Known: 1111 = 0 5231 = 0 2222 = 0 0000 = 4 3021 = 1 Known: 1111 = 0 5231 = 0 2222 = 0 0000 = 4 3021 = 1 9821 = 3 4825 = 2 4309 = 2 Q: 5889 =?

Known: 1111 = 0 5231 = 0 2222 = 0 0000 = 4 3021 = 1 Known: 1111 = 0 5231 = 0 2222 = 0 0000 = 4 3021 = 1 9821 = 3 4825 = 2 4309 = 2 Q: 5889 =?

five

What's the product of dividing 5.1 by 1.7, adding 5.12 and multiplying by 3.1?

[(5.1÷1.7)+5.12]×3.1
=(3+5.12)×3.1
=26 and 6 / 31

What is the product of 10 and the quotient of 3.5 divided by 0.7 and multiplied by 0.2?
It's additive

Product = (10-3.5 △ 0.7) × 0.2 = 1

What's the product of the sum of 4 / 5 plus the quotient of 1 / 4 divided by 3 / 4 multiplied by 1 / 3?

1 / 3 (4 / 5 + 1 / 4 divided by 3 / 4)
=1/3(4/5+1/3)
=4/15+1/9
=12/45+5/45
=17/45

Is the nine digit number of 123456789 prime?

Use the nine numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 to form prime numbers. If each number is used only once, how many prime numbers can these nine numbers form at most? Write these prime numbers
1+(9-8)+(7-6)+(5-4)+(3-2）=5

Write down the 2010 natural numbers from 1 to 2010 in turn to get a multi digit number 123456789.2010. What is the remainder of this multi digit number divided by 9?

The answer is 6
Firstly, we study the characteristics of the number divisible by 9: if the sum of the numbers in each digit can be divisible by 9, then the number can also be divisible by 9; if the sum of the numbers in each digit can not be divisible by 9, then the remainder is the remainder of the number divided by 9
Solution: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45; 45 can be divided by 9
And so on: in 1999, the sum of the digits of these numbers can be divided by 9
10~19,20~29…… 90 ~ 99 of these numbers, the number on the tenth digit appears 10 times, then the sum of the number on the tenth digit is 10 + 20 + 30 + +90 = 450, it can be divided by 9
In the same way, the sum of 100-900 digits is 4500, which is also divisible by 9
That is to say, the sum of the numbers in each bit of these continuous natural numbers from 1 to 999 can be divided by 9;
In the same way, the sum of the 100, 10, and individual digits of 1000-1999 natural numbers can be divided by 9 (the "1" on the 1000 digits has not been considered, and we need to leave 1 from the answer here
Firstly, we study the characteristics of the number divisible by 9: if the sum of the numbers in each digit can be divisible by 9, then the number can also be divisible by 9; if the sum of the numbers in each digit can not be divisible by 9, then the remainder is the remainder of the number divided by 9
Solution: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45; 45 can be divided by 9
And so on: in 1999, the sum of the digits of these numbers can be divided by 9
10~19,20~29…… 90 ~ 99 of these numbers, the number on the tenth digit appears 10 times, then the sum of the number on the tenth digit is 10 + 20 + 30 + +90 = 450, it can be divided by 9
In the same way, the sum of 100-900 digits is 4500, which is also divisible by 9
That is to say, the sum of the numbers in each bit of these continuous natural numbers from 1 to 999 can be divided by 9;
In the same way, the sum of the hundreds, tens, and ones of these continuous natural numbers from 1000 to 1999 can be divided by 9 (here, the "1" on the thousands has not been considered, and we are short of 20002001200220032004200520062007200820092010 here)
From 1000 to 1999, the sum of 1000 "1" is 1000, divided by 9 to get 1;
20002001200220032004200520062007200820092010
The sum of the numbers is 68, divided by 9 to make 5
The final answer is 5 + 1 = 6

Write down the 2009 natural numbers from 1 to 2009 in turn to get a multi digit number 123456789.2009. What is the remainder of this multi digit number divided by 9?
If title, had better have specific train of thought
It seems not quite right!

Add all the numbers first, then divide by 9 to get the remainder
#include
using namespace std;
int main()
{
int i,s,a;
s=0;
for(i=1;i

Starting from natural number 1, the multi digit number composed of continuous 2007 Natural Numbers: 123456789101112 ········ 20062007, what is the sum of all digits?

Thousand bits: 1 × 1000 + 2 × 8 = 1016
Hundred: 0 × 108 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) × 200 = 9000
Ten: 0 × 198 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) × 200 = 9000
Individuals: (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) × 200 + (1 + 2 + 3 + 4 + 5 + 6 + 7) = 9028
1016 + 9000 + 9000 + 9028 = 28044

The sum of 47 different natural numbers is 2006, and there are at most several odd numbers in the 47 natural numbers

Because 1 + 3 + 5 + 7 + 91 = (1 + 91) × 46 △ 2 = 2116 (i.e. 46 odd numbers and at least = 2116)
2116-2006=110
And 91 + 19 = 110
So there are 46-2 = 44 odd numbers at most

1 -- how many ones are there in 2006 natural numbers

The numbers between 1601.1-9, 20-29, 30-39, 40-49,..., 90-99 contain 1 each. 1,10-19 have 11 ones, so 1-99 has 9 * 1 + 11 = 20, 1.100-199 has 100 + 20 = 120, 1200-299300-399,..., 900-999 has 20 1 each. So 1-999 has 9 * 20 + 120 = 300, 1.1000-1999 has 1000 + 300 = 1300