Repeatedly Appending Any Digit to Generate Composite Numbers

TL;DR

This paper explores the existence of infinitely many integers for which appending any number of a specific digit results in a composite number, revealing new insights into number construction and digit appending properties.

Contribution

It proves the existence of infinitely many integers coprime to all digits that remain composite when any digit is appended repeatedly.

Findings

Existence of infinitely many such integers proven.

These integers are coprime to all digits.

Appending any digit repeatedly yields composite numbers.

Abstract

We investigate the problem of finding integers $k$ such that appending any number of copies of the base-ten digit $d$ to $k$ yields a composite number. In particular, we prove that there exist infinitely many integers coprime to all digits such that repeatedly appending {\it any} digit yields a composite number.

Tables

Table 1. Table 1: Divisors of s n subscript 𝑠 𝑛 s_{n} for digit d 𝑑 d using primes 11 and 37 with a seed that satisfies k ≡ 0 ( mod 11 ⋅ 37 ) 𝑘 annotated 0 pmod ⋅ 11 37 k\equiv 0\pmod{11\cdot 37} .

	$n\left(mod6\right)$
digit	$0$	$1$	$2$	$3$	$4$	$5$
$1$	$11$	$?$	11	37	$11$	$?$
3	11	?	11	37	11	$?$
7	11	?	11	37	11	$?$
9	11	?	11	37	11	$?$

Table 2. Table 2: Conditions on k 𝑘 k to guarantee that 7 or 13 divides the number obtained by appending a digit string to k 𝑘 k .

	$n\equiv 1\left(mod6\right)$	$n\equiv 5\left(mod6\right)$	$n\equiv 1\left(mod6\right)$	$n\equiv 5\left(mod6\right)$
digit 1	$k\equiv 2\left(mod7\right)$	$k\equiv 1\left(mod7\right)$	$k\equiv 9\left(mod13\right)$	$k\equiv 1\left(mod13\right)$
digit 3	$k\equiv 6\left(mod7\right)$	$k\equiv 3\left(mod7\right)$	$k\equiv 1\left(mod13\right)$	$k\equiv 3\left(mod13\right)$
digit 7	$k\equiv 0\left(mod7\right)$	$k\equiv 0\left(mod7\right)$	$k\equiv 11\left(mod13\right)$	$k\equiv 7\left(mod13\right)$
digit 9	$k\equiv 4\left(mod7\right)$	$k\equiv 2\left(mod7\right)$	$k\equiv 3\left(mod13\right)$	$k\equiv 9\left(mod13\right)$

Table 3. (a)

digit 1
classes for $n$	classes for $k$
$0\left(mod2\right)$	$0\left(mod11\right)$
$1\left(mod6\right)$	$2\left(mod3\right)$
$3\left(mod6\right)$	$0\left(mod37\right)$
$5\left(mod6\right)$	$1\left(mod13\right)$

Table 4. (a)

digit 1
classes for $n$	classes for $k$
$0\left(mod2\right)$	$0\left(mod11\right)$
$1\left(mod6\right)$	$2\left(mod3\right)$
$3\left(mod6\right)$	$0\left(mod37\right)$
$5\left(mod6\right)$	$1\left(mod13\right)$

Table 5. (b)

digit 3
classes for $n$	classes for $k$
0 (mod 2)	0 (mod 11)
1 (mod 6)	1 (mod 13)
3 (mod 6)	0 (mod 37)
5 (mod 6)	3 (mod 7)

Table 6. (a)

digit 7
classes for $n$		classes for $k$
$0$	$\left(mod2\right)$	$0$	$\left(mod11\right)$
$1$	$\left(mod6\right)$	$2$	$\left(mod3\right)$
$3$	$\left(mod6\right)$	$0$	$\left(mod37\right)$
$5$	$\left(mod30\right)$	$0$	$\left(mod271\right)$
$11$	$\left(mod30\right)$	$28$	$\left(mod41\right)$
$17$	$\left(mod30\right)$	$20$	$\left(mod31\right)$
$23$	$\left(mod30\right)$	$106$	$\left(mod211\right)$
$29$	$\left(mod30\right)$	$7$	$\left(mod241\right)$

