@actions-speak]

Note that n = 0, 1 are not solutions. Then n(n-1)(n -2)! = n! = n(n-1) + 12(n-1) + 52 and (n - 1) divides 52 = 4 * 13, so n = 2, 3, 5, 14.

[@alexanderbrinkley4332]

Why did this comment get a ❤?

[@hazem1el_abed]

I believe you mean that n should be among
2,3,5,14

[@actions-speak]

@@hazem1el_abed Yep, thanks!

[@honestadministrator]

One needs to explore Zeroes of
f ( n) = n ( ( n - 1)! - n - 11) - 40
n being a positive integer,
n divides 40
case I : (n = 5)
at n = 5,
(n - 1)! - n - 11
= (4)! - ( 5 + 11)
= 8
[ Herrby n = 5 is a feasible solution ]
Herein f ( n) being a monotonically increasing function, no more Zeroes of f ( n) would be there

[@nasrullahhusnan2289]

Note that n>4
n!=n²+11n+40
=n(n-1)+12n+40
=n(n-1)+12(n-1)+52
n(n-1)[(n-2)!-1]=12(n-1)+52
(n-1)[n(n-2)!-n-12]=52
=4×13
n-1=4 --> n=5 and n(n-2)!-n-12=13
n[(n-2)!-1]=5² --> n=5 and (n-2)!-1=5
(n-2)!=6
=3!
n-2=3
n=5
Thus the solution is n=5

[@liammoore1066]

This is great! I was looking for a more algebraic solve.

[@wooyoungkim2925]

When n is larger than 2, n! should be even, and (n-1)! is also even. In right side, n + 11 + 40/n should be even. So, n should be odd. 40/n is integer.
By n >4, only 5 is solution because only 5 is odd among 1,2,4,5,8,10,20,40..

[@dawidp4227]

But n + 11 + 40/n can be even for even values of n, specifically for n = 8 and n = 40.

[@jonsrecordcollection7172]

Once you eliminate n = 40, you have to pick 5, because f(n) when n=40 is 2080. Since 2080 is the largest amount the RHS can be while still meeting the divisibility criteria of n|40 (since RHS is a strictly increasing function) and we know 2080 < 7!, that means n must be less than 7, because the factorial on the LHS is also an increasing function. The only possible n that is greater than 4 and less than 7 is 5, which is the answer.

[@tunistick8044]

I see everyone just plugging in 5 just like that but i feel like i'm not satisfied w this solution, i seek for an analytic demonstration. Has anyone got any?

[@Qermaq]

When I put this in my graphing calculator I see at least 7 negative solutions. appx. -0.9665, -2.0501, -2.9607, -4.0205, -4.9916, 6.0021 and -6.9997. What can we do to get those values?

[@tunistick8044]

because n! streches even to negative values EXCLUDING NEGATIVE INTEGERS
for instance (-½)!=√π/2

[@aurochrok634]

we have n! >= 40 => n>=5 . OTOH, 6! = 720 > 6^2 + 11x6 + 40 = 142. So, we only need to check n=5. and it fits :)

[@hcgreier6037]

Just guessing: n=5 😂

[@phill3986]

😊😊😊👍👍👍

[@marknieuweboer8099]

n! > 40.
So n > 4.
Substitute n = 5 - yup.
Took me a few seconds.

[@benjaminvatovez8823]

n! >=n(n-1)(n-2) (obvious if n=0,1 or 2, and because the other factors of n! are >=1 if n>=3)
So n²+11n+40 >= n³-3n²+2n
thus (n+1)(n²-5n-4)

[@tunistick8044]

I don't understand the last line. Shouldn't it be (n+1)(n²-5n-4)≤40?

[@benjaminvatovez8823]

@@tunistick8044 I agree my calculation was wrong and not as useful, then the correct result is f(n)=n(n²-4n-9)-40 ≤ 0
As n is strictly increasing and n²-4n-9 is strictly increasing for all n>2 and because f(6)0, we get f(n)>0 for all n>=7, thus n must be less than 7.

[@tunistick8044]

@@benjaminvatovez8823 i would suggest an easier way:
n²+11n+40≥n³-3n²+2n
n³-4n²-9n-84≤-44
(n-7)(12+3n+n²)

[@benjaminvatovez8823]

@@tunistick8044 I agree this solution is better. I wouldn't have found it in a short time range so I ran for the "easier" way according to me.

[@tunistick8044]

@@benjaminvatovez8823 splendid

[@walterwen2975]

A Quadratic Factorial Equation: n! = n² + 11n + 40; n = ?
n! = n² + 11n + 40 = n(n + 1) + 10(n + 4), n! - n(n + 1) = 10(n + 4)
n[(n - 1)! - (n + 1)] = 10(n + 4) > 0, (n - 1)! > n + 1; n > 3
n = 4: 4[3! - (4 + 1)] = 4(6 - 5) = 4 < 10(4 + 4) = 80
n = 6: 6[5! - (6 + 1)] = 6(120 - 7) = 678 > 10(6 + 4) = 100; 6 > n > 4
n = 5: 5[4! - (5 + 1)] = 5(24 - 6) = 90 = 10(5 + 4); Proved
Answer check:
n! = (5)(4)(3)(2) = 120, n² + 11n + 40 = (5)(5) + 11(5) + 40 = 120; Confirmed
Final answer:
n = 5

[@s1ng23m4n]

4n! - 39 = (2n + 11)^2
Factorial grows much faster than a parabola so we can just brute a few n's, substitute in LHS and check the result number is a perfect square.

[@ShortsOfSyber]

Thanks for sharing!

[@redroach401]

But this is assuming n is an integer