0
How many two digit even numbers can be composed from nine digits 1 2 3....9?
Wiki User
∙ 13y ago
The first digit can be any one of the nine digits. For each of those . . .
The second digit can be any one of four digits . . . 2, 4, 6, or 8 .
Total number of possibilities = (9 x 4) = 36
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
How many even four digit numbers can be made using 02356 where the first digit is cannot be 0 and no digits are repeated?
Assuming that 2356 is a different number to 2365, then: 1st digit can be one of four digits (2356) For each of these 4 first digits, there are 3 of those digits, plus the zero, meaning 4 possible digits for the 2nd digit For each of those first two digits, there is a choice of 3 digits for the 3rd digit For each of those first 3 digits, there is a choice of 2 digits for the 4tj digit. Thus there are 4 x 4 x 3 x 2 = 96 different possible 4 digit numbers that do not stat with 0 FM the digits 02356.
How many even numbers of four digits can be formed with digits 1 2 3 4 5 6 repetition of the digit is allowed?
6 x 6 x 6 x 3 = 648 6 because the first digit can be any of the numbers 6 again, because the second digit can be any of the numbers 6, again, because the third digit can be any of the numbers 3, because the fourth/last digit can only be 2, 4, or 6
How many even 3-digit numbers can be formed from the digits 1 2 5 6 9 if each digit can be used only once?
24 of them. So we have a 2 or 6 in the unit position, therefore (2)(4)(3) = 24 even three-digit numbers, can be formed.
How many 5 digit numbers are even?
With a total of 90,000 five digit numbers, we can conclude that there is 45,000 EVEN five digit numbers.
How many 4 digit even numbers are there if numbers can be repeated?
Assuming that the second instance of "numbers" in the question means "digits", the answer can be derived as follows: For the number to be even, its last digit must be 0, 2, 4, 6, or 8, for a total of 5 choices. The remainder of the answer depends on whether numbers that begin with 0, 00, or 000 are allowed. If such initial zeroes are not allowed, there are 9 choices for the first digit. Any of the ten digits may be used for the two remaining digits. Therefore the number of instances is 9 X 102 X 5 = 4500. If initial zeroes are allowed, the number of instances is 5000.
How many two-digit even numbers can be composed from nine digits 1 2 3 9?
The second digit can be 2,4,6, or 8. There are 4 ways of choosing the second digit. The first digit can be chosen from the remaining 8 digits. 8x4 = 32 numbers. Here, it is assumed that the numbers do not repeat
How many six digit numbers are there in which no digit is repeated even digits appeared in the even places and odd digits in the odd places and the number is divisible by 4?
There are 5760 such numbers.
How many two digit even numbers are there which have different digits?
There are 41 of them.
What 5 digit number has Numbers increasing from the left Each digit is different 1st and 5th digits are square numbers 1st and 4th digits are cubed numbers 2nd 3rd digits are even?
12689 14689 12489
How many four-digit numbers consist of only even digits?
500
What is the least five-digit even numbers with no repeated digits?
24680 using only even numbers or 12346 which is an even number
How many three even digit numbers by using 12457 if each digit is used once?
None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits.
Which are the three numbers that are even 2 digit numbers where the sum of the digits is 5?
14, 32, and 50
How many even two digit numbers are there where the sum of the digits is 5?
Zero.
How many strings of six decimal digits end with an even digit?
There are 500000 such numbers.
How many 2-digit even numbers have an odd number as the sum of their digits?
25 of them
How many even 3-digit numbers can be formed from digits 0 1 2 4 5 7 9 if each digit can be used only once?
To solve this question, two cases must be considered:Case 1: The three-digit even number ends with 0.If the three-digit number ends with 0, then the number must be even. After the digit 0 is used, six digits remain. After any one of these six digits is chosen to be the first digit, five digits remain. To complete the number, any of the five remaining digits could be chosen to be the second digit of the number.___6___ * ___5___ * ___1___ = 30 three-digit even numbers ending in 0.1st digit-------2nd digit----3rddigit (0)Case 2: The three-digit even number does not end with 0.If the three-digit number does not end with 0, it can end with 2 or 4. Therefore, there are two possibilities for the third digit. The first digit could be any of the remaining digits, EXCEPT 0: if 0 were the first digit, it would not be a three-digit number. Therefore, there would be five possible digits for the first digit. 0 could be used as the second digit, along with the four remaining digits that were not previously used.___5___ * ___5___ * ___2___ = 50 three-digit even numbers not ending in 0.1st digit-------2nd digit-----3rddigit (0)The total number of three-digit even numbers is 80, since 30 three-digit even numbers ending in 0 plus 50 three-digit even numbers not ending in 0 equals 80.
