Find The Sum Of Odd Digits

from 00 to 99, total digits used = 2x100=200. Each digit from 0 to 9 is used equal number of times. i.e. 20 times. so, add 1,3,5,7,9, = 25. times 20. = 500. Add 1 for 100. Answer is 501.

Considering all integers between $1$ and $100$ inclusive, we see that:

• Separating the first $99$ numbers into $10$ sets of $10$ numbers (let the first set include $0$ ) such that each set's elements have the same digit in the ten's place, we see that $5$ of the $10$ numbers in each set contain an odd digit in the one's place (the digits are $1$ , $3$ , $5$ , $7$ , and $9$ ).
• $5$ of the $10$ sets will have an odd digit in the ten's place (10's, 30's, 50's, 70's, and 90's).
• $O(100)=1$ .

We can now add the odd digits occurring in the ten's place and those occurring in the one's place separately:

$\sum_{i=1}^{100} O(i)$ $=10(1)+10(3)+10(5)+10(7)+10(9)+10(1+3+5+7+9)+1=\boxed{501}$ .

Just to clarify, $10(1)+10(3)+10(5)+10(7)+10(9)$ is the sum of the odd ten's digits, $10(1+3+5+7+9)$ is the sum of the odd one's digits, and the lone $1$ is $O(100)$ .

Note that 1+3+5+7+9 occurs every once every 10 cycles in the last digit position

Note that 1+3+5+7+9 occurs 10 times in every 100 cycles in the second last digit position

(this can be extended... the sequence occurs 100 times in every 1000 cycles in third lasts position ... etc.)

1+3+5+7+9 = 25

There are 10 complete cycles of 10 in last position --> 10 x 25 = 250

There is 1 complete cycle of 100 in second last position --> 1 x 10 x 25 = 250

There is an additional 1 to add for O(100) --> 1

Total = 501

note that in $1-99$ , there are $20$ numbers of each digit

so , the sum is $20 *1+20*3+20*5+20*7+20*9=500$ , adding $1$ ( from $100$ ,we got $501$

Note that O(1) + O(2) + · · · + O(9) = 25, so in each group of ten, the sum of the units digits is 25. The sum of the tens digits gives us another ten groups of 25, and the hundreds digit of 100 gives us an additional 1. Thus, the sum is (10 + 10) · 25 + 1 = 501

Note that the sum of the first n odd positive integers is $n^2$ . From 1 to 9, this is $5^2 = 25$ . Using this fact, we can take the sum of the odd integers which occur in each decimal place.

• Ones place: The odd digits go from 1 to 9 and then repeat at the next factor of 10: 1 to 9, 11 to 19, 21 to 29, and so forth. There are 10 such cycles from 1 to 100, yielding $25 \times 10 = 250$ .

• Tens place: We can also think of these in sets of consecutive odd digits: the numbers ending in 1 from 11 to 91, the numbers ending in 2 from 12 to 92, and so forth. This too yields $25 \times 10 = 250$ .

• Hundreds place: $O(100) = 1.$

Adding our results together, we get $250 + 250 + 1 = \boxed{501}.$

from 00 to 99, we write 200 digits in total, (2 digits in every number, a total of 100 numbers), and also, we can be sure that we write each digit equal number of times. So, this means we write each digit 200/10, ie. 20 times. So, from O(0)...O(99) sums as 20*(1+3+5+7+9), that equals 500. Now, O(100) equals 1, so, answer for the question is 501.

First the sum of the units is 1 +3 +5 +7 +9 = 25, we see that this occurs 10 times => the sum of the odd units is 250, then the odd dozens are 1,3,5,7,9 and each appears 10 times => is 1 * 10 +3 * 10 +5 * 10 +7 * 10 +9 * 10 = 10 (1 +3 +5 +7 +9) = 250, there is only the odd digit numbers 100 which is 1 => 250 +250 +1 = 501

I may have did it the long way!

from 1 to 100 divide the nos. into 2 groups

• ODD and EVEN

• solving EVEN numbers only.

  from 0-8 , 20-28 , 40-48 , 60-68, 80-88 sum is 0

  from 10-18 , 30-38 , 50-58 , 70-78 , 90-98 sum is resp. 5, 15, 25, 35 ,45 therefore total = 125

• solving ODD numbers only.

  from 1-9 , 21-29 , 41-49 , 61-69 , 81-89 each group sum is 25 therefore total = 125

  form 11-19, 31-39, 51-59, 71-79,91-99 sum is resp. 30, 40, 50, 60, 70 therefore total = 250

  and we also have O(100) = 1

- ADDING ALL 125+125+250+1= 501

1to10 = 1+3+5+7+9 (=25)+(1in 10) 11 to 20 = 25+ 9 21 to 30= 3+25(next sum is 3 9 + 25) 31 to40= 27+25 41to50= 5+25 (next sum is 5 9 + 25) 51to60=45+25 similarly up to 100

you get 501

. . . .

