science

[Daily puzzle] Fill in the blanks

south_korea_ln

_Reasoned_ solution:

Let the blanks be filled in with $$a_0$$, $$a_1$$, $$a_2$$, $$a_3$$, $$a_{\text{ev}}$$, $$a_{\text{od}}$$, and $$a_{\text{pr}}$$, respectively. Among these, the first four entries satisfy $$a_i \geq 1$$ for $$i \leq 3$$, while the other three entries each satisfy $$a \geq 2$$. Thus, $$a_0 = 1$$, $$a_1 \geq 2$$, and $$a_{\text{od}} \geq 3$$. Given $$a_{\text{ev}} + a_{\text{od}} = 11$$, the values have opposite parity, so there’s at least one additional even and odd entry. Consequently, $$a_{\text{ev}} \geq 3$$ and $$a_{\text{od}} \geq 4$$. This limits $$(a_{\text{ev}}, a_{\text{od}})$$ to only five pairs: $$(3,8), (4,7), (5,6), (6,5),$$ or $$(7,4)$$. Each pair has exactly one prime number, hence $$a_{\text{pr}} \geq 3$$, but $$a_{\text{pr}} \neq 3$$, as this “3” would add a fourth prime to the set, implying $$a_{\text{pr}} \geq 4$$.

Now, only $$a_2$$ and $$a_3$$ are potential placements for “1,” so $$a_1 \leq 4$$. If $$a_1 = 4$$, there are three primes, including one from $$(a_{\text{ev}}, a_{\text{od}})$$, but not $$a_{\text{pr}}$$. Setting $$a_{\text{pr}} = 4$$ would then leave only three primes, and any other $$a_{\text{pr}}$$ would be too large. Thus, $$a_1 \leq 3$$. Assume $$a_1 = 3$$; then $$a_3 \geq 2$$. To assess feasibility, suppose $$a_3 = 2$$, which implies $$a_2 \geq 2$$, creating a contradiction with $$a_1 = 3$$. Thus, $$a_3 \geq 3$$. If $$a_3 \geq 4$$, $$a_{\text{pr}} = 3$$ leads to more than three primes, so $$a_3 = 3$$.

With $$a_3 = 3$$, counting primes from $$(a_{\text{ev}}, a_{\text{od}})$$ yields five primes. Setting $$a_{\text{pr}} = 5$$ results in six primes, while $$a_{\text{pr}} = 6$$ yields five primes, a contradiction. Therefore, $$a_1 = 2$$, implying $$a_2 \geq 3$$ because $$a_2 = 2$$ would introduce three “2”s. Since only $$a_3 = 2$$ allows a third 2, we conclude $$a_2 = 3$$ and $$a_3 = 2$$.

Thus, we find six primes with $$a_{\text{pr}} = 6$$ or $$a_{\text{pr}} = 7$$, resulting in two possible solutions: $$(1, 2, 3, 2, 6, 5, 6)$$ and $$
(1, 2, 3, 2, 5, 6, 7)$$.
