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Zorluk: OrtaOdd and Even Integers (Parity)

For how many ordered pairs of positive integers (a,b)(a, b), where 1≤a≤151 \le a \le 15 and 1≤b≤151 \le b \le 15, is the value of the expression (a2+a+1)(b3+b+7)(a^2 + a + 1)(b^3 + b + 7) an even integer?

Cevap: 0

Cevap

The expression evaluates to an odd integer for all positive integer pairs (a,b)(a, b), so there are 0 ordered pairs for which the value is an even integer.
The factor (a2+a+1)(a^2 + a + 1) simplifies to a(a+1)+1a(a + 1) + 1. Because a(a+1)a(a + 1) is the product of two consecutive integers, it is always even, making a(a+1)+1a(a + 1) + 1 odd for every integer aa. The factor (b3+b+7)(b^3 + b + 7) is always odd because b3b^3 and bb have matching parities (their sum is always even), so adding 7 yields an odd number. Since the product of two odd integers is always odd, the expression is never even, yielding exactly 0 ordered pairs.

Adım Adım Çözüm

1
Examine the parity of the factor a2+a+1a^2 + a + 1
a2+a+1a^2 + a + 1 is odd for all integer values of aa
The expression a2+a=a(a+1)a^2 + a = a(a + 1) represents the product of two consecutive integers, which is always even. Adding 1 to an even integer results in an odd integer.
2
Examine the parity of the factor b3+b+7b^3 + b + 7
b3+b+7b^3 + b + 7 is odd for all integer values of bb
Since b3b^3 and bb always share the same parity, their sum b3+bb^3 + b is always even. Adding 7 to an even integer results in an odd integer.
3
Determine the overall parity of the product
The product (a2+a+1)(b3+b+7)(a^2 + a + 1)(b^3 + b + 7) is odd for all inputs
The product of two odd integers is strictly an odd integer.
4
Count the number of pairs satisfying the even condition
0 pairs
Because the product is never even, zero pairs satisfy the requirement.

Anahtar Kavram

Parity rules of consecutive integer products and polynomial expressions
Tahmini Süre:1m 30s
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Konular

Odd and Even Integers (Parity)

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