The first 5 digit number is 10,000 and the last is 99,999. Thus there are 90,000 numbers between that range. Half of them are odd & half are even. 45,000 each
The last digit in the product of the first 20 odd natural numbers can be determined by looking at the pattern of the units digit in the multiplication of consecutive odd numbers. The units digit of the product of consecutive odd numbers alternates between 1 and 5. Since there are 10 odd numbers between 1 and 19, and 20 is also an odd number, the last digit in the product of the first 20 odd natural numbers is 5.
5*5*4*4 = 400
You cannot. The sum of 5 odd numbers is always odd. 30 is not odd.
To find the number of odd numbers with the middle digit 5 between 40000 and 69999 with no repeated digits, we can break down the problem. Since the middle digit is fixed at 5, we have to choose the other four digits without repetition. For the first digit (thousands place), we have 2 options (4 or 6). For the last digit (units place), we have 5 options (1, 3, 7, 9, 0). For the hundreds place and ten-thousands place, we have 8 and 7 options respectively. Therefore, the total number of odd numbers meeting the criteria is 2 * 8 * 7 * 5 = 560.
I think the question means to ask for "four-digit" odd numbers, and left out a word. Each candidate is of the form (T H N U), and we're told that (H = N + 2). ' T ' can be: 1, 2, 3, 4, 5, 6, 7, 8, 9 = 9 possibilities. For each of these ... ' H N ' can be : 97, 86, 75, 64, 53, 42, 31, and 2,0 = 8 possibilities. For each of these ... ' U ' can be: 1, 3, 5, 7, 9 = 5 possibilities. So the total number of possibilities is: (9 x 8 x 5) = 360different 4-digit odd numbers with the hundreds digit greater by two than the tens digit.
All of them. There are roughly 45,000 5-digit odd numbers. None of them end in 0.
The last digit in the product of the first 20 odd natural numbers can be determined by looking at the pattern of the units digit in the multiplication of consecutive odd numbers. The units digit of the product of consecutive odd numbers alternates between 1 and 5. Since there are 10 odd numbers between 1 and 19, and 20 is also an odd number, the last digit in the product of the first 20 odd natural numbers is 5.
To find the number of 4-digit numbers with only odd digits, we need to consider the options for each digit. Since odd digits are 1, 3, 5, 7, and 9, there are 5 choices for each digit. Therefore, the total number of 4-digit numbers with only odd digits is calculated as 5 x 5 x 5 x 5 = 625. So, there are 625 4-digit numbers that have only odd numbers.
5*5*4*4 = 400
11111
50% of numbers divisible by 5 are even, the other 50% are odd. If the single's digit of a number is 5, then the number is divisible by 5 and odd. For example, 25. However, if the single's digit of a number is 0, then the number is divisible by 5 and even. For example, 20.
1, 3, and 5.
Well, honey, the digit 3 appears in every odd number that ends in 3, 13, 23, 33, and so on up to 39. So, in the first 40 odd numbers, the digit 3 appears 4 times. Math doesn't have to be a drag, darling!
125 (5 to the third power). Each digit can be one of 5 (1/3/5/7/9).
As an example, 13579 the number 5 is in the hundreds place and has the value of 500. It does not matter whether the numbers are odd or even.
9 x 10 x 10 x 5 = 45000
oddOdd numbers are integers which are not multiples of 2 or numbers with the last digit of 1, 3, 5, 7 and 9.The last digit of 4361 is 1 so it is an odd numbers.The last digit of even numbers can be 0, 2, 4, 6, and 8.