We proceed using a counting argument. Suppose all of the integers from $001$ to $100$ are lined up vertically. First of all, consider the units digits of each of these numbers. For any given number in the tens position, there are five odd choices for the units digit: $1, 3, 5, 7,$ and $9$ . Thus, the sum of all of odd units digits is $10(1+3+5+7+9) = 250$ . The case for the tens digit is symmetric with the units digit case, so the sum of all of the odd tens digits is also $250$ . But wait! We also must acknowledge the $1$ in the hundreds position in the number $100$ . Therefore, the total sum is $1+250+250 = \boxed{501}$ .

All even places in tenths have sum 25 each(20,40,60,80) Odd ones have 25 plus the corresponding number multilplied by 10(for example,30ths have 25+(10*3) number of odd dogits) finally 100 has one odd digit 1 sum = 501

For ease, note that $\sum_{i = 0}^9 O(i) = 25$ and that this is equal to $O(1) + O(2) + O(3) + ... + O(9)$

Whenever there's an odd digit in the ten's place, we add that (That digit * 10) to 25, since having an odd ten's digit essentially just adds that digit to each value of O(n) while that digit's relevant.

From here, it's straightforward to find the sum is:

25 (0 - 9)

35 (10 - 19)

25 (20 - 29)

55 (30 - 39)

25 (40 - 49)

75 (50 - 59)

25 (60 - 69)

95 (70 - 79)

25 (80 - 89)

115 (90 - 99)

1 (100)

Which you sum up to 501.

We can look at the sum of this sequence as the sum of ten individual sequences- 1-10, 10-20, 20-30, and so on. Evaluating the terms of these mini-sequences give ten mini sums. each mini sum can be placed into one of two groups, each with a different arithmetic progression. For instance. for all intervals of ten of numbers with an even tens place, the sum progresses in the pattern of (A sub(n-1) + 2). For the intervals of ten with an odd tens place, the mini sums progress in the pattern of (A sub(n-1) + 18). Using these patterns, we can predict the mini-sums to be 26, 34, 28, 52, 30, 70, 32, 88, 34, and 106 ( To see the pattern, look at every other term). adding these sums together gives 500, but we must add the odd digit of 100. Therefore, 500 + 1 equals an answer of 501.

What we are essentially trying to find is the number of times each odd digit appears from the positive integers 1-100.

Let us consider numbers less than 100 first.

Each digit $1, 3, 5, 7, 9$ appears 10 times in the units digit and 10 times in the tens digit, for a total of 20 times. This is because for each appearance in the units digit, there are 10 digits that can occupy the tens digit (we do need to count 0, since we are including one-digit numbers - 6 can be treated as 06, for example), and vice versa.

However, we cannot forget that the number $100$ has a 1 in it. So, we must add 1 to our total.

Therefore, we have

$20\cdot(1+3+5+7+9)+1 = 20\times 25 +1 = 500+1=\boxed{501}\ \blacksquare$

From 1 to 10, we have 5 odd numbers (1,3,5,7,9) which add up to 25. From 1 to 99, we first consider all the odd numbers in units place and add them, i.e., from 1 to 10, odd nos. add up to 25, from 10 to 20, odd nos. in units place add up to 25, from 20 to 30, odd nos. in units place add up to 25 and so on. Adding we have, (1+3+5+7+9)(10)=250

Now, we must count the odd digits in the ten's place from 1 to 99. We have ten 1s, ten 2s, ten 3s.......,ten 9s which correspond to all the nos. from 10 to 19, 20 to 29, 30 to 39.......,90 to 99. Thus, adding only the odd nos. i.e., the ten 1s, ten 3s, ten 5s,....,ten 9s, we have, (1+3+5+7+9)x10=250

Lastly, the sum of odd digits in 100 is 1.

Therefore, we have 250+250+1=501 as the correct answer

Consider all numbers with 1 or 2 digits. They are on the form XY ( 0<=X,Y<=9 )

Take the digit 1. How many times did the digit 1 appear ?

It appeared either on the form:

1Y (10 options for Y)

or on the form:

X1 (10 options for X)

A total of 20 options for the digit 1.

Similarly, 20 options for each digit of the odd-digit set {1,3,5,7,9}

Then we want to add up 20 * 1 + 20 * 3 + 20 * 5 + 20 * 7 + 20 * 9 = 20 * (1 + 3 + 5 + 7 + 9) = 20 * 25 = 500

Finally the answer is 500 + O(100) = 500 + 1 = 501

That is equal to $(1 + 3 + 5 + 7 + 9)(20) + 1 = 501$ , because considering the $1$ and $2-$ digit numbers, the numbers of the set $\{1, 3, 5, 7, 9\}$ will appear $10$ times each of them, as in the tens place. Adding up the $1$ of the number $100$ , the answer is $25\cdot 20 + 1 = \boxed{501}$ .