Three missing digits in 42 factorial

When calculating $$42!=42\cdot41\cdot...\cdot2\cdot1$$ the result is $$42! = 140500611775287989854x14260624y511569936384z00000000$$
However three digits have been replaced by $$x$$, $$y$$, and $$z$$.

Can you find $$x$$, $$y$$ and $$z$$ without recalculating $$42!$$ ?

• Great puzzle! Encouraging some brain work rather than calculator work, and it's not hard with a few simple tricks. Nov 6 '21 at 20:37
• @ Rand al'Thor This problem is well and truly a cookie cutter problem for all people who dabble in a little bit of number theory or have experience in contest math.
– user58783
Nov 6 '21 at 21:07

Certainly I can find $$z$$ easily:

by calculating the number of powers of $$2$$ and of $$5$$ present in the factorisation of $$42!$$.

This can be done easily using Legendre's formula: the number of powers of a prime $$p$$ in $$N!$$ is exactly $$\sum_{k=1}^{\infty}\lfloor\frac{N}{p^k}\rfloor$$ (this is a finite sum since the summands are all zero for large enough $$k$$). With $$N=42$$ and $$p=2$$, this number is $$21+10+5+2+1=39$$; with $$N=42$$ and $$p=5$$, it is $$8+1=9$$.

So $$42!$$ is a multiple of $$10^9$$, which means $$z=0$$.

To find $$x$$ and $$y$$, it's simply a matter of

solving two simultaneous equations, by considering the fact that $$42!$$ is a multiple of both $$9$$ and $$11$$ and what this implies in terms of digit sums and alternating sums.

Since it's a multiple of $$9$$, the digit sum ($$1+4+0+5+0+0+6+1+1+\cdots$$) must be $$0$$ modulo $$9$$, which means $$x+y+2$$ must be $$0$$ mod $$9$$, so $$x+y$$ must be either $$7$$ or $$16$$ (it can't be higher since they're single digits).

Since it's a multiple of $$11$$, the alternating digit sum ($$1-4+0-5+0-0+6-1+1+\cdots$$) must be $$0$$ modulo $$11$$, which means $$-x+y-1\equiv0$$ modulo $$11$$, so $$x=y-1$$.

Therefore, the final answer is $$x=3$$ and $$y=4$$.

• I knew I'd get beaten to the punch on a question like this, take my +1. Nov 6 '21 at 20:38
• @AxiomaticSystem +1 to you too :-) Nov 6 '21 at 21:20
• That name is Legendre's formula. Nov 7 '21 at 20:29
• @aschepler Thanks! I felt like it might involve one of the L-guys (Legendre or Lagrange), but I couldn't bring it to mind. Nov 8 '21 at 5:12

First of all,

$$z = 0$$, because $$42!$$ has nine factors of five - eight from multiples of five, and a ninth because $$25$$ counts double - and many more factors of two, meaning it's divisible by $$10^9$$.
Then, realize that $$42!$$ is divisible by $$99$$. We can use a version of casting out [ninety-]nines to determine what x and y should be, and we get $$10y+x+848 \equiv 0 \text{ mod } 99$$, so $$10y+x = 43$$ or $$x=3,y=4$$.