Fill in the blanks with a single digit such that the resulting sentence is true.
"In this sentence, the number of occurences of the digit 0 is ______, of the digit 1 is ______, of the digit 2 is ______, of the digit 3 is ______, of even numbers is ______, of odd numbers is ______, and of prime numbers is ______."
1, of the digit 1 is2, of the digit 2 is3, of the digit 3 is2, of even numbers is5, <---- of odd numbers is6, and of prime numbers is7.1, of the digit 1 is2, of the digit 2 is3, of the digit 3 is2, of even numbers is6, <---- of odd numbers is5, and of prime numbers is6.a_0,a_1,a_2,a_3,a_{\text{ev}},a_{\text{od}}, anda_{\text{pr}}, respectively. Among these, the first four entries satisfya_i \geq 1fori \leq 3, while the other three entries each satisfya \geq 2. Thus,a_0 = 1,a_1 \geq 2, anda_{\text{od}} \geq 3. Givena_{\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 3anda_{\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, hencea_{\text{pr}} \geq 3, buta_{\text{pr}} \neq 3, as this “3” would add a fourth prime to the set, implyinga_{\text{pr}} \geq 4.a_2anda_3are potential placements for “1,” soa_1 \leq 4. Ifa_1 = 4, there are three primes, including one from(a_{\text{ev}}, a_{\text{od}}), but nota_{\text{pr}}. Settinga_{\text{pr}} = 4would then leave only three primes, and any othera_{\text{pr}}would be too large. Thus,a_1 \leq 3. Assumea_1 = 3; thena_3 \geq 2. To assess feasibility, supposea_3 = 2, which impliesa_2 \geq 2, creating a contradiction witha_1 = 3. Thus,a_3 \geq 3. Ifa_3 \geq 4,a_{\text{pr}} = 3leads to more than three primes, soa_3 = 3.a_3 = 3, counting primes from(a_{\text{ev}}, a_{\text{od}})yields five primes. Settinga_{\text{pr}} = 5results in six primes, whilea_{\text{pr}} = 6yields five primes, a contradiction. Therefore,a_1 = 2, implyinga_2 \geq 3becausea_2 = 2would introduce three “2”s. Since onlya_3 = 2allows a third 2, we concludea_2 = 3anda_3 = 2.a_{\text{pr}} = 6ora_{\text{pr}} = 7, resulting in two possible solutions:(1, 2, 3, 2, 6, 5, 6)and(1, 2, 3, 2, 5, 6, 7).