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分解质因数的应用是很广泛的。但在目前小学数学教材中,除了用它来求几个数的最大公约数和最小公倍数之外,几乎未作其他介绍,我觉得这很不够。下面举几个分解质因数应用于解题的例子,以供参考。例一 两个自然数的积为3315。已知其中一个数的大小在30到40之间,求这两个自然数。分析:此题若用试除法,即令k_1·K_2=3315,K_1的取值范围:30 ≤K_1≤40(K_1∈N),然后一一试除,求出K_2(K_2∈N), 从而得解,这样也未尝不可,但显然很不科学,也很麻烦。若K_1的范围较大时,甚至不能求解。本例可用分解质因数法解之。解:∵3315=3 × 5 × 13× 17根据题意,把质因数适当组合,得:3315=(3×13)×(5×17)=39×85或3315=(3×17)×(13×5):51×65。所以,所求两数为39和85。例二 某人出差四天,回家后连撕四张日历,他把这四张日历的日期数连乘起来,积得303600,试判断,这个人是那四天出差的．
The application of decomposition factor is very extensive. However, in the current primary school mathematics textbook, apart from using it to calculate the greatest common divisor and the least common multiple of several digits, there is hardly any other introduction. I think this is not enough. Here are a few examples of factorization factors applied to problem solving for reference. Example 1 The product of two natural numbers is 3315. Knowing that one of the numbers is between 30 and 40, find the two natural numbers. Analysis: If this question uses trial division, that is to say k_1 · K_2 = 3315, K_1 value range: 30 ≤ K_1 ≤ 40 (K_1 ∈ N), and then try to remove, find K_2 (K_2 ∈ N), so as to obtain Solution, this is not a bad idea, but obviously it is unscientific and it is troublesome. If the range of K_1 is large, it cannot even be solved. This example can be solved using the decomposition factor method. Solution: ∵3315=3 × 5 × 13×17 According to the meaning of the question, the proper combination of prime factors is obtained: 3315=(3×13)×(5×17)=39×85 or 3315=(3×17)× (13×5):51×65. Therefore, the two numbers required are 39 and 85. Example 2 Someone went on a business trip for four days. After returning home, he even tore four calendars. He multiplied the number of days of the four calendars and got 303,600. Try to judge that the person was on business trips for four days.