Write 146 and 99 in every possible way as the sum of three prime numbers, arranging the addends in ascending order.
The sum of three integers is even if all three numbers are even, or one is even and two are odd; the sum is odd if one or all three addends are odd.

Among prime numbers, only 2 is even (here we only consider positive prime numbers). Thus, in the representation of 146, one of the terms must be 2, and the rest involves writing 144 as the sum of two prime numbers in every possible way.

Similarly, 99 can only be broken down into three odd primes, as $99=2+2+95$ has a last term that is not prime.

To facilitate the decomposition, we list the prime numbers (less than 142) grouped by their last digits:

End in 1: $11,31,41,61,71,101,131$;

End in 3: $3,13,23,43,53,73,83,103,113$

End in 5: 5;

End in 7: $7,17,37,47,67,97,107,127,137$

End in 9: $19,29,59,79,89,109,139$.

(Prime numbers not listed are divisible by at least one of the primes $3,5,7,11$.)

The last digit of 144, 4, is produced by the following pairs of last digits: $1+3,5+9,7+7$. The first type of decomposition is obtained from the numbers in the first two rows of the list.

$$
\begin{aligned}
146 & =2+31+113=2+41+103=2+61+83= \\
& =2+71+73=2+43+101=2+13+131
\end{aligned}
$$

Similarly, the second and third type of decompositions are from the 3rd and 5th, and the 4th rows, respectively.

$$
\begin{aligned}
146 & =2+5+139= \\
& =2+7+137=2+17+127=2+37+107=2+47+97
\end{aligned}
$$

According to this, 146 has 11 valid decompositions.

The last digit of 99, 3, can be produced by the sum of three odd digits in the following seven ways:

$$
\begin{array}{llll}
1+1+7, & 1+3+5, & 1+9+9, & 3+3+3 \\
3+7+9, & 5+5+9, & 5+7+7 &
\end{array}
$$

The terms of the decompositions of each type are listed in the order of the last digits given now:

$$
\begin{aligned}
& 11+41+47=11+71+17=31+31+37=31+61+7=41+41+17 \\
& 11+83+5=41+53+5=71+23+5 \\
& 11+29+59=41+29+29=61+19+19 \\
& 3+13+83=3+23+73=3+43+53=13+13+73=13+43+43= \\
& =23+23+53 \\
& 3+7+89=3+17+79=3+37+59=3+67+29=13+7+79= \\
& =13+67+19=23+17+59=23+47+29=43+37+19= \\
& =53+17+29=73+7+19 \\
& 5+5+89 \\
& 5+47+47
\end{aligned}
$$

We found 30 decompositions, of which 21 have all three terms different.

József Csörgei (Budapest, Fazekas M. Secondary School, 2nd grade)