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What is the number of ways of forming 6 digit number whose sum of digits is even?
I look at it this way:-- There are 90,000 numbers with five digits. (100,000 minus 10,000 with less than 5 digits)-- If the sum of the first 5 digits is odd, it can be made even by using 1, 3, 5, 7, or 9for the 6th digit ... 5 choices.-- If the sum of the first 5 digits is even, it remains even by using 2, 4, 6, 8, or 0for the 6th digit ... 5 choices.-- So, for any one of the 90,000 five-digit numbers, there are 5 choices for the 6th digitthat result in an even sum of all six.-- So, there must be 450,000 suitable 6-digit numbers.
How many different positive four digit integers can be formed if the first digit must be even the last digit must be odd and all digits must be distinct?
There are 320 such numbers.
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.
What is the 3 biggest numbers using the digits 8642?
If you must use all digits precisely once then the answer is: 8642, 8624, 8462
How numbers greater than 50000 can be formed with digits 3 4 5 6 7 0 if the number are even without repetitions?
To form even numbers greater than 50,000 using the digits 3, 4, 5, 6, 7, and 0 without repetition, we need to consider the position of the digits carefully. Since the number has to be even, the last digit must be 0. The tens digit must be even to ensure the entire number is even, so it can be either 4 or 6. The thousands digit must be 5 to make the number greater than 50,000. Therefore, the only number that fits these criteria is 56,340.
Why do odd numbers must end in one of five digits?
Because if they did not, they would be divisible by 2 and so they would be even.
What are the five digits with which all odd numbers must end?
1, 3, 5, 7 or 9
What is the number of ways of forming 6 digit number whose sum of digits is even?
I look at it this way:-- There are 90,000 numbers with five digits. (100,000 minus 10,000 with less than 5 digits)-- If the sum of the first 5 digits is odd, it can be made even by using 1, 3, 5, 7, or 9for the 6th digit ... 5 choices.-- If the sum of the first 5 digits is even, it remains even by using 2, 4, 6, 8, or 0for the 6th digit ... 5 choices.-- So, for any one of the 90,000 five-digit numbers, there are 5 choices for the 6th digitthat result in an even sum of all six.-- So, there must be 450,000 suitable 6-digit numbers.
How many different three digit numbers can be formed from the digits 0 through 9 if the first digit must be odd and the last digit must be even?
5 x 10 x 5 = 250 different numbers, assuming there is no limit to each digits' use.
How many different positive four digit integers can be formed if the first digit must be even the last digit must be odd and all digits must be distinct?
There are 320 such numbers.
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.
How many five digit numbers can be formed using the digits 02345 when repetition is allowed such that the number formed is divisible by 2 or 5 or both?
There are 2000 possible five digit numbers that can be formed from the digits 02345 that are divisible by 2 or 5 or both. To be divisible by 2, the last digit must be even, namely 0, 2 or 4 (in the digits allowed). To be divisible by 5, the last digit must be 0 or 5. Thus to be divisible by 2 or 5 or both, the last digit must be 0, 2, 4 or 5 (a choice of 4). Presuming that a 5 digit number must be at least 10000, then: For the first digit there is a choice of 4 digits (2345); for each of these there is a choice of 5 digits (02345) for the second, making a total so far of 4 x 5 numbers; for each of these choices for the first and second digits there is a choice of 5 digits (02345) for the third digit making the total so far (4 x 5) x 5 numbers; for each of these choices for the first three digits there is a choice of 5 digits (02345) for the fourth digit making the total so far (4 x 5 x 5) x 5 numbers; for each of these choices for the first four digits there is a choice of 4 digits (0245 - as discussed above) for the last digit, giving a total of (4 x 5 x 5 x 5) x 4 numbers. So the total number of five digit numbers so formed is: number = 4 x 5 x 5 x 5 x 4 = 2000.
What happens to the median when you come up with two answer?
when you have an even amount of numbers while trying to find the median, you first find the two numbers that are at the median and then take all the numbers between them and find the median of that. if that amount of digits is also even, then you must have a decimal median.
What five odd numbers add up to 100?
No 5 odd numbers add up to 100. The sum of any two odd numbers is even. The sum of any two even numbers is even. So the sum of any 4 odd numbers must be even. Take that and add an odd number and the sum must be odd. Therefore, the sum of any 5 odd numbers must be odd. 100 is even. In conclusion: The sum of any odd number of odd numbers is odd.
How many five digit even palindromes are there?
Well, isn't that a happy little question! To find out how many five-digit even palindromes there are, let's break it down. A five-digit number has the form ABCBA, where A, B, and C are digits. Since the number is even, A must be even. So, there are 5 options for A (0, 2, 4, 6, 8), 10 options for B (0-9), and 1 option for C. Multiply these options together and you'll find the total number of five-digit even palindromes.
How many numbers less than a thousand have digits less than ten?
All of them. We normally count in decimal numbers and therefore all digits in decimal numbers must be less than ten.
How do you fill in forms that says must be numeric?
You only put digits or decimal points in them. If they are only taking whole numbers, then you won't even put in a decimal point in them.
