Difficulty: Medium
Correct Answer: 31
Explanation:
Introduction / Context:
This question tests prime numbers along with a common number-theory trick: using odd/even parity to restrict possibilities quickly. Since most primes are odd (except 2), the sum of three primes has predictable parity. Once we identify that one of the primes must be 2, the problem becomes finding two primes that add to 36 while making x as large as possible.
Given Data / Assumptions:
Concept / Approach:
Use parity:\n- 38 is even.\n- The sum of three odd numbers is odd.\nTherefore, all three primes cannot be odd. The only even prime is 2, so one of x, y, z must be 2. Then the remaining two primes must sum to 36. To maximize x, choose the smallest possible other prime so that the remaining number is still prime.
Step-by-Step Solution:
Since 38 is even, and odd + odd + odd = odd, at least one prime must be even
The only even prime is 2, so one of the primes is 2
Then the other two primes must sum to 38 - 2 = 36
Let x be the largest prime among the two that sum to 36
Try smallest prime for the other: 3 gives 36 - 3 = 33 (not prime)
Next 5 gives 36 - 5 = 31 (prime)
So we can take primes 2, 5, and 31, which sum to 38
This makes x = 31
Any larger x would require the partner prime to be smaller than 5, but 3 fails, and 2 would force 34 (not prime)
Verification / Alternative check:
Other valid pairs for 36 include 7 + 29, 13 + 23, 17 + 19, but these give x values 29, 23, and 19 respectively, all smaller than 31. So 31 is indeed the maximum.
Why Other Options Are Wrong:
29 occurs with (2, 7, 29).
23 occurs with (2, 13, 23).
19 occurs with (2, 17, 19).
17 is too small and cannot be the maximum given better combinations exist.
Common Pitfalls:
Forgetting that 2 is the only even prime, or assuming you can use three odd primes to make an even sum. Another mistake is not checking primality of the partner number (for example 33 or 34 are not prime).
Final Answer:
The maximum possible value of x is 31
Discussion & Comments