Seven leaves one, eight leaves two, nine leaves dead, seven leaves four. What is their remainder multiplied by?
I understand it as:
The number of seven leaves one, which is synonymous with "the number of seven leaves one". Also called "seven divided by one, seven minus one".
Among them, the number of seven leaves one ("the number of seven leaves one") and the number of seven leaves four are contradictory.
I suggest changing the title to:
There are several things, seven leaves one, eight leaves two and nine leaves four. What's the minimum?
Solution: Mark and write with number theory.
X== 1 mod 7 ==2 mod 8==4 mod 9, find X.
Solution: Ask first.
x 1 = = 1 mod 7 = = 0 mod 8 = = 0 mod 9 = = & gt； x 1 = 8 * 9k 1 = = 1 mod 7，k 1==4==-3 mod 7
x2 = = 0 mod 7 = = 1 mod 8 = = 0 mod 9 = = & gt； x2=7*9k2== 1 mod 8，k 1==- 1==7 mod 8
x3 = = 0 mod 7 = = 0 mod 8 = = 1 mod 9 = = & gt； x 1=7*8k3== 1 mod 9，k 1==5==-4 mod 9
Let's take X 1 = = 8 * 9 * 4, X2 = 7 * 9 *( 1), X3 = 7 * 8 *(4).
It is easy to see that x = =1* x1+2 * x2+4 * x3 modlcm (7,8,9) is what you want (# # #).
Lcm () is usually written as [], which means least common multiple.
This explains the solution principle of Chinese remainder theorem.
But the above formula (# # #) can also be simplified. I wrote it in the following form:
x==
1*4 @7
2*- 1 @8
4*(-4) @9 == 2 @9
==
18 @56
2 @9
== 18*9+2*56=274
This is the smallest natural number that satisfies the meaning of the question "seven numbers leave one, eight numbers leave two, and nine numbers leave four".
In fact, it can also be solved like this:
Look for it first
X1=1module7 = = 0 module8 = = 0 module9.
X2==0 modulo 7 ==2 modulo 8== 0 modulo 9.
X3==0 modulo 7 ==0 modulo 8== 4 modulo 9.
It is easy to see that x = = x1+x2+x3 mod LCM (7,8,9) is what you want.
This calculation can sometimes be simplified because of the coincidence of remainders. When the remainder is 0, the calculation can be omitted. For more simplified schemes, please see the related articles in my Baidu blog.
The above is the case where the divisor is 7, 8 and 9.
You said at the end of the topic, "Count 33, and multiply the remainder by 70; 77, the remainder is multiplied by15; 55, the remainder is multiplied by 21; When the three are added together, if it is not greater than 105 as the answer, it is necessary to subtract 105 or its multiple if it is greater than 105 ",that is, the divisor is 3, 5 and 7. Such an example is a classic example in Sun Tzu's earlier mathematical work The Art of War. In addition, it is similar to what is commonly called Han Xin's point soldier.
To solve this kind of problem, the general principle is China's remainder theorem (Sun Tzu theorem).
Please refer to my article (which also quoted some of my other articles, and I have some experience about China's remainder theorem, which I think can be greatly simplified):
/wsktuuytyh/blog/item/334 1 153 DD 73 cc 9 CB 9 e 3d 622d . html