Don't mess up with those factorials

Number Theory Level 2

( ( 3 ! ) ! ) ! 3 ! \large \dfrac{((3!)!)!}{3!}

If we know that the number above can be written as k × n ! k \times n! where k k and n n are positive integers with n n is as large as possible, find k + n . k + n.


The answer is 839.

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Parth Lohomi by Parth Lohomi , 20, India

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1 solution

Parth Lohomi
May 9, 2015

( ( 3 ! ) ! ) ! 3 ! = ( 6 ! ) ! 3 ! = 720 ! 3 ! = 720 ! 6 = 720 ⋅ 719 ! 6 = 120 ⋅ 719 ! = k ⋅ n ! \frac{((3!)!)!}{3!} = \frac{(6!)!}{3!} = \frac{720!}{3!} = \frac{720!}{6} = \frac{720 \cdot 719!}{6} = 120 \cdot 719! = k \cdot n!

We certainly can't make n n any larger if k k is going to stay an integer, so the answer is k + n = 120 + 719 = 839 k + n = 120 + 719 = \boxed{839} .

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Nice use of the recurrence relation of factorials. Bonus question: If we implement the multifactorial notation, what would the answer be if I change the fraction to ( ( 3 ! ) ! ! ) ! ! ! 3 ! ! ! ! \dfrac{((3!)!!)!!!}{3!!!!} ?

Nicely done ⌣ ¨ \ddot \smile

Sravanth C. - 6 years, 1 month ago

Using this definition of the multifactorial (it may not be the one that you are thinking of): http://mathworld.wolfram.com/Multifactorial.html,

n ! = n ( n − 1 ) ( n − 2 ) . . . ( 1 ) n! = n(n-1)(n-2)...(1) ,

n ! ! = n ( n − 2 ) ( n − 4 ) . . . n!! = n(n-2)(n-4)... up to the last positive term,

n ! ! ! = n ( n − 3 ) ( n − 6 ) . . . n!!! = n(n-3)(n-6)... up to the last positive term, etc.

Therefore, 3 ! ! ! ! = 3 3!!!!=3 , and ( 3 ! ) ! ! ) ! ! ! = 900657498850357248000 (3!)!!)!!!=900657498850357248000 (you can check it for yourself by trying 48(45)(42)(39)(36)...(3).

900657498850357248000 3 = 300219116283452416000 \frac{900657498850357248000}{3}=300219116283452416000 , of which 16 ! 16! is a factor (this can be seen by rearranging the prime factors), meaning that the answer is 16 + 14348907 = 14348922 16 + 14348907 = \boxed{14348922} is the looked for answer.

Alex Delhumeau - 6 years ago

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3 ! = 6 ⇒ ( 3 ! ) ! ! = 6 ! ! = 6 × 4 × 2 = 48 3! = 6 \Rightarrow (3!)!! = 6!! = 6 \times 4 \times 2 = 48

⇒ ( ( 3 ! ) ! ! ) ! ! ! = 48 ! = 48 × 46 × 44 × … × 2 = 2 24 ⋅ 24 ! \Rightarrow ((3!)!!)!!! = 48! = 48 \times 46 \times 44 \times \ldots \times 2 = 2^{24} \cdot 24! . you don't need to find the exact numerical value of this large number.

Highest power of 3 that divides 24 ! 24! is ⌊ 24 3 ⌋ + ⌊ 24 3 2 ⌋ = 10 \left\lfloor \frac {24}{3} \right \rfloor + \left\lfloor \frac {24}{3^2} \right \rfloor = 10

The highest power of 3 that divides the numerator is 10. Answer is simply 10 − 1 = 9 10-1=9 .

Pi Han Goh - 6 years ago

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I think you have a mistake in your second line. I don't think you can simplify ( ( 3 ! ) ! ! ) ! ! ! ((3!)!!)!!! to 48 ! ! 48!! like you did. Is it not 48 ! ! ! = 900657498850357248000 48!!! = 900657498850357248000 ?

Alex Delhumeau - 6 years ago

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@Alex Delhumeau – Oh sorry, I didn't see the triple factorial sign.

I think the (bonus) question still asked up to find k + n k+n .

48 ! ! ! = 48 × 45 × 42 × … × 3 = 3 2 4 ⋅ 24 ! 48!!! = 48 \times 45 \times 42 \times \ldots \times 3 = 3^24 \cdot 24!

k = 24 − 1 = 23 , n = 24 k = 24-1=23, n=24 .

Pi Han Goh - 6 years ago

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