Table 7. (a)

digit 7
classes for $n$		classes for $k$
$0$	$\left(mod2\right)$	$0$	$\left(mod11\right)$
$1$	$\left(mod6\right)$	$2$	$\left(mod3\right)$
$3$	$\left(mod6\right)$	$0$	$\left(mod37\right)$
$5$	$\left(mod30\right)$	$0$	$\left(mod271\right)$
$11$	$\left(mod30\right)$	$28$	$\left(mod41\right)$
$17$	$\left(mod30\right)$	$20$	$\left(mod31\right)$
$23$	$\left(mod30\right)$	$106$	$\left(mod211\right)$
$29$	$\left(mod30\right)$	$7$	$\left(mod241\right)$

Table 8. (b)

digit 9
classes for $n$		classes for $k$
$0$	$\left(mod2\right)$	$0$	$\left(mod11\right)$
$1$	$\left(mod8\right)$	$21$	$\left(mod73\right)$
$3$	$\left(mod4\right)$	$9$	$\left(mod101\right)$
$5$	$\left(mod8\right)$	$40$	$\left(mod137\right)$

Equations

$$s_n=k\cdot {10}^n+\frac{d(10^n-1)}{9}.$$

$$111\dots 11_b=\frac{b^{n+1}-1}{b-1}=\left(\frac{m^{n+1}-1}{m-1}\right)\left(\frac{m^{n+1}+1}{m+1}\right)$$

$$9k\cdot 10^n+d(10^n-1),$$

$$k\equiv 9^{-1}d(10^{-n}-1)(\mathrm{mod}p).$$

$$s_n\equiv k+d\frac{10^n-1}{9}\equiv 0(\mathrm{mod}3),$$

$$k\equiv 10^{-(6i+3)}-1={\left(10^{-6}\right)}^i\cdot 10^{-3}-1\equiv 0(\mathrm{mod}37).$$

$$d=1:$$

$$d=3:$$

$$d=7:$$

$$d=9:$$

$$1970728582053685108721(\mathrm{mod}19657858137687083324010)\text{,}$$

Full text

1 Introduction.

Recently L. Jones [5] asked about integers that yield only composites when a sequence of the same base-ten digit is appended to the right. He showed that $37$ is the smallest number with this property when appending the digit $d=1$. For each digit $d\in\{3,7,9\}$, he also found numbers coprime to $d$ that yield only composites upon appending $d$s.

In this paper we find a single integer that works for all digits simultaneously. More precisely,

Theorem**.**

There are infinitely many positive integers $k$ with $\gcd(k,2\cdot 3\cdot 5\cdot 7)=1$, such that for any base-ten digit $d$, appending any number of $d$s to $k$ yields a composite number.

Further, we investigate the question of the smallest numbers that remain composite upon appending strings of a digit for each particular digit. Jones found, for digits $3$, $7$, $9$, respectively, the examples $4070$, $606474$, and $1879711$. It appears that $4070$ is the smallest for $d=3$; for digit $7$ we found $891$, which is almost certainly minimal; and for digit $9$, the likely answer $10175$ was discovered by [14]. In the next section, we explain the obstructions to proving that these three answers are the smallest.

2 Seeds.

Given a digit $d$, let’s use the term seed for a number coprime to $d$ such that appending any number of $d$s on the right yields a composite. The smallest positive integer with this property will be referred to as a minimal seed. Only the cases $d\in\{1,3,7,9\}$ are nontrivial. Jones proved that $37$ is the minimal seed for $d=1$, and he also found the seed $4070$ for digit $3$. For every $k<4070$, except $817$, we have found a value of $n$ such that appending $n$ $3$s yields a prime or, in three cases, a probable prime. For $817$, appending up to $554789$ $3$s yielded only composites. But factorizations show no apparent obstruction to primality, so we conjecture that $4070$ is the minimal seed for digit $3$.

A key concept in this area is the notion of a covering set, a concept introduced by P. Erdős [3]. Such a set corresponds to a finite list of primes such that every member of a given sequence is divisible by one of the primes. Here the sequences are the numbers, which we call $s_{n}$, obtained by appending $n$ copies of a digit $d$ to an initial value $k$; typically the numbers are proved composite by finding a covering set. For example, when $n$ $7$s are appended to $891$, the resulting number is divisible by $11$, $37$, $11$, $3$, $11$, or $13$ according to the mod-$6$ residue of $n$ (starting at [math]).

