4 digit number > 2000. the sum of its digits is 10. only the thousands place and the hundreds place are odd numbers. the number in the ones place double the digit in the hundreds place. what number will it be?
3xxx or 5xxx
Now we have possibilities of
3x1x: 3214
3x3x: nope
3x5x: nope
3x7x: 3070
numbers over 5000 don't work - digits add up too quickly
actually, I guess if 3070 is allowed, so is 7030
If all digits must be unique, then I guess 3214 is the only choice.
To find the number that satisfies these conditions, we can break it down step by step:
1. The number is greater than 2000. So, we know that the thousands place must be 2 or greater.
2. The sum of its digits is 10. This means that when we add up all the digits of the number, we should get a total of 10. Let's try to come up with some possibilities:
- If the thousands place is 2, then we have 2 + _ + _ + _ = 10. From this, we can see that the remaining three digits must sum up to 8.
3. Only the thousands place and the hundreds place are odd numbers. This means that the thousands place can be 3 or 5, and the hundreds place can be 1, 3, 5, 7, or 9.
4. The number in the ones place is double the digit in the hundreds place. So, if the digit in the hundreds place is X, then the digit in the ones place should be 2X.
Now we can start to narrow down the possibilities:
- If the thousands place is 3 and the hundreds place is 1, then we have 3 + 1 + _ + _ = 10. From this, we can see that the remaining two digits must sum up to 6. Since the remaining digits should be odd, the only possibility is 3 and 3. So, the number would be 3132.
- If the thousands place is 3 and the hundreds place is 3, then we have 3 + 3 + _ + _ = 10. From this, we can see that the remaining two digits must sum up to 4. Since the remaining digits should be odd, the only possibility is 1 and 3. So, the number would be 3314.
- If the thousands place is 5 and the hundreds place is 1, then we have 5 + 1 + _ + _ = 10. From this, we can see that the remaining two digits must sum up to 4. Since the remaining digits should be odd, the only possibility is 1 and 3. So, the number would be 5132.
- If the thousands place is 5 and the hundreds place is 3, then we have 5 + 3 + _ + _ = 10. From this, we can see that the remaining two digits must sum up to 2. However, if the remaining digits are odd (1 and 1), the number in the ones place wouldn't be double the digit in the hundreds place. So, this is not a valid option.
Therefore, the only valid numbers that satisfy all the conditions are 3132, 3314, and 5132.