十万里加急!!! 高中数学题!!!!! 求助 大神, 希望有高人 耐心指点, thx 希望 我能早早的 被 贵人 找到! 因为 本人 打速成 很慢, 所以都用 拼音。 求解下题 并 提供 列式, 列式,因为这很重要, 能解说一下更好。?

2017-08-11 5:06 pm
Question 9

Polynomials and Multiples of 9

It is easy to find the multiples of 9 from 1 to 100. But can you easily find the multiples of 9 for numbers larger than 100 or even 10 000 without using a calculator?

In order to determine whether a number is a multiple of 9, we can add up all digits in the number. If the sum is divisible by 9, then the number is a multiple of 9.
e.g. (i) Consider 171. Sum of all digits = 1+7+1= 9 , so 171 is a multiple of 9.
(ii) Consider 123 123 123. Sum of all digits = 1 +2 +3 +1 +2 +3 +1 +2 +3 =18 since 18 is a multiple of 9. so 123123123 is a multiple of 9.
(iii) Consider 100 001. Sum of all digits = 1+0+0+0+0+1= 2
since  2 is not a multiple of 9. 
so  100 001  is not a multiple of 9.
In fact, we can use the knowledge of polynomials to prove that the    above method works.

问题见下一份档案 

回答 (1)

2017-08-11 8:50 pm
1.
f(1)=a*1^2+b*1+c=a+b+c
f(10)=a(10)^2+b(10)+c=100a+10b+c
2.
f(x)=(x-1)Q(x)+f(1)
N=f(10)=(10-1)Q(10)+f(1)
N=9Q(10)+f(1)
3.
Q(x)=px+q (p, q 是整數, 如果x是整數. Q(x)也是整數)
N=9Q(10)+f(1)
Q(10)是整數, 9Q(10)必定是9的倍數
所以只要 f(1)=a+b+c 是9的倍數, 則N是9的倍數


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