To see this, observe that $s_{n}$ is given by the formula

[TABLE]

Because $10^{6}\equiv 1$ modulo each of the four primes, easy modular arithmetic shows that $s_{6m+i}\equiv 0\pmod{p}$ for the cases $p=11$, $13$, and $37$, where $i$, depending on $p$, is [math], $2$, $4$, $5$, or $1$. The same is true for $i=3$, the case where $p=3$, because $10^{6m+3}-1$ is divisible by $27$, thus eliminating the denominator of $9$ in these cases. This proves that $891$ is a seed for digit $7$.

When a sequence of primes $(p_{0},p_{1},\dots,p_{r-1})$ divides the corresponding sequence of terms $s_{n}$ for a digit $d$ and seed $k$, we say that the primes form a prime cover for $(k,d)$. For example, $(11,37,11,3,11,13)$ is a prime cover for $(891,7)$.

We have shown that $891$ is a minimal seed for digit $7$, under the assumption that appending $11330$ $7$s to $480$, and $28895$ $7$s to $851$ yields primes. Each of these two large numbers has passed $200$ strong pseudoprime tests. For all other potential seeds below $891$, we have found primes that can be certified using elliptic curve methods with Mathematica or Primo [9]. We used Primo on the largest cases; the largest was $9777\ldots 7$ with $2904$ $7$s, which took $45$ hours.

The digit-$9$ case asks for an integer $k$ such that $(k+1)10^{n}-1$ is always composite; it is thus a variation on the classic Riesel problem [7, 11, 12, 13], which addresses the same question in base $2$. For that classic case, it is known that $509202$ is a seed, meaning that $509203\cdot 2^{n}-1$ is composite for $n\geq 0$. Participants in the Riesel project have also investigated the decimal case, and showed [14] that the expected minimal seed for digit $9$ is $10175$. To see that this is a seed, we again consider the number of appended digits modulo $6$ and find a prime cover: in this case $(11,7,11,37,11,13)$. Of the numbers smaller than $10175$, only two, $4420$ and $7018$, have not been eliminated as seeds. The Riesel project [12, 13] has checked each through the addition of $750000$ $9$s without finding a prime. In this case, primality proving for a probable prime is easy using the Lucas $n+1$ test [2].

Coverings are not the only tool in these investigations, since sometimes factorizations yield all the compositeness that is sought. Consider the situation with digit $1$ but working in base $b=m^{2}$ with $m$ odd. The minimal seed in all such cases is $1$ because, for $n$ appended $1$s to the seed $1$, with $n$ even, the factorization

[TABLE]

yields integer factors, and so the result is composite. When $n$ is odd, the total number of $1$s is even, so compositeness is clear. Similar factorization methods show that the minimal seed for digit $1$ in base $4$ is $5$, for digit $3$ in base $4$ is $8$, and for digit $8$ in base $9$ is $3$.

3 A pandigital seed.

It is not hard to find an integer that remains composite when any sequence of the form $ddd\ldots d$ is appended on the right, where $d$ is any decimal digit. We leave it as an exercise to show that $6930$ does the job; only the case $d=1$ requires a prime cover, and the one used in $\S 2$ for $891$ — $(11,37,11,3,11,13)$ — works. Some prime searching shows that $6930$ is the smallest such example (the most difficult candidate to eliminate was $6069$; $1525$ $1$s yielded a prime).

A more natural problem in our context is to consider only the digits $1$, $3$, $7$, $9$ and ask for an integer $k$ that is a seed for each of these four digits (thus $k$ is coprime to 3 and 7). We call such a positive integer $k$ a pandigital seed.

For a prime $p$ coprime to $10$, we use the term period of $p$ to mean the smallest positive integer $r$ so that, for all $n$, $s_{n+r}\equiv s_{n}\pmod{p}$. The period of $3$ is $3$, while for other primes it is simply the order of $10$ modulo $p$. If the period of a prime $p$ is small, then $p$ may divide a large proportion of the terms of the sequence $s_{n}$. In particular, if the period is $r$, then either every $r$th term of $\{s_{n}\}$ is divisible by $p$ or no terms of the sequence are divisible by $p$.

Theorem**.**

A pandigital seed exists. An example is $4942768284976776320$.

Proof.

A proof requires only checking that particular covers work, but we outline the method by which the large seed and corresponding prime covers were found. We find, for each digit, a prime cover so that the congruence conditions on $k$ arising from the four covers do not contradict each other. This method of coherent prime covers was used in [1, 4, 8] to find infinitely many values $k$ so that both $k2^{n}+1$ and $k2^{n}-1$ are composite for all $n$, and solve related problems. To find such covers, we first need to analyze the condition that a term in the sequence $\{s_{n}\}$ is divisible by a given prime $p$.

