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Zorluk: OrtaProperties of Integers and Divisibility

If pp, p+2p + 2, and p+4p + 4 are all prime numbers, what is the value of p2+5p^2 + 5?

  1. A
    6
  2. B
    9
  3. 14Cevap
  4. D
    19
  5. E
    30

Cevap

The value of p2+5p^2 + 5 is 14.
For any three consecutive odd integers pp, p+2p + 2, and p+4p + 4, exactly one of them must be a multiple of 3. Because all three expressions represent prime numbers, the term divisible by 3 must be equal to 3 (since 3 is the only prime divisible by 3). Setting p=3p = 3 gives p+2=5p + 2 = 5 and p+4=7p + 4 = 7, both of which are prime. Any choice of p>3p > 3 forces either p+2p + 2 or p+4p + 4 to be a multiple of 3 greater than 3, making it composite. Thus p=3p = 3 is uniquely determined, and evaluating p2+5p^2 + 5 gives 32+5=143^2 + 5 = 14.

Adım Adım Çözüm

1
Analyze the possible remainders when the prime pp is divided by 3.
Any positive integer pp can be written in one of three forms: 3k3k, 3k+13k + 1, or 3k+23k + 2 for some integer kk.
Dividing any integer by 3 leaves a remainder of 0, 1, or 2.
2
Test each remainder case for the expressions pp, p+2p + 2, and p+4p + 4.
If p=3k+1p = 3k + 1, then p+2=3k+3=3(k+1)p + 2 = 3k + 3 = 3(k + 1), which is divisible by 3. If p=3k+2p = 3k + 2, then p+4=3k+6=3(k+2)p + 4 = 3k + 6 = 3(k + 2), which is divisible by 3.
Among any three consecutive odd numbers of the form p,p+2,p+4p, p+2, p+4, exactly one of them must be a multiple of 3.
3
Deduce the unique value of pp.
The only way all three numbers pp, p+2p + 2, and p+4p + 4 can be prime is if the one divisible by 3 is equal to 3 itself, which forces p=3p = 3.
The only prime number divisible by 3 is 3 itself; any larger multiple of 3 is composite.
4
Evaluate the target expression p2+5p^2 + 5 using p=3p = 3.
32+5=9+5=143^2 + 5 = 9 + 5 = 14.
Substitute the uniquely determined value p=3p = 3 into the given algebraic expression.

Anahtar Kavram

Divisibility properties of consecutive odd integers and prime number definitions
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