Let E(n) denote the sum of the even digits of n. For example, `E (1243)=2 +4 =6.` What is the value of `E(1) + E(2) + E(3)+…….+ E(100)?`

Answer

To find the value of \( E(1) + E(2) + E(3) + \ldots + E(100) \), we will break down the problem into two parts: the contribution from the unit place and the contribution from the tens place.

### Step 1: Contribution from the Unit Place
We first analyze the numbers from 1 to 100 based on their unit digits. The even digits are 0, 2, 4, 6, and 8.

- For the unit place, we can consider the numbers ending in 0, 2, 4, 6, and 8 from 1 to 10:
- \( E(0) = 0 \)
- \( E(1) = 0 \)
- \( E(2) = 2 \)
- \( E(3) = 0 \)
- \( E(4) = 4 \)
- \( E(5) = 0 \)
- \( E(6) = 6 \)
- \( E(7) = 0 \)
- \( E(8) = 8 \)
- \( E(9) = 0 \)

The sum of the even digits from 1 to 10 is:
\[ E(1) + E(2) + E(3) + \ldots + E(10) = 0 + 2 + 0 + 4 + 0 + 6 + 0 + 8 + 0 + 0 = 20 \]

Since this pattern repeats for every complete set of ten numbers (i.e., 1-10, 11-20, ..., 91-100), we will have this sum repeated 10 times (for the tens from 0 to 9).

Thus, the total contribution from the unit place is:
\[ 20 \times 10 = 200 \]

### Step 2: Contribution from the Tens Place
Next, we analyze the contribution from the tens place. The even digits in the tens place are also 0, 2, 4, 6, and 8.

- The tens place contributes as follows:
- From 00 to 09: \( E(0) = 0 \)
- From 10 to 19: \( E(1) = 0 \)
- From 20 to 29: \( E(2) = 2 \) (repeated 10 times)
- From 30 to 39: \( E(3) = 0 \)
- From 40 to 49: \( E(4) = 4 \) (repeated 10 times)
- From 50 to 59: \( E(5) = 0 \)
- From 60 to 69: \( E(6) = 6 \) (repeated 10 times)
- From 70 to 79: \( E(7) = 0 \)
- From 80 to 89: \( E(8) = 8 \) (repeated 10 times)
- From 90 to 99: \( E(9) = 0 \)

Now, we calculate the contribution from the tens place:
- Contribution from 20 to 29: \( 2 \times 10 = 20 \)
- Contribution from 40 to 49: \( 4 \times 10 = 40 \)
- Contribution from 60 to 69: \( 6 \times 10 = 60 \)
- Contribution from 80 to 89: \( 8 \times 10 = 80 \)

Adding these contributions gives:
\[ 20 + 40 + 60 + 80 = 200 \]

### Final Calculation
Now, we sum the contributions from the unit place and the tens place:
\[ \text{Total} = 200 + 200 = 400 \]

Thus, the value of \( E(1) + E(2) + E(3) + \ldots + E(100) \) is \( \boxed{400} \).