If we assume that $p\notin\{2,3,5\}$, then $s_{n}\equiv 0\pmod{p}$ if and only if $p$ divides

[TABLE]

which is equivalent to

[TABLE]

If $p=3$, then we instead have the condition

[TABLE]

which, because $(10^{n}-1)/9\equiv n\pmod{3}$, reduces to $k\equiv 2dn\pmod{3}$. It is useful to observe that when $n$ is even then $10^{n}\equiv 1\pmod{11}$, so that in this case $s_{n}$ is congruent modulo 11 to the seed itself. Therefore, the condition $k\equiv 0\pmod{11}$ makes $11$ a factor of $s_{n}$ whenever $n$ is even. Hence we may focus on forcing composites for odd values of $n$.

Since the period of $p=37$ is $3$, we consider this prime next. When the number of appended digits is $n=6i+3$, equation 1 gives

[TABLE]

Application of 1 to other values of $n$ shows that $37$ divides $s_{n}$ for $n\equiv 0,1,2\pmod{3}$ provided $k\equiv 0,11d,10d\pmod{37}$, respectively. If $k\equiv 0\pmod{37}$, then $37$ may be used as a prime divisor no matter which digit is appended. Therefore we can assume $k\equiv 0\pmod{37}$, and so we have that $s_{n}$ is divisible by $11$ when $n\equiv 0$, $2$, or $4\pmod{6}$ and by $37$ when $n\equiv 0$ or $3\pmod{6}$. This leaves only the eight cases $n\equiv 1$ or $5\pmod{6}$ with digits $1$, $3$, $7$, and $9$ to be taken care of by other primes, as shown in Table 1.

To find divisors of $s_{n}$ for $n\equiv 1$ or $5\pmod{6}$, we note that the primes $7$ and $13$ have period $6$. Solving congruence 1 leads to the conditions listed in Table 2. These show that if $k\equiv 2\pmod{7}$, then two of the eight cases are divisible by $7$: the digit $1$ with $n\equiv 1\pmod{6}$ and digit $9$ with $n\equiv 5\pmod{6}$ cases. Similarly, any of $k\equiv 1$, $3$, or $9\pmod{13}$ provides divisibility for two of the cases. Each of these cases is then combined with a set of additional primes that contains $3$, $101$, $41$, $271$, $73$, and $137$, all of which have period $8$ or less. Finally, a computer search found a list of primes that handles all cases.

The smallest value of $k$ found so far uses the primes $3$, $7$, $11$, $13$, $31$, $37$, $41$, $73$, $101$, $137$, $211$, $241$, and $271$. The cover-lengths for the four digit-cases are $6$, $6$, $30$, and $8$, respectively. The prime covers for the four digits are as follows,

[TABLE]

Tables 3 and 4 show the correspondence between the values of $n$ and $k$ for each digit. For example, when we are appending $n$ $7$s to $k$ where $n\equiv 11\pmod{30}$, we see that $41$ divides $s_{n}$ whenever $k\equiv 28\pmod{41}$.

We apply the Chinese Remainder Theorem to all of the conditions on $k$ in Tables 3 and 4 to find the pandigital seed $k=4942768284976776320$.∎

Because $k$ is not divisible by $3$ or $7$, we can add $k\equiv 1\pmod{10}$ to the conditions used in the proof, which then gives us

[TABLE]

a value of $k$ that satisfies the theorem as stated in $\S 1$. This yields infinitely many such values.

4 Open problems.

We conclude with some unsolved problems.

Find a number of $3$s which can be appended to $817$ to obtain a probable prime, thus completing the proof, modulo probable primes, that $4070$ is minimal for the digit $3$. 2. 2.

Find a number of $9$s which can be appended to $4420$ or $7018$ to produce a prime. 3. 3.

Certify primality of $480$ with $11330$ $7$s appended and $851$ with $28895$ $7$s. Doing so would complete the digit-$7$ case. 4. 4.

Data for all bases up to $10$ can be found at [10]. Similar problems exist for these bases. 5. 5.

Find a base-ten pandigital seed that is smaller than $4942768284976776320$. 6. 6.

Investigate for various bases the situation where the appended digits come from a fixed sequence, as was done by Jones and White [6] for base ten.

The authors thank the referee for many valuable comments that helped improve the exposition.