To determine how many times the digit 4 appears in numbers from 1 to 623, we can consider the units, tens, and hundreds place separately. In the units place, the digit 4 appears once in numbers ending in 4 (4, 14, 24, ..., 614). This happens 62 times. In the tens place, the digit 4 appears 10 times in every hundred numbers (40-49, 140-149, ..., 540-549), so it appears 6 times within the range from 1 to 623. In the hundreds place, the digit 4 appears once in every hundred numbers (400-499), occurring 1 time. Therefore, the digit 4 appears a total of 62 + 6 + 1 = 69 times in numbers from 1 to 623.
How many times does the digit 1 occur in ten place in the numbers from 1 to 1000?
The numbers from 1 to 39 include both single-digit and double-digit numbers. There are 9 single-digit numbers (1 to 9) and 30 double-digit numbers (10 to 39). Therefore, the total number of digits is 9 (from single-digit numbers) + 60 (from double-digit numbers, as each has 2 digits) = 69 digits in total.
With 123 digits you can make 123 one-digit numbers.
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
There are 238 - 1 = 237 numbers between 1 and 238. To find the number of digits, we need to consider the range of numbers from 1 to 9 (1-digit numbers), 10 to 99 (2-digit numbers), and 100 to 238 (3-digit numbers). There are 9 one-digit numbers, 90 two-digit numbers, and 139 three-digit numbers between 1 and 238, totaling 9 + 90 + 139 = 238 digits.
1
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
Count them
How many times does the digit 1 occur in ten place in the numbers from 1 to 1000?
999-111=888 888 3 digit numbers can be made with numbers between 1 - 9
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
the right answer is 10 cause there is o,1,2,3,4,5,6,7,8,9 so there 10 numbers in total which have 1 digit
There are 900 three-digit whole numbers between 1 and 1000
The numbers from 1 to 39 include both single-digit and double-digit numbers. There are 9 single-digit numbers (1 to 9) and 30 double-digit numbers (10 to 39). Therefore, the total number of digits is 9 (from single-digit numbers) + 60 (from double-digit numbers, as each has 2 digits) = 69 digits in total.
10,000
With 123 digits you can make 123 one-digit numbers.
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.