AMC 8 2007 Problem 19
AMC8 2007 年第 19 题
奇偶性★★★☆☆
Pick two consecutive positive integers whose sum is less than 100 . Square both of those integers and then find the difference of the squares. Which of the following could be the difference?
选择两个连续的正整数，使得它们的和小于 100。将这两个整数平方后再相减，则下面哪个可能是所得到的差？
- A.2
- B.64
- C.79
- D.96
- E.131
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
Let the smaller of the two numbers be . Then, the problem states that (x+1)+x<100. . 2x+1 is obviously odd, so only answer choices C and E need to be considered. 2x+1=131 contradicts the fact that 2x+1<100, so the answer is
Since for two consecutive numbers and , the difference between their squares is , which equals to a+b, because and are consecutive. Because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and the sum of and is less than 100, you can eliminate all answers except for .
From the question we can make and we can let be b-1 where a+b<100. First we factor to (a+b)(a-b) and then plug in a=b-1 to have: (a+a+1)(a-a+1)=(2a+1)(1)=2a+1=a+b. since a+b will always be an odd number, we can narrow the answers down to (C) and (E). Since 131 would be too big for our range of a+b<100, we would choose . note that this solution would be better if the question was asking for the values of a and b
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