You can only use the digits 0-9. I need an odd number that is a multiple of five with no repeated digits. Half the digits are odd and the other half is even. The largest digit in the number is in the tens place (but it is not the largest digit). The digit in the hundreds place is half the digit in the tens place. The sum of the digits is greater than 20.
123485
To: bobpursley - How do you teach a fifth grader how to figure out the answer?
(I think 7485 will also do it: an explanation of the logic is supplied in response to Amy's posting of this question.)
To find a number that satisfies all the given conditions, we can go through each condition and manipulate the digits accordingly.
1. An odd number: The number must end with 5 or 9. Let's choose 5.
2. No repeated digits: We need to choose different digits for each place. We already have 5 in the units place. Let's select the other digits: 2, 4, 6, 8 in ascending order.
3. Half the digits are odd and half are even: So, we already have one odd digit (5). We need one more odd digit. Let's choose 7.
4. The largest digit is in the tens place (but it is not the largest digit): The largest digit we have is 8, so we place it in the tens place.
5. The digit in the hundreds place is half the digit in the tens place: The tens place has 8, so the hundreds place will have 4.
Now, our number looks like this: 4875.
6. The sum of the digits is greater than 20: Let's calculate the sum: 4 + 8 + 7 + 5 = 24. As this sum is indeed greater than 20, the number we found satisfies all the given conditions.
Therefore, a number that satisfies all the given conditions is 4875.