0,2,4,6,8,

There are two ways to solve:

METHOD 1 - CRUDE METHOD:

Since the number has to be greater than 10000, the first digits (ten thousandth place) can only be 2 or 4 or 6 or 8. We take turns. (i) First digit is 2: The remaining four digits can be arranged in 4 remaining places in 4! ways. Therefore there are 24 such cases. Each of the remaining numbers 4,0,6,8 would have appeared in 2nd, 3rd, 4th and 5th digit exactly 6 times each respectively. Therefore the sum of all the 5th digit (Units place) will be 6 * 4 + 6 * 0 + 6 * 6 + 6 * 8 = 108. The sum of 4th digit (Tens place) is also same 108, and so will be 3rd digit and 2nd digit.

So, the sum of all 24 numbers is 108 * 1 + 108 *10 + 108 * 100 + 108 * 1000 + 24 * 20000. The first digit is always 2 in this case. Sum = 599,988

(ii) First digit is 4: Similar to previous case, here also , the remaining four digits can be arranged in 4 remaining places in 4! ways. Therefore there are 24 such cases. Each of the remaining numbers 2,0,6,8 would have appeared in 2nd, 3rd, 4th and 5th digit exactly 6 times each respectively. Therefore the sum of all the 5th digit (Units place) will be 6 * 2 + 6 * 0 + 6 * 6 + 6 * 8 = 96. The sum of 4th digit (Tens place) is also same 96, and so will be 3rd digit and 2nd digit.

So, the sum of all 24 numbers is 96 * 1 + 96 * 10 + 96 * 100 + 96 * 1000 + 24 * 40000. The first digit is always 4 in this case. Sum = 1,066,656

(iii) First digit is 6: Similar to previous cases, here also , the remaining four digits can be arranged in 4 remaining places in 4! ways. Therefore there are 24 such cases. Each of the remaining numbers 2,0,4,8 would have appeared in 2nd, 3rd, 4th and 5th digit exactly 6 times each respectively. Therefore the sum of all the 5th digit (Units place) will be 6 * 2 + 6 * 0 + 6 * 4 + 6 * 8 = 84. The sum of 4th digit (Tens place) is also same 84, and so will be 3rd digit and 2nd digit.

So, the sum of all 24 numbers is 84 * 1 + 84 * 10 + 84 * 100 + 84 * 1000 + 24 * 60000. The first digit is always 6 in this case. Sum = 1,533,324

(iv) First digit is 8: Similar to previous cases, here also , the remaining four digits can be arranged in 4 remaining places in 4! ways. Therefore there are 24 such cases. Each of the remaining numbers 2,0,4,6 would have appeared in 2nd, 3rd, 4th and 5th digit exactly 6 times each respectively. Therefore the sum of all the 5th digit (Units place) will be 6 * 2 + 6 * 0 + 6 * 4 + 6 * 6 = 72. The sum of 4th digit (Tens place) is also same 72, and so will be 3rd digit and 2nd digit.

So, the sum of all 24 numbers is 72 * 1 + 72 * 10 + 72 * 100 + 72 * 1000 + 24 * 80000. The first digit is always 8 in this case. Sum = 1,999,992

Answer = Case (i) + Case (ii) + Case (iii) + Case (iv) = 599,988 + 1,066,656 + 1,533,324 + 1,999,992 = 5,199,960

METHOD 2 - SHORTCUT:

Sum of all numbers (formed using 2,0,4,6,8) greater than 10000 = Sum of all numbers (formed using 2,0,4,6,8) - Sum of all numbers (formed using 2,0,4,6,8) lesser than 10000

(i) Sum of all numbers formed using 2,0,4,6,8: These 5 numbers can be placed in 5 different digits in 5! ways. (120). Therefore there are 120 numbers in total. Each of the five numbers 0,2,4,6,8 will appear 24 times in each of the five digits respectively. Therefore the sum of all the units place is 24 * 0 + 24 * 2 + 24 * 4 + 24 * 6 + 24 * 8 = 480. The sum of tenth place, 100th place, 1000th place and 10000th place will also be 480. So, the sum of all the 120 numbers is 480 * 10000 + 480 * 1000 + 480 * 100 + 480 * 10 + 480 * 1 = 5,333,280

(ii) Sum of all numbers formed using 2,0,4,6,8 which is less than 10000 : For the value to be lesser than 10000, the first digit (10000th place) has to be 0. Therefore, the remaining 4 numbers (2,4,6,8) can be placed in the remaining 4 different digits in 4! ways. (24). Therefore there are 24 numbers in total. Each of the four numbers 2,4,6,8 will appear 6 times in each of the four digits respectively. Therefore the sum of all the units place is 6 * 2 + 6 * 4 + 6 * 6 + 6 * 8 = 120. The sum of tenth place, 100th place and 1000th place will also be 120. So, the sum of all the 24 numbers is 120 * 1000 + 120 * 100 + 120 * 10 + 120 * 1 = 133,320

Answer = Case (i) - Case (ii) = 5,333,280 - 133,320 = 5,199,960

LONGER, EASIER TO UNDERSTAND (for most) VERSION

We know any permutation of the 5 digits in which 0 isn't the first digit is a valid number, since the smallest of these is 20468.

