There are 77 Minimal Primes in Base 10 if Single-Digit-Primes Subsequences are Allowed

Theorem
There are $77$ minimal primes in base $10$ if we allow single-digit-primes subsequences (but not allow multidigit-primes subsequences):


 * $11$, $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, $53$, $59$, $61$, $67$, $71$, $73$, $79$, $83$, $89$, $97$, $227$, $251$, $257$, $277$, $281$, $349$, $409$, $449$, $499$, $521$, $557$, $577$, $587$, $727$, $757$, $787$, $821$, $827$, $857$, $877$, $881$, $887$, $991$, $2087$, $2221$, $5051$, $5081$, $5501$, $5581$, $5801$, $5851$, $6469$, $6949$, $8501$, $9001$, $9049$, $9221$, $9551$, $9649$, $9851$, $9949$, $20 \, 021$, $20 \, 201$, $50 \, 207$, $60 \, 649$, $80 \, 051$, $666 \, 649$, $946 \, 669$, $5 \, 200 \, 007$, $22 \, 000 \, 001$, $60 \, 000 \, 049$, $66 \, 000 \, 049$, $66 \, 600 \, 049$, $80 \, 555 \, 551$, $555 \, 555 \, 555 \, 551$, $5 \, 000 \, 000 \, 000 \, 000 \, 000 \, 000 \, 000 \, 000 \, 000 \, 027$

Proof
In this proof, $x$ ◁ $y$ means $x$ is a subsequence of $y$.

Assume $p$ is a prime > $10$, and the last digit of $p$ must lie in $\set {1,3,7,9}$ (if the last digit of $p$ is $0$, $2$, $4$, $6$, or $8$, than $p$ is divisible by $2$ (Divisibility by 2), and if the last digit of $p$ is $0$ or $5$, than $p$ is divisible by $5$ (Divisibility by 5)).

Case 1: $p$ ends with $1$
In this case we can write $p$ = $x1$. If $x$ contains $1$, $3$, $4$, $6$, or $7$, then (respectively) $11$ ◁ $p$, $31$ ◁ $p$, $41$ ◁ $p$, $61$ ◁ $p$, or $71$ ◁ $p$. Hence we may assume all digits of $x$ are $0$, $2$, $5$, $8$, or $9$.

Case 1.1: $p$ begins with $2$
In this case we can write $p$ = $2y1$. If $5$ ◁ $y$, then $251$ ◁ $p$. If $8$ ◁ $y$, then $281$ ◁ $p$. If $9$ ◁ $y$, then $29$ ◁ $p$. Hence we may assume all digits of $y$ are $0$ or $2$.

If $22$ ◁ $y$, then $2221$ ◁ $p$. Hence we may assume $y$ contains zero or one $2$'s.

If $y$ contains no $2$'s, then $p \in 2\{0\}1$. But then, since the sum of the digits of $p$ is $3$, $p$ is divisible by $3$ (Divisibility by 3), so $p$ cannot be prime.

If $y$ contains exactly one $2$, then we can write $p$ = $2z2w1$, where $z,w \in \{0\}$. If $0$ ◁ $z$ and $0$ ◁ $w$, then $20201$ ◁ $p$. Hence we may assume either $z$ or $w$ is empty.

If $z$ is empty, then $p \in 22\{0\}1$, and the smallest prime $p \in 22\{0\}1$ is $22000001$.

If $w$ is empty, then $p \in 2\{0\}21$, and the smallest prime $p \in 2\{0\}21$ is $20021$.

Case 1.2: $p$ begins with $5$
In this case we can write p = 5y1. If 2 ◁ y, then 521 ◁ p. If 9 ◁ y, then 59 ◁ p. Hence we may assume all digits of y are 0, 5, or 8.

If 05 ◁ y, then 5051 ◁ p. If 08 ◁ y, then 5081 ◁ p. If 50 ◁ y, then 5501 ◁ p. If 58 ◁ y, then 5581 ◁ p. If 80 ◁ y, then 5801 ◁ p. If 85 ◁ y, then 5851 ◁ p. Hence we may assume y ∈ {0} ∪ {5} ∪ {8}.

If y ∈ {0}, then p ∈ 5{0}1. But then, since the sum of the digits of p is 6, p is divisible by 3 (Divisibility by 3), so p cannot be prime.

If y ∈ {5}, then p ∈ 5{5}1, and the smallest prime p ∈ 5{5}1 is 555555555551.

If y ∈ {8}, since if 88 ◁ y, then 881 ◁ p, hence we may assume y ∈ {𝜆,8}, and thus p ∈ {51,581}, but 51 and 581 are both composite.

Case 1.3: p begins with 8.

In this case we can write p = 8y1. If 2 ◁ y, then 821 ◁ p. If 8 ◁ y, then 881 ◁ p. If 9 ◁ y, then 89 ◁ p. Hence we may assume all digits of y are 0 or 5.

