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To form a 6-digit even number, the leftmost digit can be any digit from 1 to 9 (9 options), while the last digit must be even (0, 2, 4, 6, or 8, giving 5 options). The remaining four digits can be any digit from 0 to 9 (10 options each). Therefore, the total number of 6-digit even numbers is calculated as: (9 \times 10^4 \times 5 = 450,000).
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How many five-digit odd numbers are possible if the leftmost digit cannot be zero?
To find the number of five-digit odd numbers where the leftmost digit cannot be zero, we consider the structure of a five-digit number: ABCDE. The leftmost digit (A) can be any digit from 1 to 9 (9 choices). The last digit (E) must be odd, which gives us 5 choices (1, 3, 5, 7, or 9). The middle three digits (B, C, D) can each be any digit from 0 to 9 (10 choices each). Thus, the total number of such five-digit odd numbers is (9 \times 10 \times 10 \times 10 \times 5 = 45000).
How many three-digit area codes are possible if the first digit cannot be a 0?
There are 900 possible three-digit numbers not beginning with 0. (Note, however, that this question does not accurately describe the restrictions on numbers that can be used as area codes.)
How many three digit numbers are possible if the left digit cannot be zero?
a lot * * * * * Possibly, though that answer is relative. The correct answer is 900.
How many even 3-digit positive integers can be written using the digits 1 3 4 5 and6?
If a digit can be repeated there are 5 x 5 x 5 = 125 possible numbers If a digit cannot be repeated there are 5 x 4 x 3 = 60 possible numbers.
How many four digit numbers are possible if the pin cannot begin with a 0 and the pin must be an odd number?
There are 4,500 combinations.
How many three digit even numbers are possible if the left digit cannot be zero?
450
How many four digits numbers can be formed using 0123456789?
Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.
What is the leftmost nonzero digit of a number?
It is the leading digit.
How many three-digit area codes are possible if the first digit cannot be a 0?
There are 900 possible three-digit numbers not beginning with 0. (Note, however, that this question does not accurately describe the restrictions on numbers that can be used as area codes.)
How many three digit numbers are possible if the left digit cannot be zero?
a lot * * * * * Possibly, though that answer is relative. The correct answer is 900.
What is leftmost nonzero digit?
Oh, dude, the leftmost nonzero digit is basically the first digit in a number that isn't zero. Like, it's the number that kicks off the party and gets things going. So, if you see a number like 503, the leftmost nonzero digit would be 5. Hope that clears things up for ya!
What is the leading digit of 473?
It is the first or leftmost digit: 4.
How many different 7 digit phone numbers are possible if the first digit cannot be 0 or 1?
hi wazup hi wazup
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 three-digit odd numbers are possible if the left digit cannot be zero?
450... There are 500 odd numbers from 000 to 999 inclusive. From 000 to 099, there are 50 odd numbers. 500 - 50 = 450.
How many four-digit odd numbers are possible if the left digit cannot be zero?
9 x 10 x 10 x 5 = 45000
How many seven-digit telephone numbers are possible if the first two digits cannot be ones or zeros?
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.
