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For natural numbers $n\ge m$, let $n\underset{m \text{ times}}{\underbrace{!!!\dots!}}=n(n-m)(n-2m)(n-3m)\dots$ where all factors are natural numbers (we exclude $0$ and negative factors).

Question:

What is the units digit of $1!+2!+3!+4!!+5!!+\dots+k\underset{\left \lfloor \sqrt{k} \right \rfloor \text{ times}}{\underbrace{!!!\dots!}}+\dots+1992\underset{44 \text{ times}}{\underbrace{!!!\dots!}}$? ($\left \lfloor \cdot \right \rfloor$ denotes the floor funtion).


My Attempt (Is wrong as Peter Foreman commented below):

Consider the first $9$ terms:

$1!+2!+3!+4!!+5!!+6!!+7!!+8!!+9!!!$

$=1+2+6+8+15+48+105+384+162=731$

Each of the remaining terms includes at least on factor that ends with $0$. Therefore, the each term ends with $0$.

Hence the units digit of the given expression is equal to the units digit of the sum of the first $9$ terms. So, $1$ is the units digit of the given expression.


Peter Foreman said: "$17!!!!=9945$". This showed me that my attempt is wrong. Thanks Peter Foreman.


Any help would be appreciated. THANKS.

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  • $\begingroup$ @PeterForeman ,, yeah , you are right that I am wrong. $\endgroup$ – Hussain-Alqatari Aug 22 at 8:25
  • $\begingroup$ It's also going to be non-zero whenever the number of ! is a multiple of 5 and n isn't. $\endgroup$ – Matthew Daly Aug 22 at 8:38
  • $\begingroup$ @MatthewDaly Right. So can not we solve this problem by ordinary methods? $\endgroup$ – Hussain-Alqatari Aug 22 at 8:42
  • $\begingroup$ I'd write a program that spits out the first hundred unit digits and stare at it for a while. The pattern will be easier to prove when you know what it is. $\endgroup$ – Matthew Daly Aug 22 at 8:51
  • $\begingroup$ It's $1$. (I don't know how to prove it without computation) $\endgroup$ – Vepir Aug 22 at 9:58
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When $k \geq 25$ and the floor part $p$ of $k^{0.5}$ is coprime with $10$, the unit digit of $k’=k! \ldots !$ is $0$ (there is an even number in $k,k-p$ and one divisible by $5$ in $k!!!!!$).

When $k \geq 25$ and $p \wedge 10=2$, there is going to be a number divisible by $5$ in $k,k-p,k-2p,k-3p,k-4p$, and $k’$ is divisible by $5$ and congruent to $k$ mod $2$, so the unit digit of $k’$ is $5$ if $k$ is odd and $0$ if $k$ is even.

When $k \geq 25$ and $p \wedge 10=5$, then $k(k-p)$ is even, so $k’$ is even. The congruence mod $5$ is trickier: $k’$ is congruent to $k^r$ mod $5$, where $r$ is the number of factors in he product (ie $r-1$ is the floor part of $(k-1)/p$, so $r$ is either $p$, $p+1$ or $p+2$).

When $10 | p$, as above, $k’$ is congruent to $k^r$ mod $10$.

Note that everything depends only on the unit digit of $k$, $p$ and $r$: when $p$ is set, and $p \wedge 10=2$, the sum of any four $k’$ corresponding to consecutive $k \geq 25$ vanishes mod $10$.

When $p,r$ are set and $p \wedge 10=5$, the sum of any five $k’$ corresponding to consecutive $k \geq 25$ is always divisible by $10$.

When $p,r$ are set and $p \wedge 10=10$, the sum of any ten $k’$ corresponding to consecutive $k \geq 100$ is congruent to $3$ mod $10$ if $4|r$ and $5$ mod $10$ otherwise.

So all we need now is time for processing all integers from $1$ to $1992$.

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  • $\begingroup$ Why the downvotes? $\endgroup$ – Mindlack Aug 22 at 9:33
  • $\begingroup$ Can you please clarify what do you mean by the notations $k'$, $p \wedge 10=10$,...? Thank you very much. $\endgroup$ – Hussain-Alqatari Aug 24 at 10:52
  • $\begingroup$ $\wedge$ is the gcd. $p$ is the floor part of $\sqrt{k}$. $r-1$ is the floor part of $(k-1)/p$. $k’=k(k-p)(k-2p)\ldots(k-(r-1)p)$. $\endgroup$ – Mindlack Aug 24 at 11:08
  • $\begingroup$ So this problem is not solvable unless we find the units digit of all $1992$ terms, right? Is not there any method to solve it? $\endgroup$ – Hussain-Alqatari Aug 24 at 11:19
  • $\begingroup$ The answer outlines many possible shortcuts, but why would you expect there to be an instant solution without any calculation? $\endgroup$ – Mindlack Aug 24 at 13:34

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