Sum of even digits

Question

Let $E(n)$ denote the sum of the even digits of $n$.For example,$E(1243)=2+4=6$.

What is the value of $E(1)+E(2)+E(3)+....+E(100)$?

I got the answer after literally adding like $2+4+6+8+2...+6+8=400$ (for all $100$ numbers) Is there an easy tricky method for solving this?

How many times does each of $2$, $4$, $6$, $8$ appear as first digit? As second digit? — Robert Israel, Aug 14 '16 at 06:05
okay..so 2,4,6,8 appears as first digit for 10 times each and as second digit also for 10 times.. Therefore , 10(2+4+6+8)+10(2+4+6+8)=400. — Ushosee, Aug 14 '16 at 06:09
Yeah that's correct. By the way, if you want to ping a specific user with your comment you can include @(user's name), for example @Ushosee — Ovi, Aug 14 '16 at 06:12

Adrito Pal RKMV · Answer 1 · 2021-05-18T06:10:24.177

The sum of all the even digits in the numbers $1$ to $100$ is required and both inclusive. Observe that $100$ only two null even digits which doesn't add any significant value. So we remove them from our addition list. We make a list with the following numbers:-

$01, 02, 03, ..., 10, 11, ..., 98, 99$

There are $2\cdot 100 = 200$ in our list considering all, and each digit appears $\frac{200}{10} = 20$ times.[By H&T] or, [Formula]

Thus, each even digit appears $20$ times, and the sum of all the even digits is $0 \cdot 20 + 2\cdot 20 + 4\cdot 20 + 6\cdot 20 + 8\cdot 20 = (0 + 2 + 4 + 6 + 8)\cdot 20 = 400$, and the correct answer is $20\cdot 20 = 400$

$ THANK YOU$

Sum of even digits

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