0
How many seven digit numbers contain the number seven at least once?
Wiki User
∙ 15y ago
4,748,472
Confirmed using the following C# function
string sTemp;
int total = 0;
for (int i = 1000000; i < 10000000; i++)
{
sTemp = i.ToString();
if (sTemp.Contains("7"))
{
total++;
}
}
label1.Text = total.ToString();
Wiki User
∙ 15y ago
Add your answer:
Earn +20 pts
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 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 are the greatest and least 6-digit numbers in which the digit in the thousands place is twice the digit in the tens place?
The largest six digit number period is 999999. The largest number meeting your criteria is 998949.
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 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 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 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.
What are the greatest and least 6-digit numbers in which the digit in the thousands place is twice the digit in the tens place?
The largest six digit number period is 999999. The largest number meeting your criteria is 998949.
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?
