Use 1234 to form an integer without repetition

Use 1234 to form an integer without repetition

1234 1243 1324 1342 1423 1432
2134 2143 2314 2341 ``````
And so on, there are five at the beginning of 1, and five at the beginning of 2, a total of 20

How many four digits can 1234 make up? What's the rule? For example, 1230

1234
There are four possibilities for the thousand. At this time, the hundred 4-1 = three, the ten 3-1 = two, one for each
4*3*2*1=24
one thousand two hundred and thirty
Thousand bits can't be 0, so we need to subtract several 3 * 2 * 1 from 0 = 6
So 4 * 3 * 2 * 1-3 * 2 * 1 = 18

C programming language, input a positive integer, calculate and display the sum of the numbers

#include
int main()
{
int a,sum = 0;
scanf("%d",&a);
while (a)
{
sum += a % 10;
a /= 10;
}
printf("sum = %d\n",sum);
return 0;
}

The complete square of (x-1) square

The complete square of (x-1) square
=(x-1)(x-1)
=x²-x-x+1
=x²-2x+1;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
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If the square of X + 2 (a + 4) x + 36 is a complete square, then a=

The square of X + 2 (a + 4) x + 36 is a complete square
Square of X + 2 (a + 4) x + 36 = (x ± 6) square = square of X ± 12x + 36
2(a+4)=±12
A = - 4 ± 6 = - 10, or 2

Why is the Square-1 of X under the root of X-2 meaningful

X & # 178; - 1 > = 0 under root sign
x=1
Denominator X-2 ≠ 0
x≠2
So x ≤ - 1, X ≥ 1 and X ≠ 2

Define the operation "*" as follows: a * b = a (a > = b), B ^ 2 (a)

First of all, this is a piecewise function with the domain R
Imagine a number axis with dots 1 and 2 on it, and the domain of definition is divided into three parts
That is, X ≤ 1, 1 < x ≤ 2, x > 2
1-2=-1 x≤1
x^2-2 1＜x≤2
y=﹛x^2-x^2=0 x＞2
Looking at the second function, we can find that the maximum value is 2
Comparing 2 with - 10, the final answer is 2

The new operation "*" is defined as follows: when a ≥ B, a ⁃ B = AB + B, when a ≤ B, a ⁃ B = AB-A, if (2x-1) ⁃ x + 2) = 0, then X=______ ．

When 2x-1 ≥ x + 2, that is, X ≥ 3, (2x-1) * (x + 2) = (2x-1) (x + 2) + X + 2 = 0, the solution is: x = 0 or x = - 2, ∵ x ≥ 3 ∵ x = 0 or x = - 2 are all rounded off; when 2x-1 ≤ x + 2, that is, X ≤ 3, (2x-1) * (x + 2) = (2x-1) (x + 2) - (2x-1) = 0, the solution is: x = - 1 or x = 12

The new operation "⊕" is defined as follows: when a ≥ B, a ⊕ B = AB + B, when a ⊕

When 2x-1 ≥ x + 2, X ≥ 3
Then the original formula = (2x-1) (x + 1) + (x + 2) = 2x ^ 2 + 2x-x-1 + X + 2 = 2x ^ 2 + 2x + 1
When 2x-1 < x + 2, x < 3
This is the original formula = (2x-1) (x + 2) - (2x-1) = 2x ^ 2 + 2x-x-1-2x + 1 = 2x ^ 2-x
So the answer is when x ≥ 3, the original formula = 2x ^ 2 + 2x + 1
When x ＜ 3, the original formula 2x ^ 2-x
If there is any mistake, please understand

The new operation "+" is defined as follows: a + B = AB-A + B, then (x-1) + (2x + 3)

Don't use the + sign, or it will go on in an infinite loop