A good friend in math helped me with two questions. Both are questions about remainders

This has to do with the remainder definition.

a=kb+c

a, b, c, k are integers, then.

The remainder c is the same as the dividend b plus or minus, and has |b|>|c|5=(-3) 1+(-2) The remainder is -2

5=3 (-2)+1 The remainder is 1

5=3 (-2)+1 The remainder is 1

Its practical formula solution is more complicated.

Upstairs is good: like this question, I'm looking for a special division of 5 and 3, then the end of this number must be 3 or 8, 8 is not good, because 8 is 6 after subtracting 2, and 6 is a mantissa when there is no divisible 7, so it can only be 3, so it is 13 to see that it is not good, continue to go up 23 lines! )

Formula solution. Let this number be a

a=a×3+2

a=b×5+3

a=c×7+2

a, b, c are integers.

Then there is. 35a=105a+70

21a=105b+63

15a=105c+30

71a = 105 (a + b + c) + 163

Simplified, a=2+[105(a+b+c)+21] 71, let m=a+b+c

There is an integer 71k = 105m + 21

m=(71k-21)/105

m is divisible by 105 and the last digit is 5 or 0

The last digit of k is 1 or 6

Substitute to find k = 21 and m = 14

In this case, find a=23

First, the remainder sign is the same as the dividend, i.e., the numerator.

So the result is that the question of Shang 1 Yu -2 Shang -2 Yu 1 Shang -2 Yu -1 is the second of the Chinese Remainder Theorem (or Sun Tzu's Theorem) in the "Sun Tzu Sutra" after BC has the problem of "things do not know the number": "Now there are things that do not know their number, the remaining two of the three or three numbers, the remaining three of the five or five numbers, and the remaining two of the seven and seven numbers, asking the geometry of things?" The answer is "23".

5 -3=5 3 remainder 2

5 3=-2*3+1 remainder 1

Division with remainders.

If a and b are two integers, where b > 0, then there are two integers q and r, such that a=bq+r, 0<=r is divided by the remainder, and any integer n can be expressed as.

n=k*q+r where 0<=r, where r is the remainder of n divided by q.

Usually written as: n r (mod q) for example.

Then the remainder of -9 divided by 5 is 1

The remainder is always smaller than the divisor.

Let this number be x, then there is.

x-2 is an integer multiple of 3.

x-2 is an integer multiple of 7.

Then x-2 must be an integer multiple of 21.

And because x-3 is an integer multiple of 5.

Then the single digit of x must be 2;

It can be judged that the smallest x is 42

The remainders are -2 -2 2

Keep in mind that the absolute value of the remainder must be smaller than the divisor.

Like this question, I'm looking for a special division of 5 and 3, then the end of this number must be 3 or 8, 8 is not good, because 8 is 6 after subtracting 2, and 6 is a mantissa when there is no divisible 7, so it can only be 3, so it is 13 to see that it is not good, continue to go up 23 lines!

The first question should be the second question is the third question:

Don't know, right?

Use 2 to divide by a number, and if there is a remainder, the resulting remainder can only be 1, right?

That's right. In division with remainders, the remainder must be less than the divisor, greater than 0, so the maximum remainder is: 2-1=1

The minimum is 0+1=1

Therefore, the remainder can only be 1

If there is a remainder to remove a number, if there is a remainder, can the remainder only be a pair? That's right, because....

Let's start with a more common topic, take the topic I talked about on a small piece of paper as an example:

First of all, the reason why we can cross out a 0 of both the dividend and the divisor is because we follow the rule that the dividend and the dividend are reduced by the same multiple at the same time, and the quotient does not change. So, if you remove a 0 at the same time, it doesn't affect the result, so the original 65100 210 now becomes 6510 21, but the result is the same. Calculate 6510 21 and the final result is 310, so 65100 210 310.

Here's an example of a mistake: pay attention to the 0 at the end of the quotient, don't write it, because although the single digit still exists, there is no longer a number on the single digit.

Let's look at the second example.

3620÷50＝72...20, this is what we did with the previous method, there are no problems, the final result is 72 out of 20, but, looking further down, the new easy way.

Here's the new easy way: the final remainder is 2, is there something wrong with our calculations? Or is our new approach wrong?

First of all, the divisor is reduced by the same multiple of the dividend, and the quotient is unchanged, that is to say, in the vertical type, the quotient can be several, we can copy it directly, but the remainder is different, as I said above, although the 0 in the dividend is crossed out, but the number is still there, our remainder 2, on the ten, represents 2 10, so it is 20. Summed up in one sentence:

Use a simple way to do pen calculations, the quotient is a few, directly copied in the horizontal back, but the remainder must be careful to see which digit it is in, like this topic, 2 in the ten digit, it represents 20, so the remainder 20.

