How many integers from 1 to 10000 have the sum of their digits equal to 5

How many integers from 1 to 10000 have the sum of their digits equal to 5

One digit: 5 [1] two digit: composed of 1, 4 or 2, 3 or 5, 0, three digit: composed of 1, 2, 2 or 1, 1, 3 or 0, 1, 4 or 0, 2, 3 or 0, 0, 5 [2.3 take 2 combination number + 2.2 take 1 combination number + 1]

If there is an integer greater than l, divide it by 300, 262 and 205 to get the same remainder

Divide the number by 300 and 262 to get the same remainder, so divide the number by 300-262 = 38. Similarly, divide the number by 262-205 = 57, so it is the common divisor 19 of 38 and 57

The sum of five consecutive integers is 60
Write all five numbers,

10,11,12,13,14
(60 / 5 = 12 is the middle number)

Five consecutive integers, if the middle one is a, then the sum of the five integers is?

(a-1)+(a-2)+a+(a+1)+(a+2)=5a

The sum of five consecutive integers is 55, and the five numbers are?

9\10\11\12\13

For five consecutive integers, the sum of squares of the first three numbers equals the sum of squares of the last two numbers

Let these five consecutive integers be n, N + 1, N + 2, N + 3, N + 4, N2 + (n + 1) 2 + (n + 2) 2 = (n + 3) 2 + (n + 4) 2, and the solution is n = 10 or n = - 2. When n = 10, these five numbers are 10, 11, 12, 13, 14. When n = - 2, these five numbers are - 2, - 1, 0, 1, 2

Three consecutive integers, the sum of the squares of the first two numbers is equal to the square of the last number. Can you try to find these three consecutive integers?

3,4,5
Let three numbers be X-1, x, x + 1
Set up an equation (x-1) 2 = x2 + (x + 1) 2
Just work out x = 4

Write nine consecutive integers so that the sum of squares of the first five numbers equals the sum of squares of the last four numbers

-4,-3,-2,-1,0,1,2,3,4

There is an integer greater than 1. Divide 300, 262 and 205 to get the same remainder. This integer is ()
Fill in the blanks

Let the number be m and the remainder be n
300-n is divisible by M
262-n can be divisible by M
205-n can be divisible by M
(300-n) - (262-n) = 38 divisible by M
(262-n) - (205-n) = 57 can be divisible by M
The common factor of M is 38 and 57 is 19

There is an integer that is not equal to 1. The remainder obtained by dividing it by 300262205 is the same. What is the integer?

300-262=38
262-205=57
This number can be divided by 38 and 57 at the same time
This number is 19