Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862. Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862.


My Approach:

single digit no formed = 2,4,6,8

sum= 2+4+6+8= 20

two digit= 24,26,28,42,46,48,62,64,68,82,84,86

sum= 660

three digit no=246+264+426+462+624+642=2664

268+286+628+682+826+862=3552

248+284+428+482+824+842=3108

468+486+648+684+846+864=3996

sum of all 3 digit nos =13310

Similarly for all 4 digit numbers.
There are:


four $1$-digit numbers, whose average is $5$: sum = $20$
twelve $2$-digit numbers, whose average is $55$: sum = $660$
twenty-four $3$-digit numbers, whose average is $555$: sum = $13320$
twenty-four $4$-digit numbers, whose average is $5555$: sum = $133320$


To see why these averages are correct, you can pair off each number uniquely with its 'complement' obtained by exchanging $2$ with $8$ and $4$ with $6$. The average of each of these pairs is $5\dots5$.
Consider an example by taking digits as $3,4,6,8$.
total combinations possible$=4\times 3\times 2=24$ numbers possible
no of digits$=4$
hence $\dfrac{24}{4}$ each digit comes $6$ times in ones, tens and hundred's place
at ones place=$6\times (3+4+6+8)=126\equiv6$.
at ten's place$=6\times (3+4+6+8)+12(\text{carry})=138\equiv 8$.
At hundred's place$=6\times (3+4+6+8)=139\equiv 9$
so, $3$ will come at thousands place and $1$ at ten thousands place,
Hence, the number is $139864$.