What is the sum of all 3-digit positive integers such that all the dig

hundred place = 2 4 6 8 4 number
ten place = 0 2 4 6 8 5 number
one place = 0 2 4 6 8 5 number
total number = 4*5*5 = 100
smallest number = 200
largest number = 888
average = (200+888)/2 = 544
sum = 544*100 = 54400

­Total possible such numbers could be 4*5*5 = 100
So at unit's and ten's place, each of 0, 2, 4, 6, 8 can come (100/5) 20 times. At hundred's place, each of 2, 4, 6, 8 can come (100/4) 25 times.

Numbers will include the 200s, 400s, 600s and 800s. Within each 00s tens digit can be 0, 2, 4, 6 and 8, and this is the same for ones digit

So there are 5*5=25 possible numbers for each 00s, so 100 total numbers

The avg. value for the set is largest number (888) plus smallest number (200) = 1088/2 = 544

100 total numbers * avg value of 544 = 54400­

Can you explain why each of 2,4,6,8 will appear 25 times? Sorry just not able to grasp it..

__PRESENT

Here are the numbers that satisfy the condition:

{200, 202, 204, 206, 208} {220, 222, ..., 228} {240, 242, ..., 248} {260, 262, ..., 268} {280, 282, ..., 288} - total of 25 numbers, 5 in each of 5 groups of numbers

• The sum of the hundreds digits: 25 * 2
• The sum of the tens digits: 5 * (0 + 2 + 4 + 6 + 8)
• The sum of the units digits: 5 * (0 + 2 + 4 + 6 + 8)

{400, 402, 404, 406, 408} {420, 422, ..., 428} {440, 442, ..., 448} {460, 462, ..., 468} {480, 482, ..., 488} - total of 25 numbers, 5 in each of 5 groups of numbers

• The sum of the hundreds digits: 25 * 4
• The sum of the tens digits: 5 * (0 + 2 + 4 + 6 + 8)
• The sum of the units digits: 5 * (0 + 2 + 4 + 6 + 8)

{600, 602, 604, 606, 608} {620, 622, ..., 628} {640, 642, ..., 648} {660, 662, ..., 668} {680, 682, ..., 688} - total of 25 numbers, 5 in each of 5 groups of numbers

• The sum of the hundreds digits: 25 * 6
• The sum of the tens digits: 5 * (0 + 2 + 4 + 6 + 8)
• The sum of the units digits: 5 * (0 + 2 + 4 + 6 + 8)

{800, 802, 804, 806, 808} {820, 822, ..., 828} {840, 842, ..., 848} {860, 862, ..., 868} {880, 882, ..., 888} - total of 25 numbers, 5 in each of 5 groups of numbers

• The sum of the hundreds digits: 25 * 8
• The sum of the tens digits: 5 * (0 + 2 + 4 + 6 + 8)
• The sum of the units digits: 5 * (0 + 2 + 4 + 6 + 8)

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• The sum of the hundreds digits:

25 * 2 + 25 * 4 + 25 * 6 + 25 *8 = 25(2 + 4 + 6 + 8)

• The sum of the tens digits:

5 * (0 + 2 + 4 + 6 + 8) + 5 * (0 + 2 + 4 + 6 + 8) + 5 * (0 + 2 + 4 + 6 + 8) + 5 * (0 + 2 + 4 + 6 + 8) = 20 * (0 + 2 + 4 + 6 + 8)

• The sum of the units digits:

5 * (0 + 2 + 4 + 6 + 8) + 5 * (0 + 2 + 4 + 6 + 8) + 5 * (0 + 2 + 4 + 6 + 8) + 5 * (0 + 2 + 4 + 6 + 8) = 20 * (0 + 2 + 4 + 6 + 8)

Hope this helps.

But this method of calculating the average only works when the entire set is evenly spaced, and this set is evenly spaced for five integers at a time. Am I missing something here?

The method works here because we are considering only 100 numbers, let's suppose we considered all even integers from 200 to 888, we will have 345 numbers and their sum would be 544*345 = 187680, now if we remove the 100 numbers (the ones with all even digits), we will have 245 numbers, 544*245 = 133280. If we subtract 133280 (245 nos. avg.) from 187680 (345 nos. avg.) we get => 187680 - 133280 = 54400 => The average of 100 numbers all whose digits are even! So the numbers are present in a symmetry, hence Avg. / AP is valid here.

"f we remove the 100 numbers (the ones with all even digits), we will have 245 numbers, 544*245 = 133280." How did you arrive at 544 being the average? Did you assume again the set is symmetric? Even if you did, won't the average be (886+200)/2 =543 as 888 will be removed from the set?

I didn't follow why 888 is removed, as 886 and 200 are also all even digits. However, the point I was trying to convey was if we consider all digits (345), and find the average i.e. 544 and then when we selectively consider digits be it 100 all even digit no.s or the remaining 245 no.s keeping the avg. same, they give us sums that end up correct, indicating there is a symmetry. I won't recommend to calculate the average this way, as you have correctly mentioned that the method is only for evenly spaced sets, but it is working out in this problem due to some symmetry.