Two Irrational Numbers That Give the Last Non-Zero Digits of
n! and n
n
.
Gregory P. Dresden
Washington & Lee University
Lexington, VA 24450
Authors Note: This is a slightly revised version of the article that appeared in print in
Mathematics Magazine in 2001. The original proof of Theorem 2 was incorrect; Ive xed that
mistake here. My thanks to Antonio M. Oller-Marcen and Jose Mara Grau for pointing out the
error. Also, a sequel to this paper appeared in 2008.
We begin by looking at the pattern formed from the last (i.e. unit) digit of n
n
.
Since 1
1
= 1, 2
2
= 4, 3
3
= 27, 4
4
= 256, and so on, we can easily calculate the
rst few numbers in our pattern to be 1, 4, 7, 6, 5, 6, 3, 6 . . .. We construct a decimal
number N = 0.d
1
d
2
d
3
. . . d
n
. . . such that the n
th
digit d
n
of N is the last (i.e. unit)
digit of n
n
; that is, N = 0.14765636 . . .. In a recent paper [1], R. Euler and J. Sadek
showed that this N is a rational number with a period of twenty digits:
N = 0.14765636901636567490.
This is a nice result, and we might well wonder if it can be extended. Indeed, Euler
and Sadek in [1] recommend looking at the last non-zero digit of n! (If we just looked
at the last digit of n!, we would get a very dull pattern of all 0s, as n! ends in 0 for
every n 5.)
With this is mind, lets dene lnzd(A) to be the last nonzero digit of the positive
integer A; it is easy to see that lnzd(A) = A/10
i
mod 10, where 10
i
is the largest
power of 10 that divides A. We wish to investigate not only the pattern formed by
lnzd(n!), but also the pattern formed by lnzd(n
n
). In accordance with [1], we dene
the factorial number F = 0.d
1
d
2
d
3
. . . d
n
. . . to be the innite decimal such that
each digit d
n
= lnzd(n!), and we dene the power number P = 0.d
1
d
2
d
3
. . . d
n
. . .
to be the innite decimal such that each digit d
n
= lnzd(n
n
), and we ask whether
these numbers are rational (i.e. are eventually-repeating decimals) or irrational.
Although the title of this article gives away the secret, wed like to point out
that at rst glance, our factorial number F exhibits a suprisingly high degree of
Dresden 2
regularity, and a fascinating pattern occurs. The rst few digits of F are easy to
calculate:
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880 . . .
10! = 3628800
11! = 39916800
12! = 479001600
13! = 6227020800
14! = 87178291200
...
Reading the underlined digits, we have:
F = 0.1264 22428 88682 . . .
Continuing along this path, we have (to forty-nine decimal places):
F = 0.1264 22428 88682 88682 44846 44846 88682 22428 22428 66264 . . .
It is not hard to show that (after the rst four digits) F breaks up into ve-digit
blocks of the form x x 2x x 4x, where x {2, 4, 6, 8}, and the 2x and 4x are taken
mod 10. Furthermore, if we represent these ve-digit blocks by symbols (
2 for 22428,
4 for 44846,
6 for 66264,
8 for 88682, and
1 for the initial four-digit block of 1264),
we have:
F = 0.
1
2
8
8
4
4
8
2
2
6 . . .
Grouping these symbols into blocks of ve and then performing more calculations
(with the aid of Maple) give us F to 249 decimal places:
F = 0.
4
4
6
2
8
4
6
4
6
8
2
2
8
6
4
2
8
2
8 . . .
The reader will notice additional patterns in these blocks of ve symbols (twenty-ve
digits). In fact, such patterns exist for any block of size 5
i
. However, a pattern is
dierent from a period, and doesnt imply that our decimal F is rational. Consider
the classic example of 0.1 01 001 0001 00001 000001 . . ., which has an obvious pattern
but is obviously irrational. It turns out that our decimal F is also irrational, as the
following theorem indicates:
Theorem 1. Let F = 0.d
1
d
2
d
3
. . . d
n
. . . be the innite decimal such that each
digit d
n
= lnzd(n!). Then, F is irrational.
Dresden 3
As for our power number P, it too might seem to be rational at rst glance.
P is only slightly dierent from Euler and Sadeks rational number N, as seen here:
N = 0.14765 63690 16365 67490 14765 63690 16365 67490 . . .
and P = 0.14765 63691 16365 67496 14765 63699 16365 67496 . . .
