The hundred thousands digits of a six-digit even number is 3 more than the thousand digit, which is twice the ones digit. Give at least four numbers that satisfy the given condition.

754838

I dont know

To find the four numbers that satisfy the given condition, we can break down the problem into the different digits of a six-digit even number.

Let's assign variables to represent each digit of the number:
- Let A represent the hundred thousands digit.
- Let B represent the ten thousands digit.
- Let C represent the thousands digit.
- Let D represent the hundreds digit.
- Let E represent the tens digit.
- Let F represent the ones digit.

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, which means that the ones digit (F) must be even.

Now, with these conditions, let's find the four numbers:

Number 1:
Let's assume A = 6, C = 3, F = 2. From condition 1, we know that the hundred thousands digit (A) is 3 more than the thousands digit (C), so A = C + 3, which gives us A = 3 + 3 = 6. From condition 2, we know that the thousands digit (C) is twice the ones digit (F), so C = 2 * F, which gives us C = 2 * 2 = 4. The remaining digits (B, D, E) can be any even number. Therefore, a possible number is 644246.

Number 2:
Let's assume A = 7, C = 4, F = 2. Similar to Number 1, we use the given conditions to find the values of the other digits. A = C + 3, so A = 4 + 3 = 7. C = 2 * F, so C = 2 * 2 = 4. The remaining digits (B, D, E) can be any even number. Therefore, a possible number is 744246.

Number 3:
Let's assume A = 8, C = 5, F = 4. Using the given conditions, A = C + 3, so A = 5 + 3 = 8. C = 2 * F, so C = 2 * 4 = 8. The remaining digits (B, D, E) can be any even number. Therefore, a possible number is 855448.

Number 4:
Let's assume A = 9, C = 6, F = 6. Using the given conditions, A = C + 3, so A = 6 + 3 = 9. C = 2 * F, so C = 2 * 6 = 12. Since the thousands digit cannot be greater than 9, we need to adjust the value of C. We can make C = 8 and adjust F accordingly, so C = 2 * 4 = 8 and F = 4. The remaining digits (B, D, E) can be any even number. Therefore, a possible number is 966848.

Note that there are other possible numbers that satisfy the given conditions, but these four numbers demonstrate the process of finding numbers that meet the requirements.

Possible cases:

3A0AA0
7A4AA2 , where A can be any number from 0-9
since those positions have not been defined

If we make the unit digit 4, remember your number has to be even, then the thousand digit is 8 and the hundred thousand digit would be 11 ???? )

Now fill in any digits for A that you like.

Your school subject is MATH , not the name of your school