If there is a 1994 digit a which can be divided by 9, the sum of its digits is B, and the sum of its digits is C, then C =?

C = 9. A number divisible by 9 must be a multiple of 9

If the three digits of a three digit number are a, B, C, and (a + B + C) can be divisible by 9

It is proved that if the hundred digit is a, the ten digit is B, and the single digit is C, then the three digit is 100A + 10B + C = 99A + 9b + (a + B + C), ∵ 99A, 9B, (a + B + C) can be divisible by 9, and the three digit must be divisible by 9

A three digit number is exactly five times the product of its digits. What's the number
There should be specific steps

The three digit number must be a multiple of 5, and each digit is not 0. So the individual digit number is 5
Let 100 bits be a and 10 bits be B, then 100A + 10B + 5 = 5 (a * b * 5)
The result is: 20A + 2B + 1 = 5ab (1)
5ab must be a multiple of 5, and its bit is not 0. There is a 1 and 20A on the left side of the equation, which must be a multiple of 5, then the bit of 2B should be 4, and B should be 2 or 7. If B = 2 is substituted into Formula 1, 20A + 5 = 20a, then B = 7
Substituting Formula 1, a = 1
So, the number is 175

Any nine digit number from 1 to 9 must be divisible by 3

This statement is correct, because as long as the sum of all numbers of a number can be divided by 3, then the number can be divided by 3
1+2+3+4+5+6+7+8+9=45
45 is divisible by three
Then any nine digit number from 1 to 9 must be divisible by 3

It consists of 9 digits 1-9, the first digit can be divided by 1. The 9 digits can be divided by 9
The first digit can be divided by 1; the first two digits can be divided by 2; the first three digits can be divided by 3; the first four digits can be divided by 4; the first five digits can be divided by 5; the first six digits can be divided by 6; the first seven digits can be divided by 7; the first eight digits can be divided by 8, This nine digit number can be divided by nine

Let this number be abcdefghi. In fact, the 5th bit e must be 5. Further analysis shows that the even bit must be even (bdfh = {2,4,6,8}), and the odd bit must be odd (acgi = {1,3,7,9}). Further analysis shows that 4 can divide 10 * C + D, so D = 2 or 6. In addition, 8 can divide 10 * G + H, so D, H = {2,6}, so B, f = {4,8}. Next analysis shows that 3 can divide 100 * D + 10 * 5 + F, so def = {258,654}, ABC, If def = 258, then ABC = {147741}, GHI = {369963}, but 1472589742589 cannot be divisible by 7, which does not meet the conditions, so def = 654, B = 8, H = 2. If 7 can be divisible by a8c654g, then 7 is divisible (a + 4C + G), and g = {3,7}, if G = 3, ABC {189789981987} does not meet the conditions, so g = 7, at this time, only 381 in ABC = {183189381981} meets the conditions, so abcdefghi = 381654729

Fill in the appropriate number in () so that the six digit () 1991 () can be divided by 66
Fill in the appropriate number in () so that the five digit 3 () 24 () can be divided by 9
Fill in the appropriate number in () so that the six digit 19 () 88 () can be divided by 35

19910 / 66 = 301 + 44
100000 / 66 = 1515 + 10
10*2+44+2 = 66*1
10*8+44+8 = 66*2
Therefore, six digits can be 219912, 819918
30240/9 = 3360
Therefore, the five digit number can be 30240, 30240, 30240, 30283, 30283, 30284
190880 / 35 = 5453 + 25
1000 / 35 = 28 + 20
20*2+25+5= 35*2 ,20*4+25+0= 35*3 ,20*9+25+5= 35*6
Therefore, six digits can be 192885, 194880, 199885

Fill in the appropriate number so that the six digit 43217 can be divided by 4 and 25?

There should be no solution
First of all, if we look at the division by 25, that is, if we divide by 5 and then by 5, only 0 meets the conditions, but if we look at the division by 4, that is, if we divide by 2 and then by 2, only 4 and 8 have no intersection, so there is no solution to this problem

Fill in the appropriate number in () so that six digits 19 () 88 () can be divided by 45

Fill in the appropriate number in () so that the six digit 19 (1) 88 (0) can be divided by 45
191880 divided by 45 = 4264

Any three digit number can be divided by seven if it is changed into six digits twice. I want to know why it can be divided by seven and what other numbers it can be divided by?

If any three digit is written twice to six digit, the six digit is 1001 times of the original three digit
So break down the factor, this six digit number
=Original three digits * 1001
=Original three digits * 7 * 143
=Original three digits * 7 * 11 * 13
So this number can be divided not only by 7, but also by 11 and 13

If six digits () 1991 () can be divided by 45, what is the six digit number?

719910 or 219915
Divisible by 5 indicates that the end is 0 or 5
It can be divided by 9, which means that the sum of each digit is a multiple of 9
So for the above two answers

If there is a 1994 digit a which can be divided by 9, the sum of its digits is B, and the sum of its digits is C, then C =?