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How many seven digit numbers contain the number seven at least once?
To find the total number of seven-digit numbers that contain the number seven at least once, we can use the principle of complementary counting. There are a total of 9,999,999 seven-digit numbers in total. To find the number of seven-digit numbers that do not contain the number seven, we can count the number of choices for each digit (excluding seven), which is 9 choices for each digit. Therefore, there are 9^7 seven-digit numbers that do not contain the number seven. Subtracting this from the total number of seven-digit numbers gives us the number of seven-digit numbers that contain the number seven at least once.
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.
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 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 ten thousands place is twice the digit in the tens place?
Greatest: 989949 Least: 100000
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
How many seven digit numbers contain the number seven at least once?
To find the total number of seven-digit numbers that contain the number seven at least once, we can use the principle of complementary counting. There are a total of 9,999,999 seven-digit numbers in total. To find the number of seven-digit numbers that do not contain the number seven, we can count the number of choices for each digit (excluding seven), which is 9 choices for each digit. Therefore, there are 9^7 seven-digit numbers that do not contain the number seven. Subtracting this from the total number of seven-digit numbers gives us the number of seven-digit numbers that contain the number seven at least once.
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 five digit counting numbers contain at least one 6?
i think alot like 100 at least
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?
If the houses on main street are numbered 1 to 150 how many house numbers contain a least one digit 91 and 4?
10293330201309i31841481389
What is the least six digit numbers?
100,000
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 of the integers from 1 to 200 contain the digit 1 at least twice?
There are twelve instances where the integers from 1 to 200 contain the digit 1 at least twice:-11,101,110,111,121,131,141,151,161,171,181,191.
What is the ones digit of the product of 5 consecutive natural numbers?
Zero. Any five consecutive natural numbers will contain at least one multiple of 2 and at least one multiple of 5, meaning that the product will be a multiple of 10.
