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今年是2013年,我们看一个与2013有关的问题.题把2013写成若干正整数的和,试求这些正整数的积的最大值.解首先,由于不同形式的和的个数有限,因此积的个数也是有限的,积一定有最大值.设2013=a_1+a_2+…+a_n且使积p=a_1a_2…a_n最大,那么正整数a_i(i=1,2,…,n)应该具有下列性质:①a_i>1(i=1,2,…,n).如若不然,不妨设a_1=1,令2013=(a_1+a_2)+a_3+…+a_n,考察q=(a_1+a_2)a_3…a_n,由于a_1+a_2>a_2,这样就得至q>p,而这和p最大矛盾.②a_i<5(i=1,2,…,m).如若不然,不妨设a_1
This year is 2013. Let’s look at a 2013-related question. Let 2013 be the sum of a number of positive integers, and try the maximum of the product of these positive integers. Solution First, because of the finite number of sums of different forms, The number of positive integers a_i (i = 1, 2, ..., n) should have the following values: a = a_1 + a_2 + ... + a_n and maximizes the product p = a_1a_2 ... a_n Properties: ① a_i> 1 (i = 1,2, ..., n). Otherwise, let a_1 = 1, let 2013 = (a_1 + a_2) + a_3 + ... + a_n, investigate q = (a_1 + a_2) a_3 ... a_n, since a_1 + a_2> a_2, so that we get q> p, which is the biggest contradiction with p. ②a_i <5 (i = 1,2, ..., m)