There is a four digit number, and the sum of the numbers on each bit of it can be divided by 17. If you add 1 to the four digit number, the sum of the numbers on each bit of the sum will be equal

There is a four digit number, and the sum of the numbers on each bit of it can be divided by 17. If you add 1 to the four digit number, the sum of the numbers on each bit of the sum will be equal

Let this four digit number be ABCD,
Namely
A + B + C + D = 17
or
A + B + C + D = 34
According to the meaning of the title, after adding 1 to four digits, the sum of each digit has this rule:
(1) If carry does not occur, then sum of all digits = sum of original digits + 1 = 18 or 35, which cannot be divisible by 17
(2) If a carry occurs, the sum of each digit = the sum of original digits + 1 - (10 - 1) = 9 or 26, which cannot be divisible by 17
(3) If carry occurs three times, then sum of each digit = sum of original digits + 1 - (10 - 1) * 2 = 0 or 17, which can be divided by 17 and conform to
So the sum of the original number ABCD is 17 or 34
Moreover, when the sum of digits in ABCD is 17, the most likely carry digit (such as 1079) added with 1 can not generate twice carry, so the sum of digits in ABCD can only be 34, and only twice carry digit occurs when added with 1
Therefore, ABCD may be:
⑧8⑨9
⑨7⑨

A 4-digit number, the sum of each digit can be divided by 4, how many such numbers?
The answer is 2249, who knows how?

A total of 2249, write a small program can solve
#include
#include
void main()
{ fstream fsOutFile("klm1.txt",ios::out);
int i=0;
int count=0;int a,b,c,d;
for(i=1000;i

There is a four digit number, and the sum of its digits can be divided by 17. If you add 1 to the four digit number, the sum of its digits can also be divided by 17. The minimum number of the four digit number is 1______ ．

According to the meaning of the question, the sum of four digits after adding 1 has the following rules: (1) if there is no carry, then the sum of four digits = the sum of original digits + 1 = 18 or 35, which cannot be divisible by 17; (2) if there is one carry, then the sum of four digits = the sum of original digits + 1 - (10-1) = 9 or 26, which cannot be divisible by 17; (3) if there are two carry, then the sum of each digit = the sum of original digits The sum of numbers + 1 - (10-1) × 2 = 0 or 17 can be divided by 17, so the sum of all digits of the original four digits is 17 or 34. When the sum of all digits of the original four digits is 17, the number (such as 1079) which is most likely to carry by adding 1 can not carry twice, so the sum of all digits of the original four digits can only be 34, and only carry twice when adding 1 But: 88999799. The minimum is 8899. So the answer is: 8899

Is the number of three ones and seven tenths 7.3 right or wrong

Wrong, 3.7