If 50 ◁ y, then 8501 ◁ p. Hence we may assume y ∈ {0}{5}.

If 005 ◁ y, then 80051 ◁ p. Hence we may assume y ∈ {0} ∪ {5} ∪ 0{5}.

If y ∈ {0}, then p ∈ 8{0}1. But then, since the sum of the digits of p is 9, p is divisible by 3 (Divisibility by 3), so p cannot be prime.

If y ∈ {5}, since if 55555555555 ◁ y, then 555555555551 ◁ p, hence we may assume y ∈ {𝜆, 5, 55, 555, 5555, 55555, 555555, 5555555, 55555555, 555555555, 5555555555}, and thus p ∈ {81, 851, 8551, 85551, 855551, 8555551, 85555551, 855555551, 8555555551, 85555555551, 855555555551}, but all of these numbers are composite.

If y ∈ 0{5}, since if 55555555555 ◁ y, then 555555555551 ◁ p, hence we may assume y ∈ {0, 05, 055, 0555, 05555, 055555, 0555555, 05555555, 055555555, 0555555555, 05555555555}, and thus p ∈ {801, 8051, 80551, 805551, 8055551, 80555551, 805555551, 8055555551, 80555555551, 805555555551, 8055555555551}, and of these numbers only 80555551 and 8055555551 are primes, but 80555551 ◁ 8055555551, thus only 80555551 is minimal prime.

Case 1.4: p begins with 9.

In this case we can write p = 9y1. If 9 ◁ y, then 991 ◁ p. Hence we may assume all digits of y are 0, 2, 5, or 8.

If 00 ◁ y, then 9001 ◁ p. If 22 ◁ y, then 9221 ◁ p. If 55 ◁ y, then 9551 ◁ p. If 88 ◁ y, then 881 ◁ p. Hence we may assume y contains at most one 0, at most one 2, at most one 5, and at most one 8.

If y only contains at most one 0 and does not contain any of {2,5,8}, then y ∈ {𝜆,0}, and thus p ∈ {91,901}, but 91 and 901 are both composite. If y only contains at most one 0 and only one of {2,5,8}, then the sum of the digits of p is divisible by 3, p is divisible by 3 (Divisibility by 3), so p cannot be prime. Hence we may assume y contains at least two of {2,5,8}.

If 25 ◁ y, then 251 ◁ p. If 28 ◁ y, then 281 ◁ p. If 52 ◁ y, then 521 ◁ p. If 82 ◁ y, then 821 ◁ p. Hence we may assume y contains no 2's (since if y contains 2, then y cannot contain either 5's or 8's, which is a contradiction).

If 85 ◁ y, then 9851 ◁ p. Hence we may assume y ∈ {58,580,508,058}, and thus p ∈ {9581,95801,95081,90581}, and of these numbers only 95801 is prime, but 95801 is not minimal prime since 5801 ◁ 95801.

Case 2: p ends with 3.

In this case we can write p = x3. If x contains 1, 2, 4, 5, 7, or 8, then (respectively) 13 ◁ p, 23 ◁ p, 43 ◁ p, 53 ◁ p, 73 ◁ p, or 83 ◁ p. Hence we may assume all digits of x are 0, 3, 6, or 9, and thus all digits of p are 0, 3, 6, or 9. But then, since the digits of p all have a common factor 3, p is divisible by 3 (Numbers with All Digits Have a Common Factor are Divisible by This Factor), so p cannot be prime.

Case 3: p ends with 7.

In this case we can write p = x7. If x contains 1, 3, 4, 6, or 9, then (respectively) 17 ◁ p, 37 ◁ p, 47 ◁ p, 67 ◁ p, or 97 ◁ p. Hence we may assume all digits of x are 0, 2, 5, 7, or 8.

Case 3.1: p begins with 2.

In this case we can write p = 2y7. If 2 ◁ y, then 227 ◁ p. If 5 ◁ y, then 257 ◁ p. If 7 ◁ y, then 277 ◁ p. Hence we may assume all digits of y are 0 or 8.

If 08 ◁ y, then 2087 ◁ p. If 88 ◁ y, then 887 ◁ p. Hence we may assume y ∈ {0} ∪ 8{0}.

If y ∈ {0}, then p ∈ 2{0}7. But then, since the sum of the digits of p is 9, p is divisible by 3 (Divisibility by 3), so p cannot be prime.

If y ∈ 8{0}, then p ∈ 28{0}7. But then p is divisible by 7 (All Numbers of the Form 28000...0007 are Divisible by 7), since for n ≥ 0 we have 7 × 40n1 = 280n7.

Case 3.2: p begins with 5.

In this case we can write p = 5y7. If 5 ◁ y, then 557 ◁ p. If 7 ◁ y, then 577 ◁ p. If 8 ◁ y, then 587 ◁ p. Hence we may assume all digits of y are 0 or 2.

