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What is the last digit in the product of the first 20 odd natural numbers?
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.
How many four digit whole numbers are there such that the thousand digit is odd the hundreds digit is even and all four digits are different?
5*5*4*4 = 400
How do you divide 30 into 5 odd numbers?
You cannot. The sum of 5 odd numbers is always odd. 30 is not odd.
How many odd numbers with middle digit 5 are there between 40000 and 69999 with no digits repeated?
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.
How many four odd numbers are possible such that the hundred digit is two more than tens digit?
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.
