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Home » Conclusion: the remainder is constant \( -1 \) - Dyverse

Conclusion: the remainder is constant \( -1 \) - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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Try including 2: then we need sum of distinct odd primes â¡ 37 mod 2...
So check: 39 = 2 + 37? Yes, 2 and 37 are distinct primes.
Is 37 â¤ 39 and prime? Yes.
And 2 + 37 = 39, and distinct â valid.
Check if 39 is the maximum possible. Can we get 39? Yes.
Is 40 possible? No, upper bound is less than 40.
Is 39 the maximum? Yes, since 38, 37, etc., are smaller.
But confirm: could 38 be written as sum of distinct primes?
is even. Use multiple odd primes.
Try largest primes < 38: 37 + 1 â 1 not prime.

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So \( a + b = 6 \) or \( a + b = 15 \) (since \( a \geq 1 \), \( b \le...
First, count solutions to \( a + b = 6 \), \( a \geq 1 \), \( b \geq 0...
\( a \) ranges from 1 to 6 → \( b = 6 - a \geq 0 \), so 6 solutions.
Now, \( a + b = 15 \):
\( a \geq 6 \) (since \( b \leq 9 \)), and \( a \leq 9 \). So \( a = 6...
\( a = 6 \), \( b = 9 \)
\( a = 7 \), \( b = 8 \)
\( a = 8 \), \( b = 7 \)
\( a = 9 \), \( b = 6 \) → 4 solutions.
Total solutions: \( 6 + 4 = 10 \)

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