the hundred thousands digit of a six-digit even number is 3 more than the thousands digit,which is twice the ones digit.give at least four nuumbers that satisfy the given condition
let's start plugging in numbers as we figure them out:
xxxxxxx
the last digit is even, so we have
xxxxx0 xxxxx2 ... xxxxx8
the 1000's is twice the ones', so we can't have the last digit more than 4:
xx0xx0 xx4xx2 xx8xx4
the 100,000 is the 1000's+3, so we are left with
3x0xx0 7x4xx2
Now you can fill in the x's with any other digits you like
3742
To find the numbers that satisfy the given condition, let's break down the conditions step by step:
1. The hundred thousands digit is 3 more than the thousands digit.
2. The thousands digit is twice the ones digit.
3. The number is even.
Let's start by finding the possible values for the ones digit.
The ones digit can range from 0 to 9. Let's try each possibility and apply the conditions to check if they are satisfied.
1. If the ones digit is 0, then the thousands digit is 2 x 0 = 0. But the condition states that the thousands digit cannot be 0. So, this doesn't satisfy the condition.
2. If the ones digit is 1, then the thousands digit is 2 x 1 = 2. The hundred thousands digit is 3 more than the thousands digit, which gives us 2 + 3 = 5. So, one possible solution is 105,201.
3. If the ones digit is 2, then the thousands digit is 2 x 2 = 4, and the hundred thousands digit is 4 + 3 = 7. So, another possible solution is 207,402.
4. If the ones digit is 3, then the thousands digit is 2 x 3 = 6, and the hundred thousands digit is 6 + 3 = 9. So, another possible solution is 309,603.
By using the same approach, you can find more numbers that satisfy the given conditions.