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How many three digit numbers contain at least two sevens?
27: 077,177,277,377,477,577,677,877,977,717,727,737,747,757,767,777,787,797
What proportion of the first 10000 natural numbers contain a 5?
271 of the first 1000 natural numbers contain at least one digit 5. That is 27.1 % of them.
How many integers from 1 to 100000 contain the digit 6 at least once?
Oh, isn't that a happy little question! To find the number of integers from 1 to 100,000 that contain the digit 6 at least once, we can think about it like painting a beautiful landscape. We can first count the numbers that do not have the digit 6, which are the numbers from 1 to 9,999. There are 10,000 numbers that do not have the digit 6. Then, we subtract this from the total numbers from 1 to 100,000 to find the answer. Happy counting!
How many numbers between 1 and 999999 contain at least one 1?
We are to calculate the number of 1-digit to 6-digit numbers (i.e. numbers less than one million) that have at least one 1. Let's divide this into six cases by the number of digits in the number.For one digit numbers, the only number is 1 - 1 such number.For two digit numbers, we can find the number which do not contain a 1 more easily. There are 8 possible digits for the tens place (0 and 1 are not allowed), and 9 digits for the ones place (1 is not allowed). Thus we have 8 * 9 = 72 which do not work. There are 90 two digit numbers, so 90 - 72 = 18 such numbers.For three digit numbers, we will proceed in the same manner. There are 900 total 3 digit numbers, of which 8 * 9 * 9 = 648 numbers which do not contain a 1. The remaining 900 - 648 = 252 numbers contain at least 1 one - 252.For four digit numbers, continuing in the same manner we have 8 * 9 * 9 * 9 = 5832 which are not good out of the 9000 total. So, 9000 - 5832 = 3168.For five digit numbers, we have 8 * 9 * 9 * 9 * 9 = 52488. The remaining 90000 - 52488 = 37512 are good.For six digit numbers, we have 8 * 9 * 9 * 9 * 9 * 9 = 472392. The reamining 900000 - 472392 = 427608 are good.Finally, adding all of the totals, we have 1 + 18 + 252 + 3168 + 37512 + 427608 = 468,559 numbers between 1 and 999999 which contain at least 1 one.
What is the least possible sum of two 4-digit numbers?
what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?
How many 3 digit numbers contain the digit 2 at least once?
252
How many 3 digit numbers are there that contain at least 1 seven?
252
What is the least number in single digit numbers?
0
How many three digit numbers contain at least two sevens?
27: 077,177,277,377,477,577,677,877,977,717,727,737,747,757,767,777,787,797
How many five-digit counting numbers contain at least one 6?
46000
What proportion of the first 10000 natural numbers contain a 5?
271 of the first 1000 natural numbers contain at least one digit 5. That is 27.1 % of them.
How many integers from 1 to 100000 contain the digit 6 at least once?
Oh, isn't that a happy little question! To find the number of integers from 1 to 100,000 that contain the digit 6 at least once, we can think about it like painting a beautiful landscape. We can first count the numbers that do not have the digit 6, which are the numbers from 1 to 9,999. There are 10,000 numbers that do not have the digit 6. Then, we subtract this from the total numbers from 1 to 100,000 to find the answer. Happy counting!
How many numbers between 1 and 999999 contain at least one 1?
We are to calculate the number of 1-digit to 6-digit numbers (i.e. numbers less than one million) that have at least one 1. Let's divide this into six cases by the number of digits in the number.For one digit numbers, the only number is 1 - 1 such number.For two digit numbers, we can find the number which do not contain a 1 more easily. There are 8 possible digits for the tens place (0 and 1 are not allowed), and 9 digits for the ones place (1 is not allowed). Thus we have 8 * 9 = 72 which do not work. There are 90 two digit numbers, so 90 - 72 = 18 such numbers.For three digit numbers, we will proceed in the same manner. There are 900 total 3 digit numbers, of which 8 * 9 * 9 = 648 numbers which do not contain a 1. The remaining 900 - 648 = 252 numbers contain at least 1 one - 252.For four digit numbers, continuing in the same manner we have 8 * 9 * 9 * 9 = 5832 which are not good out of the 9000 total. So, 9000 - 5832 = 3168.For five digit numbers, we have 8 * 9 * 9 * 9 * 9 = 52488. The remaining 90000 - 52488 = 37512 are good.For six digit numbers, we have 8 * 9 * 9 * 9 * 9 * 9 = 472392. The reamining 900000 - 472392 = 427608 are good.Finally, adding all of the totals, we have 1 + 18 + 252 + 3168 + 37512 + 427608 = 468,559 numbers between 1 and 999999 which contain at least 1 one.
How many thousands are there in Ten Thousands?
There are 9000 4 digit numbers Find 4 digit numbers that are all different from 1 find 4 digit numbers that are all different from 1 The thousands digit has 8 ways to choose, hundreds, tens, and units all have 9 ways => there are 8 x 9 x 9 x 9 = 5832 digits => 4 digit numbers contain at least 1 digit 1 = number of 4 digit numbers , the number of 4 digit numbers is different by digit 1 =9000 - 5832 = 3168 numbers
How many five digit counting numbers contain at least one 6?
i think alot like 100 at least
What is the least sum you can get when adding two four-digit numbers?
2000.The smallest four-digit number there can be is 1000. So, add 1000 and 1000 to get 2000, then least sum you can get when adding two four-digit numbers.
How many 3 digit numbers contain the digit 3 at least once?
172==============================Another contributor was amazed:Wow ! That's a neat little problem.Let's see . . . . .-- There are three ways to have a single 3 in the number: 3xx, x3x, and xx3.For 3xx, there can be 81 different numbers. (2nd and 3rd digits can't be a 3. 9x9=81.)For x3x, there can be 72 numbers. (1st can't be 0 or 3. 3rd can't be a 3. 8x9=72.)For xx3, there are also 72. (1st can't be 0 or 3. 2nd can't be a 3. 8x9=72.)So far, we have 81+72+72 = 225 possibilities, with a single 3 in each one.-- There are three ways to have two 3s in the number: 33x, 3x3, and x33.For 33x, there can be 9 different numbers. (3rd digit can't be a 3.)For 3x3, there can be 9 numbers. (2nd digit can't be a 3.)For x33, there are 8 numbers. (1st digit can't be a 0 or a 3.)So we have 9+9+8 = 26 more possibilities.-- There is only one way to have three 3s in the number: 333.That's one more possibility.So the total number of 3-digit numbers where at least one of the digits is a 3 is225 + 26 + 1 = 252 all together.