(Again, calculations were performed by Maple.) Despite this striking similarity
between P and N, it turns out that P, like F, is irrational:
Theorem 2. Let P = 0.d
1
d
2
d
3
. . . d
n
. . . be the innite decimal such that each
digit d
n
= lnzd(n
n
). Then, P is irrational.
Before we begin with the (slightly technical) proofs, let us pause and see if we
can get a feel for why these two numbers must be irrational. There is no doubt
that both F and P are highly regular, in that both exhibit a lot of repetition.
The problem is that there are too many patterns in the digits, acting on dierent
scales. Taking P, for example, we note that there is an obvious pattern (as shown
by Euler and Sadek in [1]) repeating every 20 digits with 1
1
, 2
2
, 3
3
, . . . , 9
9
and
11
11
, 12
12
, . . . , 19
19
, but this is broken by a similar pattern for 10
10
, 20
20
, . . . , 90
90
and 110
110
. . . 190
190
, which repeats every 200 digits. This, in turn, is broken by
another pattern repeating every 2000, and so on. A similar behaviour is found for
F, but in blocks of 5, 25, 125, and so on, as mentioned above. So, in vague terms,
there are always new patterns starting up in the digits of P and of F, and this is
what makes them irrational.
Are there some simple observations that we can make about P and F which
might help us to prove our theorems? To start with, we might notice that every
digit of F (except for the rst one) is even. Can we prove this? Yes, and without
much diculty:
Lemma 1. For n 2, then lnzd(n!) is in {2, 4, 6, 8}.
Proof: The lemma is certainly true for n = 2, 3, 4. For n 5, we note that the prime
factorization of n! contains more 2s than 5s, and thus even after taking out all the
10s in n!, the quotient will still be even. To be precise, the number of 5s in n! (and
thus the number of trailing zeros in its base-10 representation) is e
5
=
i=1
n/5
i
,
which is strictly less than the number of 2s, e
2
=
i=1
n/2
i
(here, [] represents
the greatest integer function). Hence, n!/10
e
5
is an even integer not divisible by 10,
and so lnzd(n!) = n!/10
e
5
mod 10, which must be in {2, 4, 6, 8}. This completes the
Dresden 4
proof.
Another helpful observation is to note that the lnzd function appears to be
multiplicative. For example,
lnzd(12) lnzd(53) = 2 3 = 6,
and lnzd(12 53) = lnzd(636) = 6.
However, we note that at times this rule fails:
lnzd(15) lnzd(22) = 5 2 = 10,
yet lnzd(15 22) = lnzd(330) = 3.
So, we can only prove a limited form of multiplicativity, but it is useful none the
less:
Lemma 2. Suppose a, b are integers such that lnzd(a) = 5, lnzd(b) = 5. Then,
lnzd is multiplicative; that is, lnzd(a b) = lnzd(a) lnzd(b) mod 10.
Proof: Let x
denote the integer x without its trailing zeros; that is, x
= x/10
i
,
where 10
i
is the largest power of 10 dividing x. (Note that lnzd(x) = x
mod 10.)
By hypothesis, a
and b
are both = 0 mod 5, and so (a b)
= 0 mod 5 and so
(a b)
= a
. Thus,
lnzd(a b) = lnzd((a b)
) = lnzd(a
) = a
mod 10
= (a
mod 10) (b
mod 10) = lnzd(a
) lnzd(b
) = lnzd(a) lnzd(b).
This completes the proof.
We are now ready to supply the proof of Theorem 1, in which we show that F
is irrational. The proof is a little technical, but it relies rst on assuming that F
has a repeating decimal expansion, then on choosing an appropriate multiple of the
period
0
and choosing an appropriate digit d, in order to arrive at a contradiction.
Proof of Theorem 1: We argue by contradiction. Suppose F is rational. Then F
is eventually periodic; let
0
be the period (i.e. for every n suciently large, then
d
n
= d
n+
0
). Write
0
= 5
i
K such that 5| K (we acknowledge that K could be 1)
and let = 2
i
0
= 10
i
K. Then, lnzd() = lnzd(K), and since 5| K, then 10| K
and so lnzd(K) = K mod 10. Note also that lnzd(2) = 2K mod 10. Choose M
suciently large so that both of the following are true: lnzd(10
M
+) = lnzd() (this
can easily be done by demanding that 10
M
> ), and for all n M, then d
n
= d
n+
0
,
which of course would then equal d
n+
. Finally, let d = lnzd((10
M
1)!). By Lemma
1, d {2, 4, 6, 8}, and since 10
M
! = (10
M
1)! 10
M
, then d also equals lnzd(10
M
!).
