11.1.30
How many three-digit even numbers are possible if the leftmost digit cannot be zero?

There are 450 three-digit even numbers where the leftmost digit cannot be zero

The smallest even three-digit is 100, and the largest even three-digit is 998.

The difference between consecutive even numbers is 2.

So, the count of even numbers is calculated using the following arithmetic progression formula

Where:

-- the last term

-- the first term

-- the common difference

So, we have:

Subtract 100 from both sides

Divide both sides by 2

Add 1 to both sides

Hence, there are 450 three-digit even numbers

Read more about arithmetic progressions at:

link

The total number of 3 digit even numbers is 450

Step-by-step explanation:

Step 1 :

We need to determine the  number of 3 digit numbers which is even and the leftmost digit is not zero .

The  3 digits with left most digit not equal to zero starts from 100 and goes up to 998

If we consider only the even 3 digit numbers in this interval, this would form an arithmetic progression with the first number a = 100 and the common difference d = 2.

Step 2:

The last number l  in this series is 998 .

So we have

a = 100

l = 998

d = 2

The nth term in the arithmetic progression is given by a + (n-1) d

so substituting the above values we get

100 + (n-1) 2 = 998

2n = 900  => n = 450

Step 3 :

Answer :

The total number of 3 digit even numbers is 450

its d. the first coordintaes are 10,32

step-by-step explanation:

thats not a system of equations, that is a solution