Reading the number from left to right, let A be the first non-zero number encountered, B the second, C the third and D the fourth.

So given the permutation A,B,C,D = 4,2,8,6

The only numbers we can make are

• 42860

• 42806

• 42086

• 40286

First, we will consider the sum of all values for a unique permutation of ABCD. The numbers we can form are (with _ denoting a 0):

• ABCD_

• ABC_D

• AB_CD

• A_BCD

So the sum consists of:

• (A x 10,000) 4 times

• (B x 1000) 3 times

• (B x 100) 1 time

• (C x 100) 2 times

• (C x 10) 2 times

• (D x 10) 1 time

The sum of the above four numbers is:

(4 x A x 10000) +

(3 x B x 1000) + (1 x B x 100) +

(2 x C x 100) + (2 x C x 10) +

(1 x D x 10) + (3 x D x 1)

=

40,000A + 3100B + 220C + 13D

Which is the sum for a particular permutation of A,B,C,D.

However, when considering all permutations of A,B,C,D, the distribution of each number among each letter is equal (A is equally likely to be 2,4,6 or 8; same for the other letters). So when considering all permutations, each letter has an average value of (2 + 4 + 6 + 8)/4 = 5.

So coming back to the above sum for a given permutation, when we extend this to be for all permutations, A = B = C = D = 5, so on average , each permutation will sum to:

40,000A + 3100B + 220C + 13D

= 5*(40,000 + 3100 + 220 + 13)

= 5*(43333)

Since that's the sum of one permutation, we multiply by the total number of permutations for 2,4,6,8, which is 24. So the answer is 24 5 43333 = 5,199,960.

‌ ‌

SHORTER VERSION, more general:

Ignore the restriction that the number must be > 10,000. So we now consider all permutations of the digits 0,2,4,6,8.

Let A,B,C,D,E be the digits from leftmost (A) to rightmost (E) Now consider all the possible values each letter can be.

Letters A,B,C,D,E can be any number from {0,2,4,6,8} with equal representation. Thus on average A = B = C = D = E = the average value of the possible digits each of these letters can be (in this case 4). Call this average value X

Thus, when considering the sum, each number can be simplified into: XXXXX.

Since there are 5! = 120 different numbers of the form XXXXX, the sum is 120*XXXXX

which, in this case, is 120*44444.

Repeat the above to calculate the sum of all numbers below 10,000, to get 24*5555.

So the sum for all values above 10,000 is (120 x 44444) – (24 x 5555)

= 5,199,960.

We have to find the sum of all the numbers > 10000 formed by digits 0,2,4,6,8(with no repetition), so 0 cannot be the 1st digit.

Then the possible numbers can be :

$20468, 20486, 20648......$

We can write $20468 = 2 \times {10}^{4} + 0 \times {10}^{3} + 4 \times {10}^{2} + 6 \times 10 + 8$

Hence, SUM = $2 \times {10}^{4} + 0 \times {10}^{3} + 4 \times {10}^{2} + 6 \times 10 + 8 + 2 \times {10}^{4} + 0 \times {10}^{3} + 4 \times {10}^{2} + 8 \times 10 + 6.......$

1 Now, there will be $4! 2's$ , similarly, there will be $4! 4's, 6's, 8's$ for the first digit.

$= 2 \times 4! \times {10}^{4} + 4 \times 4! \times {10}^{4} + 6 \times 4! \times {10}^{4} + 8 \times 4! \times {10}^{4}......$

$= 20 \times 4! \times {10}^{4} +........$

2 For the 2nd digit, there will be $3! \times 3$ each term. This is because

[let 4 be the digit we are dealing with here] with 2 as the first digit, there will be $3!$ , with 6 as the first digit, there will be $3!$ , with 8 as the first digit, there will be $3!$ but with 4 as the first digit, there will be 0

As a result,

SUM = $20 \times 4! \times {10}^{4} + 20 \times 3! \times 3 \times {10}^{3} + .....$

3 Now, for the 3rd digit, there will be $3! \times 3$ each term too. Because:

[let 4 be the digit we are dealing with here] With 2 as the first digit and

with 0 as the 2nd digit, there will be $2$ , with 6 as the 2nd digit, there will be $2$ , with 8 as the 2nd digit, there will be $2$ as well. So, there will be $3 \times 2$

Hence, SUM = $20 \times 4! \times {10}^{4} + 20 \times 3! \times 3 \times {10}^{3} + 20 \times 3! \times 3 \times {10}^{2}.....$

4 Similarly, for the 4th and fifth digit too, there will be $3! \times 3$ each term as well.

As a result, SUM = $20 \times 4! \times {10}^{4} + 20 \times 3! \times 3 \times {10}^{3} + 20 \times 3! \times 3 \times {10}^{2} + 20 \times 3! \times 3 \times 10 + 20 \times 3! \times 3 \times 1$

= $\boxed{5199960}$