Dresden 5
Since is a multiple of the period
0
, then if we let A = 10
M
1 + and
B = 10
M
1 + 2, then:
d = lnzd((10
M
1)!) = lnzd(A!) = lnzd(B!)
and d = lnzd(10
M
!) = lnzd((A+ 1)!) = lnzd((B + 1)!)
Lets now look at the last two terms in the above equation; it is here we will nd
our contradiction. Note that since lnzd(A!) = d, then lnzd(A!) = 5. Also, since
lnzd(A+1) = lnzd(10
M
+) = lnzd() = K mod 10, we know that lnzd(A+1) = 5.
Thus, we can apply Lemma 2 to lnzd(A! (A+ 1)) to get:
d = lnzd((A+ 1)!) = lnzd(A!) lnzd(A+ 1) = d K mod 10.
Likewise, working with B, we nd:
d = lnzd((B + 1)!) = lnzd(B!) lnzd(B + 1) = d 2K mod 10.
Combining these two equations, we get:
d = dK mod 10 d = 2dK mod 10,
and this becomes d(1 K) = 0 = d(1 2K) mod 10. Since d is even, this implies
that 1 K = 0 mod 5 and 1 2K = 0 mod 5, which is a contradiction. Thus, there
can be no period
0
and so F is irrational. This completes the proof.
We now turn our attention to the power number P derived from the last non-
zero digits of n
n
. This part was more dicult, but a major step was the discovery
that the sequence lnzd(100
100
), lnzd(200
200
), lnzd(300
300
) . . . was the same as the
sequence lnzd(100
4
), lnzd(200
4
), lnzd(300
4
) . . .. This relies not only on the fact that
4|100 but also on the fact that a
b
= a
b+4
mod 10 for b > 0, as used in the following
lemma:
Lemma 3. Suppose 100 | x. Then, lnzd(x
x
) = (lnzd x)
4
mod 10.
Proof: As in Lemma 2, let x
denote the integer x without its trailing zeros; that is,
x
= x/10
i
, where 10
i
is the largest power of 10 dividing x. Now,
lnzd(x
x
) = lnzd((10
i
x
)
10
i
x
)
= lnzd((10
i10
i
x
)(x
)
10
i
x
)
= lnzd((x
)
10
i
x
).
Since 10| x
, then 10| (x
)
10
i
x
, and so:
lnzd(x
x
) = (x
)
10
i
x
mod 10.
Dresden 6
Since 100 | x, then 4 | 10
i
x
, and since (x
)
n
= (x
)
n+4
mod 10 for every positive
n, we have:
lnzd(x
x
) = (x
)
4
mod 10
= (lnzd x)
4
mod 10.
This completes the proof.
With Lemma 3 at our disposal, the proof of Theorem 2 is now fairly easy.
Proof of Theorem 2: Again, we argue by contradiction. Suppose P is rational. Let
0
be the period, and choose j suciently large such that 10
j
> 100(
0
+ 1)! and
such that lnzd((10
j
+ n
0
)
10
j
+n
0
) = lnzd((10
j
)
10
j
) for every positive n. Choosing
n = 100(
0
+ 1)(
0
1)!, we get:
lnzd((10
j
+ 100(
0
+ 1)!)
10
j
+100(
0
+1)!
) = lnzd((10
j
)
10
j
).
We reduce the left side of the above equation by Lemma 3 and the right side is
obviously 1, so we have:
(lnzd(10
j
+ 100(
0
+ 1)!))
4
mod 10 = 1,
but since 10
j
> 100(
0
+1)! and lnzd(100(
0
+1)!) = lnzd((
0
+1)!), we can rewrite
the above equation as:
(lnzd (
0
+ 1)!)
4
mod 10 = 1.
Note that by Lemma 1, the only values of lnzd((
0
+ 1)!) are 2, 4, 6, and 8, and
raising these to the fourth power mod 10 gives us:
6 = 1,
which is a contradiction. Thus, P is irrational. This completes the proof.
We close by asking the obvious (and very dicult) question: Are F and P
algebraic or transcendental? I suspect the latter, but it is only a hunch, and I hope
some curious reader will continue along this interesting line of study.
References
[1] R. Euler and J. Sadek, A number that gives the unit digit of n
n
, Journal of
Recreational Mathematics, 29 (1998) No. 3, pp. 2034.
