10a + b = 7(a + b)
10a + b = 7a + 7b
3a = 6b
a = 2b
(2,1), (4,2), (6,3), and (8,4) meet these requirements.
So the numbers that are solutions are 21, 42, 63, and 84.
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
If the number can contain repeated digits, the answer is 800000. Without repetition, there are 483840.
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.
Phone numbers are seven digits long in some countries but differing lengths in others. It purely depends upon what number of phone numbers the country needs .
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
Any pair of digits (not including 0), can be used to generate 14 four-digit numbers. If one of the digits is 0, only seven will start with a non-zero digit.
In the numbers 1-9 each number has 1 digit and there are 9 of them, so that's 9.In 10-99 each number has 2 digits, and there are 90 of them: 2x90 = 180There are 900 three digit numbers [100 through 999]: 2700 digits.There are 9000 four digit numbers: 36000 digits.90,000 numbers with five digits: 450,000 digits.900,000 numbers with six digits: 5,400,000 digits.Then 1 number with seven digits: 7 digits.Add them up and you have 5,888,896 digits.
1,111,111 since 17 = 1
No, a 10-digit phone number that is missing the last three digits can not be a legitimate number. In order for a number to work, it must have 3 numbers for an area code and seven digits for the phone number itself. Some telemarketers are now using fake numbers to hide from the people they're calling.
There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
All of the numbers between one and two million have seven digits.
9,999,876 is the greatest seven-digit number using four different digits.
That makes:* 8 options for the first digit * 8 options for second digit * 10 options for the third digit * ... etc. Just multiply all the numbers together.
It's not entirely clear from the wording of your question what it is you want to know. Seven is a one-digit number. The sum of its digits is 7. Seven has two factors: 1 and 7. That sum is 8.
Assuming all digits are equally likely, then it is 0.9395.However, that assumption is flawed since some numbers begin with 0. Some sequences are reserved and so on. So the distribution of numbers is NOT truly random and the above answer depends on all seven digit numbers being equally likely..
9,876,543 is.