If a 1004 15s and 2008 3s are multiplied, then what is the sum of the digits on all digits of the in

Let l be the number of digits of a, then l = [lg (a)] 1, where [m] denotes the integer portion of m.

Let k = lg a;

k = lg 15^1004 * 3 ^ 2008 = lg 5^1004 * 3^（2008+1004）

lg 5 1004 + lg 3 30121004*lg 5 + 3012*lg 3 and lg3=,lg5=.

So k = 1004*

l = [k] +1 = 2139

Finding the number of bits l of a decimal number a can be solved by using a logarithmic function, l = [log k (a)] 1, this lz can be known by thinking about it yourself, and it is not proved here.

Suppose the three-digit number is Huitong 121, and the integer number of tens is 10,121 10=1210;

Therefore, three digits are multiplied by whole tens, and there is at least 1 rolling circle 0 at the end of the big collapse of the product;

Therefore, a

According to the characteristics of the number divisible by 11, we get:

The sum of the numbers on the number of the odd brothers: a+a+a+a+a+a=5a, the sum of the numbers on the even digits: 5+4+3+2+1=15, because 5a-15=0, so the solution is a=3

Therefore, this integer attendant is 1323334353

abc is any three integers, and according to the principle that integers are not even or odd, three integers are nothing more than the following four combinations:

Three even numbers, two even numbers, one odd number, one even number, two odd numbers, three odd numbers.

The even numbers of a+b, b+c, and c+a they are composed of are:

Case 1: 3 even numbers.

Case 1: 1 even number.

Case 1: 1 even number.

Case 1: 3 even numbers.

The product of three numbers, the product is odd only if all three numbers are odd. In other cases, as long as there is at least one even number of the three numbers, their product will be even.

According to the four cases analyzed above, at least one of the three numbers is even, so no matter how the original three numbers are given, the final product must be even.

is an even number, a+b, b+c, c+a, as long as one of the three results is even, the product is determined to be even, because the product of odd and even numbers is even.

Regardless of the parity of a, b, c, or three numbers, their sums cannot all be odd.

a+b)(b+c)(a+c) is an even number.

a+b, b+c, a+c must have an even number.

For example, the odd number 3, 3 5 = 15, and 15 is the odd number buried;

3+1=4, 4 is an even number;

3 3 = 1, 1 is an odd number;

3-2=1, 1 is an odd number;

So if you add 1 to an odd number in a medium, the result is an even number

Therefore, b

From the analysis, it can be seen that if an even number is added to 1, the result is an odd number;

Therefore, a

An odd number if (), the result is even.

Select b plus 1

Analysis: Odd + Odd = Even.

1515…15×333…3

50505…05×999…9(505…05 total of 2007 digits, 999....9 total of 2008 digits).

50505…05×（1000…000-1）=50505…05000…000-50505…05=50505…050494949…49495 (front 50505....0504 has a total of 2007 bits, 9 bits in the middle, 1 bits in the middle, and 4949 bits in the end....49495 out of 2007).

5050505 ahead....04 plus last 4949....49495 is exactly 2007 9s, and then it is a 9 in the middle, so the sum of all digits is 9*2008=18072

a= 1515...15 (1004 15s) multiplied by 333....33 (2008 3, i.e. 1004 33).

15×0101…01 (1004 01) 33 0101....01 (1004 01).

45×0101…01 (1004 01) 11 0101....01 (1004 01).

45×0101…01 (1004 01) 9999....99 (1004 99) 9

0505…05 (1004 05) 9999....99 (1004 99).

5050…50 (1004 50) [10000.]00 (1004 00)-1】 10

5050…50 (1003 01) 4949....49 (1004 49) 50 10

5050…50 (1003 50) 4949....49 (1004 49) 5

So the sum of all digits = (5 + 4 + 9) 1004 = 18072