Soru

Zorluk: OrtaOdd and Even Integers (Parity)

If pp, qq, and rr are integers such that 3p+2q+r2=173p + 2q + r^2 = 17, and pp is an odd integer, which of the following statements MUST be true?

  1. rr is an even integerCevap
  2. B
    qq must be an odd integer
  3. C
    p+rp + r is an even integer
  4. D
    r2+1r^2 + 1 is an even integer
  5. E
    pqrp \cdot q \cdot r must be an odd integer

Cevap

rr is an even integer
Because pp is odd, 3p3p is odd. The term 2q2q is always even for any integer qq. Adding an odd and an even number gives an odd sum (3p+2q3p + 2q). Since the total sum 3p+2q+r23p + 2q + r^2 equals 1717, which is an odd number, r2r^2 must be even. An integer whose square is even must be even itself, so the statement that rr is an even integer must be true.

Adım Adım Çözüm

1
Analyze the parity of the term 3p3p
Since pp is given as an odd integer, odd×odd=odd\text{odd} \times \text{odd} = \text{odd}, so 3p3p is an odd integer.
Multiplying an odd integer by an odd integer yields an odd integer.
2
Analyze the parity of the term 2q2q
2q2q is an even integer for any integer qq.
Any integer multiplied by 22 produces an even integer.
3
Determine the parity of the sum 3p+2q3p + 2q
3p+2q=odd+even=odd3p + 2q = \text{odd} + \text{even} = \text{odd}.
The sum of an odd integer and an even integer is always odd.
4
Evaluate the equation (3p+2q)+r2=17(3p + 2q) + r^2 = 17 to find the parity of r2r^2 and rr
odd+r2=17 (odd)    r2\text{odd} + r^2 = 17\ (\text{odd}) \implies r^2 must be even     r\implies r must be an even integer.
For the sum of an odd integer and r2r^2 to equal an odd integer (1717), r2r^2 must be even. An integer whose square is even must itself be even.

Anahtar Kavram

Parity rules of addition and multiplication of integers: odd+even=odd\text{odd} + \text{even} = \text{odd} and even×any integer=even\text{even} \times \text{any integer} = \text{even}.
Tahmini Süre:1m 30s
Bu soruyu puanla