Numbers With No Even Digits

We are looking for the percentage of the first 100 positive integers that don't contain any even digit. For a number to satisfy this condition, none of its digits can be even, therefore, all of its digits must be odd. So what we're going to do now is to find how many of the first 100 positive integers have digits that are all odd.

We know that the first 100 positive integers are 1 to 100, so these are the only numbers that we will take into consideration. However, we can automatically remove 100 from our list since it clearly has two even digits (0) and does not satisfy the condition. So we are left with 99 positive integers that are either 1 digit or 2 digit numbers.

Furthermore, we also know that there are 5 different odd digits, namely 1,3,5,7 and 9. Therefore, we have five different 1 digit numbers that satisfy the given condition.

For a 2 digit number to satisfy the given condition, both the tens and ones digits have to be odd. We have 5 possible odd numbers to fill up the tens place and another 5 possible odd numbers to fill up the ones place. Therefore, we have 5x5=25 possible 2 digit numbers that will satisfy the given condition.

Adding the number of possible 1 digit numbers and 2 digit numbers that will satisfy the given condition, we get 5+25=30, so in total, we have 30 possible numbers out of the first 100 positive integers that will satisfy the given condition. In percentage, this is 30/100 which equals 30%

Therefore, our final answer is 30%.

There are five single-digit odd numbers (1, 3, 5, 7 and 9). First and second digit of a double-digit number can be 1, 3, 5, 7 and 9 only. Therefore, 5 x 5 = 25, 25 + 5 = 30.

Since the first $100$ integers are: $1, 2, 3, ... , 100.$

If we check the first 9 1-digit integer five of them are contains no even digit, namely, $1,3,5,7,9.$

for two digit numbers, we can use fundamental principle of counting. Thus, there are five odd digit for the choice of the tens digit and five for the choice of the unit digit. Therefore we have $5 \times 5 = 25$ choices for two digit number that contains no even digit. In addition, 100 is excluded since $0$ is even.

Therefore we have 30 positive integers. These yields 30% of the first 100 positive integers.

100 contains even digits. Among the integers from 1 to 9, 5 (the odd ones) contain no even digits. So we only need to take care of the two-digit numbers. The first digit is an arbitrary odd digit from 1 to 9, and the same holds for the second digit. In total there are 5 + 5 * 5 = 30 such numbers, i.e., 30% of the first 100 positive integers.

1,3,5,7,9,11,13,15,17,19,31,33,35,37,39,51,53,55,57,59,71,73,75,77,79,91,93,95,97,99=30

The odd numbers from 1-10 are, 1,3,5,7, and 9.

The only possible 2 digit numbers are, 11,13,15,17,19,31,33,35,37,39,51,53,55,57,59,71,73,75,77,79,91,93,95,97, and 99

There are 25 possiblities , but we havent counted the 1 digit numbers, which are 1,3,5,7 and 9.

25 (2-digit numbers) + 5 (1-digit numbers) = 30 possible numbers

30 numbers divided by the total number of numbers from 1 to 100.

30 divided by 100 = 0.30

0.30 times 100 to find what percent of it of 100. Equals 30%

There are 5 usable digits (1,3,5,7,9) to form the nos. with no even digits. Now, since there is no restriction on repetition of digits so, out of the first 100 positive integers , i.e from 1 to 100 , we have $5\times 5=25$ such nos. that are of 2-digit. Also, we have 5 other such nos (1,3,5,7,9), so in total we have $=25+5=\boxed{30}$ such nos.

So, the answer is $=\boxed{30}$

single digits, 5 numbers: 1, 3, 5, 7, 9 double digits, 5x5 = 25 numbers

so total 30 numbers with odd digits only.

There are 5 one digit numbers that have 0 even numbers There is 5/9 chance that the first number of a two digit number is odd. There is a 5/10 chance the the second number of a two digit number is odd The chance that both numbers are odd is 5/9 5/10=25/90. Since there are 90 two digit numbers, we have 25/90 90=25 two digit numbers that have 0 digit numbers. We sum the 2 digit numbers with the 1 digit numbers that satisfy the rules, and we get 25+5=30 numbers. There a total of 100 cases, so the percentage is 30/100 * 100% hence the answer is 30.

for two digit numbers there are 5 x 5 = 25 numbers that possible and 5 numbers for one digit number (1,3,5,7,9) thus, there are 30 numbers so, the percentage is \frac{30}{100} x 100% = 30%

First digit can be selected in 6 ways (1,3,5,7,9,0). Second digit can be selected in 5 ways (1,3,5,7,9). Hence $6\times 5$ = $\boxed{30}$

numbers with all digits odd... 1 digit numbers are 1,3,5,7,9 = 5 and 2 digit number are {1 or 3 or 5 or 7 or 9 at both places} =5*5 = 25... So 5+25= 30 and 30/100...

