AMC12 2012 B

AMC12 2012 B · Q5

AMC12 2012 B · Q5. It mainly tests Casework, Parity (odd/even).

Two integers have a sum of $26$. when two more integers are added to the first two, the sum is $41$. Finally, when two more integers are added to the sum of the previous $4$ integers, the sum is $57$. What is the minimum number of even integers among the $6$ integers?

两个整数的和为 $26$。当再把两个整数加到前两个整数上时，和为 $41$。最后，当再把两个整数加到前面 $4$ 个整数的和上时，和为 $57$。这 $6$ 个整数中偶数的最小个数是多少？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (A)

正确答案：（A）

Solution

Since, $x + y = 26$, $x$ can equal $15$, and $y$ can equal $11$, so no even integers are required to make 26. To get to $41$, we have to add $41 - 26 = 15$. If $a+b=15$, at least one of $a$ and $b$ must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning from $26$ to $41$. Finally, we have the last transition is $57-41=16$. If $m+n=16$, $m$ and $n$ can both be odd because two odd numbers sum to an even number, meaning only $1$ even integer is required. The answer is $\boxed{\textbf{(A)}}$. ~Extremelysupercooldude (Latex, grammar, and solution edits)

由于 $x + y = 26$，可以取 $x=15$，$y=11$，因此凑出 $26$ 不需要偶数。要得到 $41$，必须再加 $41 - 26 = 15$。若 $a+b=15$，则 $a$ 与 $b$ 中至少有一个必须是偶数，因为两个奇数之和为偶数。因此从 $26$ 到 $41$ 的这一步需要 $1$ 个偶数。最后一步为 $57-41=16$。若 $m+n=16$，则 $m$ 和 $n$ 可以都为奇数，因为两个奇数之和为偶数，这意味着只需要 $1$ 个偶数。答案是 $\boxed{\textbf{(A)}}$. ~Extremelysupercooldude (Latex, grammar, and solution edits)

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