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Zorluk: OrtaOdd and Even Integers (Parity)

If mm and nn are integers such that 5m+3n5m + 3n is an even integer and m−2nm - 2n is an odd integer, which of the following expressions must be an even integer?

  1. A
    mn+2mn + 2
  2. B
    m+n+1m + n + 1
  3. m2+n2m^2 + n^2Cevap
  4. D
    mn+n+1mn + n + 1
  5. E
    2m+n2m + n

Cevap

The expression m2+n2m^2 + n^2 must be an even integer.
Deduce the parities of mm and nn: 5m+3n=2(2m+n)+(m+n)5m + 3n = 2(2m + n) + (m + n), so m+nm + n must be even, implying mm and nn have the same parity. Next, m−2nm - 2n is odd, and since 2n2n is even, mm must be odd. Therefore, nn is also odd. Evaluating m2+n2m^2 + n^2 with mm and nn both odd yields odd2+odd2=odd+odd=even\text{odd}^2 + \text{odd}^2 = \text{odd} + \text{odd} = \text{even}.

Adım Adım Çözüm

1
Analyze the parity of the expression 5m+3n5m + 3n.
m+nm + n is an even integer.
Rewrite 5m+3n5m + 3n as 2(2m+n)+(m+n)2(2m + n) + (m + n). Since 2(2m+n)2(2m + n) is always even, 5m+3n5m + 3n has the same parity as m+nm + n. Because 5m+3n5m + 3n is even, m+nm + n must be even, meaning mm and nn share the same parity (both even or both odd).
2
Analyze the parity of the expression m−2nm - 2n.
mm is an odd integer.
Since 2n2n is always even, m−2nm - 2n has the same parity as mm. Given that m−2nm - 2n is odd, mm must be odd.
3
Determine the parity of nn and evaluate the options.
Both mm and nn are odd integers, so m2+n2=odd+odd=evenm^2 + n^2 = \text{odd} + \text{odd} = \text{even}.
Since mm is odd and m+nm + n is even, nn must also be odd. The square of an odd integer is odd, so m2m^2 and n2n^2 are both odd, making their sum m2+n2m^2 + n^2 an even integer.

Anahtar Kavram

Parity rules for integer addition, subtraction, and multiplication
Tahmini Süre:1m 30s
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Konular

Odd and Even Integers (Parity)

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