Write five 7-digit numbers that are even. The sum of the digits in each number is 29.
all you have to do is come up with 1 such number, then shuffle the digits around, leaving the last (even) digit alone.
6152438
now shuffle the digits, leaving 8 where it is. There are 6! = 720 ways to do it.
It was easy to come up with, since 29-8=21, so I just picked 3 pairs of digits that add up to 7, and stuck 'em in front of the 8.
Thank you for your response. But, I just don't understand what you wrote. It seems completely different than what the question is asking. Because the statement says all 7 numbers are even?
Thank you Eric!! I understand after thinking about your solution. You are 100% correct!!! :=)
good job. Thanks for puzzling it out on your own.
To find five 7-digit numbers that are even and have a sum of digits equal to 29, we can use a step-by-step approach. Here's how you can find these numbers:
1. Start with the fact that the sum of the digits in each number is 29. Since we need an even number, the sum of the first six digits must be an odd number to ensure the last digit can be even.
2. Begin by selecting the last digit of the number. Since it needs to be even, it can be 0, 2, 4, 6, or 8.
3. Distribute the remaining sum of 29 - [last digit] over the first six digits. To achieve this, we can consider partitioning this sum into six parts. One way to divide 29 - [last digit] into six parts is [last digit] + 1 + 1 + 1 + 1 + 1 + (22 - [last digit]).
4. Choose a combination of digits for the first six digits that satisfies the partitioning in step 3. For example, if the last digit is 2, we have 2 + 1 + 1 + 1 + 1 + 1 + (22 - 2) = 29. So one possible number is 1111112.
5. Repeat steps 2-4 for the remaining possible values of the last digit (0, 4, 6, and 8) to find the remaining four 7-digit numbers.
Here are five 7-digit numbers that are even and have a sum of digits equal to 29:
1. 1111112
2. 1111140
3. 1111164
4. 1111186
5. 1111108