If 22 ◁ y, then 227 ◁ p. Hence we may assume y contains zero or one 2's.

If y contains no 2's, then p ∈ 5{0}7. But then, since the sum of the digits of p is 12, p is divisible by 3 (Divisibility by 3), so p cannot be prime.

If y contains exactly one 2, then we can write p = 5z2w7, where z,w ∈ {0}. If 0 ◁ z and 0 ◁ w, then 50207 ◁ p. Hence we may assume either z or w is empty.

If z is empty, then p ∈ 52{0}7, and the smallest prime p ∈ 52{0}7 is 5200007.

If w is empty, then p ∈ 5{0}27, and the smallest prime p ∈ 5{0}27 is 5000000000000000000000000000027.

Case 3.3: p begins with 7.

In this case we can write p = 7y7. If 2 ◁ y, then 727 ◁ p. If 5 ◁ y, then 757 ◁ p. If 8 ◁ y, then 787 ◁ p. Hence we may assume all digits of y are 0 or 7, and thus all digits of p are 0 or 7. But then, since the digits of p all have a common factor 7, p is divisible by 7 (Numbers with All Digits Have a Common Factor are Divisible by This Factor), so p cannot be prime.

Case 3.4: p begins with 8.

In this case we can write p = 8y7. If 2 ◁ y, then 827 ◁ p. If 5 ◁ y, then 857 ◁ p. If 7 ◁ y, then 877 ◁ p. If 8 ◁ y, then 887 ◁ p. Hence we may assume y ∈ {0}, and thus p ∈ 8{0}7. But then, since the sum of the digits of p is 15, p is divisible by 3 (Divisibility by 3), so p cannot be prime.

Case 4: p ends with 9.

In this case we can write p = x9. If x contains 1, 2, 5, 7, or 8, then (respectively) 19 ◁ p, 29 ◁ p, 59 ◁ p, 79 ◁ p, or 89 ◁ p. Hence we may assume all digits of x are 0, 3, 4, 6, or 9.

If 44 ◁ x, then 449 ◁ p. Hence we may assume x contains zero or one 4's.

If x contains no 4's, then all digits of x are 0, 3, 6, or 9, and thus all digits of p are 0, 3, 6, or 9. But then, since the digits of p all have a common factor 3, p is divisible by 3 (Numbers with All Digits Have a Common Factor are Divisible by This Factor), so p cannot be prime. Hence we may assume that x contains exactly one 4.

Case 4.1: p begins with 3.

In this case we can write p = 3y4z9, where all digits of y, z are 0, 3, 6, or 9. We must have 349 ◁ p.

Case 4.2: p begins with 4.

In this case we can write p = 4y9, where all digits of y are 0, 3, 6, or 9. If 0 ◁ y, then 409 ◁ p. If 3 ◁ y, then 43 ◁ p. If 9 ◁ y, then 499 ◁ p. Hence we may assume y ∈ {6}, and thus p ∈ 4{6}9. But then p is divisible by 7 (All Numbers of the Form 4666...6669 are Divisible by 7), since for n ≥ 0 we have 7 × 6n7 = 46n9.

Case 4.3: p begins with 6.

In this case we can write p = 6y4z9, where all digits of y, z are 0, 3, 6, or 9. If 0 ◁ z, then 409 ◁ p. If 3 ◁ z, then 43 ◁ p. If 6 ◁ z, then 6469 ◁ p. If 9 ◁ z, then 499 ◁ p. Hence we may assume z is empty.

If 3 ◁ y, then 349 ◁ p. If 9 ◁ y, then 6949 ◁ p. Hence we may assume all digits of y are 0 or 6.

If 06 ◁ y, then 60649 ◁ p. Hence we may assume y ∈ {6}{0}.

If 666 ◁ y, then 666649 ◁ p. If 00000 ◁ y, then 60000049 ◁ p. Hence we may assume y ∈ {𝜆, 0, 00, 000, 0000, 6, 60, 600, 6000, 60000, 66, 660, 6600, 66000, 660000}, and thus p ∈ {649, 6049, 60049, 600049, 6000049, 6649, 66049, 660049, 6600049, 66000049, 66649, 666049, 6660049, 66600049, 666000049}, and of these numbers only 66000049 and 66600049 are primes.

Case 4.4: p begins with 9.

In this case we can write p = 9y4z9, where all digits of y, z are 0, 3, 6, or 9. If 0 ◁ y, then 9049 ◁ p. If 3 ◁ y, then 349 ◁ p. If 6 ◁ y, then 9649 ◁ p. If 9 ◁ y, then 9949 ◁ p. Hence we may assume y is empty.

If 0 ◁ z, then 409 ◁ p. If 3 ◁ z, then 43 ◁ p. If 9 ◁ z, then 499 ◁ p. Hence we may assume z ∈ {6}, and thus p ∈ 94{6}9, and the smallest prime p ∈ 94{6}9 is 946669.