A number has four digits. the sum of the first and last digits is twice the second digit. The second digit is 2 less than the third digit. The last digit is twice the first. Some of the digits are alike. what is the number??

This is a hard one. What helps me is to read each part of the problem carefully. Write the numbers you think on a piece of paper. When you find out if it is wrong put an x next to it. This will narrow down the process.

i got the equations from my sister

a+d= 2b
b= c-2
d=2a

but what do i do next?

To find the number, we need to analyze the information given step by step. Let's break it down:

1. "A number has four digits." This means the number can be represented as ABCD, where A is the thousands place, B is the hundreds place, C is the tens place, and D is the units place.

2. "The sum of the first and last digits is twice the second digit." This can be expressed as A + D = 2B.

3. "The second digit is 2 less than the third digit." This can be expressed as B = (C - 2).

4. "The last digit is twice the first." This can be expressed as D = 2A.

5. "Some of the digits are alike." This means that at least two of the four digits are the same.

Now, let's find the number by solving these equations:

From equation 4, we can substitute D with 2A:
A + 2A = 2B
3A = 2B

From equation 3, we can substitute B with (C - 2):
3A = 2(C - 2)

From equation 2, we can substitute 2B with (A + D):
3A = A + D
3A = A + 2A
3A = 3A

Notice that we have the equation 3A = 3A, which is always true. This implies that the value of A can be any number.

Therefore, there isn't a unique solution to this problem. The number can be any four-digit number with matching digits.