Question

Find the sum of all 3-digit natural numbers which contain at least one odd digit
and at least one even digit.

Answer 1

Its a very hard maths problem
Lets see who can solve this

Answer 2

let me give it a shot, find 100- 199 using AP with common difference 1, then 300-399, 500 -599, 700-799, 900-999..

Answer 3

ohh no that won't work .. crap! sorry :/

Answer 4

dont worry

Answer 5

try again

Answer 6

you could do something similiar to that though

Answer 7

except the difference would be 2

Answer 8

also there would be a few more cases to consider.

Answer 9

yea there will too many cases to consider with that approach! :/

Answer 10

ohh I dont think so

Answer 11

yea, find 1 to 99 sum using AP with common difference 2.. that ll always have one odd, then keep adding, 200, 400, 600, 800 to it.. next find 0 to 99 sum using AP with common difference 2.. that ll always have one even, so keep adding, 100, 300, 500, 700, and 900.. would that do?

Answer 12

or is this total bul pellet?!

Answer 13

dont understand sry

Answer 14

my logic was, divide 100 to 999 into set of two groups, ones with hundreths place even and one with hundreths place odd.. next for the ones with hundreths place even, we find ap, such that those numbers will always have one even in them.. for example.. if we consider from 100 - 200.. this has 1 in hundredths place, so we need even so consider 102, 104, 106 ..

Answer 15

try it
wats the answer

Answer 16

dere are two choices in ques
370775
646943

Answer 17

damn, i have no clue. and my logic is flawed too.. ! :/

Answer 18

dont worry

Answer 19

I started by finding that the sum of all the numbers from 100 - 999 is
\[\frac{(999)(1000)}{2}-\frac{(99)(100)}{2}=494500\]
Next, I figured that it would be easier to count how many numbers DON'T have one even and one odd. This would mean that ALL 3 digits would need to be even or odd.

Evens:
Start with the interval 200-208. We have 200+202+204+206+208=1020
Consider the interval 220-228. Each number is 20 more than before and there are 5 numbers, which means the sum will be 100 more than before. The sum of the numbers in this interval is 1120.
Consider the interval 240-248. We have 1120+100=1220.
Consider the interval 260-268. We have 1220+100=1320.
Consider the interval 280-288. We have 1320+100=1420.
So, from 200 to 299, the sum of all the numbers where all 3 digits are even is 1020+1120+1220+1320+1420=6100.

Consider the interval 400-499. Each number is 200 larger than before, which gives us an extra 1000 on top of each total.
5*1000+6100=11100.
Consider the interval 600-699. 5000+11100=16100.
Consider the interval 800-899. 5000+16100=21100.
So from 100 to 999, the sum of the numbers with all even digits is 6100+11100+16100+21100=54400.


Odds:
Do the same thing.
111-119: Sum is 575
191-199: Sum is 975.

111-199: Sum is 3875.
311-399. Sum is 8875.
511-599. Sum is 13875.
711-799. Sum is 18875.
911-999. Sum is 23875.

Sum of all the numbers with odd digits from 100 to 999 is 69375.


The sum of all the numbers with all odd or all even digits from 100 to 999 is 54400+69375=123775.

Subtract this from 494500 to get all the numbers that are NOT composed of all odd or all even digits.

370725
Check my math. I caught a lot of mistakes since I started...

Answer 20

Very very good

Answer 21

haaa