the hundred thousands digit of a six-digit even numbers is 3 more than the thousand digit,which is twice the ones digit.give at least four numbers that satisfy the given condition.
the ones digit is 0,2,4,6,8
the 1000's digit is twice the one's, so must be 0,4,8
so far we have
xx0xx0 or xx4xx2 or xx8xx4
now add 3 and we have
3x0xx0 7x4xx2
Now fill in the x's with whatever you want
To find the four numbers that satisfy the given condition, we need to break down the given information and analyze each digit of the six-digit even number.
Let's assume the six-digit number can be represented as ABCDEF, where A represents the hundred thousands digit, B represents the ten thousands digit, C represents the thousands digit, D represents the hundreds digit, E represents the tens digit, and F represents the ones digit.
Now, according to the given condition:
1. The hundred thousands digit (A) is 3 more than the thousands digit (C).
2. The thousands digit (C) is twice the ones digit (F).
3. The number is even, so the ones digit (F) must be an even number.
We can start by selecting a suitable value for the ones digit (F) from the set of even numbers. Let's consider F as 2.
Now, we can use the second condition to find the value of the thousands digit (C). Since C is twice the ones digit, we have C = 2 * 2 = 4.
Next, we can use the first condition to find the hundred thousands digit (A), which is 3 more than the thousands digit. A = C + 3 = 4 + 3 = 7.
Thus, our first number is 742002.
To find three more numbers that satisfy the given condition, we can repeat this process by considering different even numbers for the ones digit (F), finding the value of the thousands digit (C), and then the value of the hundred thousands digit (A).
Here are four numbers that satisfy the given condition:
1. 742002
2. 944004
3. 1462006
4. 3482008
Please note that these are just some examples, and there could be other valid numbers that satisfy the given condition as well.