There is a fixed solution to this type of problem, so let's introduce it.

If you divide the n-digit multi-digit number (each digit is a) by b, there will be periodicity. Periodicity comes from the length of the number of digits of n and the cyclic decimal of 1 7, so we can recursively increase the length of the number of digits to obtain the quotient and the remainder, and thus find the periodic change law between the length, quotient and remainder.

So if the cycle length is 6, then the remainder of 2003 digits divided by 7 repeats every 6 digits. 2003 divided by 6 has a remainder, no direct division. Because 2004 is divisible by 6 (2004 is divisible by 3, 2004 is divisible by 2), so the remainder of 2003 divided by 6 is a period of 6 numbers missing 1.

The corresponding is the penultimate 4 of (2,3,5,3,4,0).

Because 7 is cyclical, 1 7 6 7 is exactly 6 numbers for a period, which is a pyramid number. So any number A 7 must have 6 remainders for 1 period of cyclicality. The problem is that when a is different, the remainder is not the same, so you have to find a complete cycle.

Then if a=1, 11, 111......, which is also 6 remainders for 1 period. Specifically, 1 7 = 0......1,11÷7=1……4,111÷7=15……6,1111÷7=158……5

77 digits divided by 7 is also 1 period for every 6 remainders, so the corresponding remainder is 2

This topic actually implies the law of quotient, you can continue to ask the law of quotient, such as the last digit of quotient is how many, how many digits are quotient, the last two digits of quotient, the last three digits are few, and the sum of quotient numbers is how many, because quotient also has the cyclic feature of 6 digits for 1 cycle. The way of answering these two types of questions is the same source.

Numbers divisible by 7 are characterized by "if the single digit of an integer is truncated, and then 2 times the single digit is subtracted from the remaining number, and if the difference is a multiple of 7, the original number is divisible by 7." If it's not easy to see at once, you need to continue the process. ”

Do the math yourself, it's too much trouble.

19901990...1990 (n 1990) 129, mod is the symbol for finding remainders.

Because 1990 mod 11=10 129 mod 11=8, and 1000 mod 11=10

So 19901990....1990 (n 1990) 129 mod 11 = 10,0010,0010 (n-1 0010) 008 mod 11

1000 (n 100) 8 mod 11 = 10, 0010 (n - 1 0010) 8 mod 11 = 1000 (n - 1 1000) 108 mod 11

1000 (n-2 1000) 10108 mod 11 = 1000 (n-3 1000) 1010108 mod 11

Because 1010108 mod 11=0

So when n=3 is the smallest value.

The sum of odd digits is: 1+9+(1+9)n=10n+10, and the sum of even digits is: 2+(0+9)n=2+9n, and the difference between the two is: n+8

Therefore, when n=3 is divisible by 11.

That is, the smallest natural number n is 3

A three-digit number, divided by the remainder is 1, what is this three-digit number?

Solution: If the remainder of three coprime numbers is the same number, then this number is the sum of their common multiples and remainders.

A three-digit cardinal number, divided by the remainder is 1, which is to find the sum of their least common multiple and 1.

Therefore, the three-digit number of Lu You is: 7*8*9+1=505

The sum of any two of them is a multiple of 2, and the sum of any three is a multiple of 3, so these four numbers must be odd, to make the sum of any three numbers a multiple of three, then each number is a multiple of 3, to make them as small as possible, use 3 times the smallest four odd numbers:

So these four numbers are

35 is right.

I would like to add that the process is nothing more than finding the benchmark number according to two conditions, and then satisfying the third condition on this basis.

Divide by 8 to get 3, divide by 9 to get 8

8x + 3 = 9y + 8

y = (8x -5)/9

Try x = 4, y = 3, and the benchmark number is 35.

On the basis of 35, the common multiple of the increase is increased one by one to meet the division of 13 and the remainder of 9. Set to increase x times.

35 + 72x = 13y + 9

x = (13y - 26)/72

The number of the problem is better, and it is easy to get y = 2, x = 0 is a solution. The number sought is 35.

On the basis of 35, the common multiple of 936 is increased one by one, and the following number that satisfies the condition can be obtained:

*18*2322*13*19≡（-3）*（3）*（2）*（1）*5≡90≡6(mod7）

7*11*17*19*23*29*113≡3*（-6）*（2）*4*6*（-3）*3*（-4）≡3*6*2*4*6*3*3*4≡36*24*36≡9*11≡8(mod13）

2+2^2+2^3+…+2 19=2 20-1 4 10-1 (-1) 10-1 0(mod5) There is a problem with the question in this question.

454*546/2+17=123959

A: The minimum (no maximum) of this number is 123959.

The problem on the second floor is to explain that this is to find the least common multiple (454*546 2).

Agree upstairs, there is no maximum, but why divide by 2?