Answer to Question #331116 in Combinatorics | Number Theory for ash

Question

For how many three digit numbers (100 to 999) is the sum of the digits even? (For example, 343 has an even sum of digits: 3 + 4 + 3 10 which is even.) Find the answer and explain why it is correct in at least two different ways.

Answer 1

1.An odd amount will be if you add an even number and an odd number.

An even sum is formed by adding two even numbers, or two odd ones.

Thus, we are satisfied with all three-digit numbers in which any two digits are even and one is odd.

We are also satisfied with numbers in which all digits are odd.

You can already count.

Numbers in which any two digits are even and one is odd.

hhh+hhh+hhn (taking into account that the number cannot start from zero)

5*5*5 + 4*5*5 + 4*5*5 = 125+100+100 = 325

Numbers where all digits are odd

5*5*5 = 125

Total 125+325 = 450

2.Exactly half, i.e. 450.

Explanation.

Let's break the whole series 100-999 into dozens of numbers like NN0-NN9, i.e. 100-109, 110-119, ..., 200-209,...,990-999. Each such interval contains exactly 5 numbers, the sum of which is even: in those intervals where the sum of the digits of NN0 is even, these are the numbers NN0,NN2,NN4,NN6,NN8, and where it is odd - NN1,NN3,NN5,NN7,NN9.

And since the number of such intervals is (999-99) / 10 = 90 pieces, then all these numbers are 90 * 5 = 450 (well, or easier: the sum of all halves is equal to half of everything).

3.And the second for the rules of combinatorics.

Let's divide three-digit numbers into pairs of consecutive ones: ... In each pair, the sum of the digits of the second number is 1 more than the sum of the digits of the first, which means that exactly one of them is even. Therefore, numbers with an even sum of digits are exactly half the number of all three-digit numbers, and their

45*10²=450