求助两道gmat网络课堂数学题两道

DAYUEYUE2000 · 发表于 2011-5-22 14:33:47

a平方+b平方=c平方，其中a，b，c为整数，下面哪个不能是a+b+c的值 A.-2 B.-1 C.2 D.4 E.6求一下计算过程，黄欣那个讲的真是费解!

If n is an integer greater than 6, which of thefollowing must be divisible by 3 ?
(A) n(n+1)(n-4)
(B) n(n+2)(n-1)
(C) n(n+3)(n-5)
(D) n(n+4)(n-2)
(E) n(n+5)(n-6)

这个黄欣讲的是 n-4 和n-1一样，也没说明白，在网上查查是套数，这个也太简单了，能按黄欣那个讲法做这题吗？

sdcar2010 · 发表于 2011-5-22 20:08:28

categories: a b c (a+b+c)
1 0 1 -2, 2 or 0
3 4 5 -6, -4, -2, 2, 4, 6

Apparently B -1 is not possible.

Another way to look at this problem is the following:
If the sum of a, b, and c is odd (like -1), then among a, b, and c, either 1 or 3 numbers are odd. However, since the square of an odd number is still odd, therefore there are either 1 or 3 odd numbers in a^2, b^2, and c^2. However, it would require that the sum of two even numbers to be odd, or the sum of an even and an odd number to be odd, or the sum of two odd numbers to be odd. Impossible.

sdcar2010 · 发表于 2011-5-22 20:13:34

If n is an integer greater than 6, which of thefollowing must be divisible by 3 ?
(A) n(n+1)(n-4)
(B) n(n+2)(n-1)
(C) n(n+3)(n-5)
(D) n(n+4)(n-2)
(E) n(n+5)(n-6)

First, you should realize among three consecutive integers, there must be one number that is a multiple of 3.
Second, in terms of the property of being a multiple of 3, n and n+3 and n + 6 are the same.

Based on the above rules, in terms of the property of being a multiple of 3,
n(n+1)(n-4) = n(n+1)(n-4+6) = n(n+1)(n+2) = the product of 3 consecutive numbers.

rontsai · 发表于 2011-5-23 01:19:42

第一题：由于a^2+b^2＝c^2，所以(a+b+c)^2＝a^2+b^2+c^2+2ab+2bc+2ca＝c^2+c^2+2ab+2bc+2ca＝2(c^2+ab+bc+ca)，则(a+b+c)^2为偶数；根据偶数×偶数＝偶数，显然有(a+b+c)为偶数
第二题：I、临场用特殊值法，然后逐个排除；
II、基本上可把(n-4)等价(n-1)，原因是：
n(n+1)(n-4)＝n(n+1)(n-1-3)=n(n+1)(n-1)－n(n+1)3=(n-1)n(n+1)－3n(n+1)，
由于n是整数，而连续3个整数相乘，乘积显然为3的倍数。

求助两道gmat网络课堂数学题两道

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