My nephew is in the 4th grade and asked me to help him answer this question: Two 6 digit numbers have three 8's and three 7's. no two digits are the same. write the numbers in standard form. and i am not sure how to solve it.
I'm not sure I understand the problem -- but this is possibly the answer.
787,878 and 878,787
it would be 878,787 and 787,878 because you go back and forth or it could be 777,888 and 888,777.
To solve this problem, we need to find two 6-digit numbers that meet the given conditions: each number must contain three 8s and three 7s, and no two digits in each number can be the same.
Let's break down the problem step by step:
Step 1: We know that each number has three 8s and three 7s. This means that one number could be 888777, where there are three 8s and three 7s in that order.
Step 2: Now we need to find the other number. Since no two digits can be the same in each number, we must arrange the remaining digits (1, 2, 3, 4, 5, 6, 9, and 0) in a way that satisfies this condition.
Step 3: One way to arrange the remaining digits would be to start with any digit, say 1, and then place it in one of the remaining three spots in the second number. So we have:
- Number 1: 888777
- Number 2: 1_____ (We have one 1, two remaining 7s, two remaining 8s, and four remaining digits: 2, 3, 4, 5, 6, 9, and 0)
Step 4: Now we need to fill in the remaining spots in Number 2. We can begin with the digit 1, and since Number 1 already has three 1s, we can skip to the next digit, 2. We can place the first 2 in the thousands place (2_____) and then continue with placing the remaining digits:
- Number 1: 888777
- Number 2: 182___
Step 5: Continuing this process, we can fill in the remaining spots in Number 2:
- Number 1: 888777
- Number 2: 182735
Step 6: Finally, we have our two 6-digit numbers in standard form:
- Number 1: 888777
- Number 2: 182735
Therefore, the two 6-digit numbers that meet the given conditions are 888777 and 182735.