1, 3, 5, 7, and 9 is an odd number or not even number. For the first 100 digit: 1 digit number : 5 2 digit number : 5 x 5 = 25 So, the answer is 30

My thinking was from 1-20 theres (10 ) that have none then (5 )in each the 30s 50s 70s & 90s which is(10+20) equalling 30 which is of course 30% of 100

Look at 1,3,5,7,9......1) then 11,13,15,17,19........2) then 21,23,25,27,29.........3) In 1)&2) there are 5 numbers in each of them which have no odd digit.But in case of 3),all the odd numbers from 20 have even numbers in tens place,so we count 1)&2) which implies there are ten such numbers upto now.Then we count from 30,40,50...90 and by this logic we are having 5*6=30 such numbers.

We can divide the first 10 numbers in to two category even and odd.Even(0,2,4,8,6) and odd(1,3,5,7,9) So from this I can conclude that at least all even numbers except 0 will not be considered as all are itself a even number. So we can decrease 40 numbers from 100. 60 numbers are left and we can observe that each odd number will occur in 10th place digit Ex (10..15..19, 30..31..39,...99) So each odd number will occur 10 times in tenth place and every odd number will have 5 even number when they are at tenth position. so total number will be.. 5*5=25 number and 5 from all the odd numbers below 10 that is 1,3,5,7,9. So total number is 25+5=30.

There are 5 odd numbers in between two multiples of 10. There are 50 numbers with an odd number in the tenth place and 50 numbers with an odd number in the unit's place. So there are 25 odd numbers between 10 to 100 and 5 odd numbers between 1 to 10. So totally there are 30 odd numbers.

no . of single digit numbers which do not have even digit =5 In two digit number, tens's place canbe filled with 5 digits and one's place can also be filled with 5 digits . so no.of numbers formed are 5*5=25 so, total no.of numbers in first 100 positive integers which do not have even digit=25+5=30

It reminds me of a question that you would see on some sort of brain test online.. First you figure out how many numbers without an even number are in 10. That was 5. Then figure out how many numbers in the tens place were not even . The answer is also 5. Remember that before you get into the tens place you must start off in the ones place, because of this that added another 5. 5x6=30

The numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 31, 33, 35, 37, 39, 51, 53, 57, 59, 71, 73, 75, 77, 79, 91, 93, 95, 97 and 99. There are 30 numbers, out of 100.

Required percentage = 30 / 100 * 100% = 30%

No. of 1 digit numbers with no even digits will be (1,3,5,7,9)= 5

No. of 2 digit numbers with no even digits will be= 5.5=25

Total= 25+5= 30

% of +ve integers contains no even digits= 30/100 *100=30%

Numbers with odd digit at units place are 1,3,5,7,9,11,13,15....,97,99. Total = 50. Excluding numbers with even digit at ones place. Total = 20. Remaining numbers = 30

no even digits....therefore....both the digits should be odd which are 1,3,5,7,9,11,13,15.........and not 21,23,25.........so the answer comes out to be 30....

1. 1-digit numbers: $5 {1,3,5,7,9}$
2. 2-digit numbers: $25$ (rule of product)
3. 1-digit + 2-digit = $5+25 = 30$
4. $P = \frac{3}{100}$

Numbers contains no even digits are the numbers which do not contains 2,4,6,8,0 as its unit and tens digit's place. So, numbers which do not contains even digits are 1,3,5,7,9,11,13,15,17,19,31,33,35,37,39,51,53,55,57,59,71,73,75,77,79,91,93,95,97,99. So totals numbers are 30. Percentage is (30/100)*100=30

first we have 5 odd no between 1-10

then 20-30,40-50,60-70,80=90 this digits are discard bcz these are starts with even number.

only others r remains and every 10 range has 5 numbers which has only odd numbers. here we have 25 numbers

so total 30 numbers.

For all the 2 digit numbers among the first 100 positive integer, ab be the numbers that contain no even digits. So, b and a may take values 1, 3, 5, 7 and 9. Therefore, there are (5 x 5) = 25 two-digit numbers that contain no even digits. And, among 1-digit numbers, we have 5 integers that contain no even digit that are 1, 3, 5, 7 and 9. Therefore, Number of positive integers with no even digit among first 100 positive integers = (25 + 5) = 30 i.e. equal to 30% of the first 100 positive integers.

The possible combinations that can be made are from 1,3,5,7 and 9. We keep ten's digit common and continue with one's digit one by one(eg-11,13,15,17,19 then 31,33,35,37,39 and so on). From each Ten's digit we get 5 combinations + 1,3,5,7 and 9. So, 5 x 5 = 25+5= 30

Number of One-Digit Odd Numbers- 1,3,5,7,9 i.e 5 Number of 2-Digit Odd Looking Numbers(Numbers which do not contain any even digit) = 5 . 5 = 25. Total Number of Odd Looking Numbers = 5 + 25 = 30. Hence % of Numbers which do not contain any Even Digit = 30 / 100 * 100 = 30%

First 100 positive integers- 1-100 No of odd numbers in each set of 10 numbers(1-10,11-20 and so on)-5 No of sets not considered- 4 (20-30,40-50,60-70,80-90) No of sets considered- 6 Answer= 5*6=30