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could be:

8,5,8,4

fits all conditions:

$4*2 = 8 implies 8-4 = 4 implies 8+5+8+4 = 25 implies 1st=3rd=8$

by putting all conditions except "Large and small are different by 4" as letters below

$2a$,$b$,$2a$,$a$ where sum is $25$

should be these numbers. so we can conclude that

First condition:

$2a$ could be biggest while $b$ could be lowest, so $2a-b=4$

Second condition:

$b$ could be biggest, $b-a=4$ since $2a$ is greater than $a$.

and Last condition:

$2a$ could be biggest and $a$ could be lowest. $a=4$ then.

In other words,

$5a+b=25$ where $2a-b=4$ or $a=4$ or $b-a=4$

so using first condition;

$7a-4=25$ so $a$ is not integer.

or using second condition;

$6a+4=25$ so $a$ is not integer again.

and using the last condition

$a=4$, $2a=8$ and $b=5$

it seems last condition is the only solution we could have.

The first is 2 times the fourth

Third is equal to first

We define the numbers as 2a, b, 2a, a

The sum of the whole is 25

We have 2a + b + 2a + a = 25

Large and small are different by 4

We have max(a, b, 2a) - min(a, b, 2a) = 4

Solving the 2 equations...

We got 2 solutions.

8, 5, 8, 4
7, 7.5, 7, 3.5

To expand on a process to get to @Preet's answer.



We know that the biggest number must be bigger than 6. Since $25 / 4 = 6.25$

So the smallest number must be greater than 3.

We know that some number is double the other. If the smallest was 5 than the smallest double is 10, $10 - 5 = 5 > 4$ and it can be easily shown that the difference gets bigger as the small number gets bigger.

We now know that 4 must be the smallest, and the biggest is 8.

Since 8 is double of 4 we know that 4 is the fourth, and 8 is the first, because no other doubles can fit between these numbers

Since the third is the same as the first, we end up with 8, _, 8, 4

We have shown that this is the only integer solution. Using the same logic it may be possible to show that there is only an integer solution.

A proof of whether or not a non-integer solution is possible:

We know that for any value of the smallest number greater than 4, it is impossible for one number to be doubled of another.

We know that $2.25$ is a naive lower bound for the smallest as well. Since $25 / 4 = 6.25$ and $6.25 - 4 = 2.25$

So the lowest number is between 2.25 and 4.

2.25 is only possible if all 4 values were 6.25 (an immediate contradiction).

If we refine the equation such that 3 are the same, and the 4th is 4 less:
$4x - 4 = 25$
$x = 29 / 4 = 7.25$

So the largest number is greater than 7.25 so the lowest must be larger than 3.25

The maximum largest number is just under 8 (limit) if one of the two remaining numbers is less than 4. than the other must be greater than 9 to reach 25 as a sum. 9 is impossible so this is a contradiction.

So one of the middle numbers must be double the smallest. (For any number other than 4, the largest will not be double that of the smallest).

Furthermore we know that one number must exist twice, and it cannot be the smallest. If the largest was copied, than the other must be double of the smallest. $x + 2x + z + z = 25$
$3x + z + z = 25$
$z = x + 4$
$5x + 8 = 25$
$5x = 17$
$x = 3.4$
So we found another solution.
3.4, 6.8, 7.4, 7.4

Finally the last case is if the 2 non-extrema are the same. They must also be double the smallest so we get:

$x + 2x + 2x + x+4 = 25$
$6x+4 = 25$
$6x = 21$
$x = 3.5$
so another solution is
7, 7.5, 7, 3.5

It is straightforward:

    1) $~ x + y + z + w = 25$

    2) $~ x - w = 4$

    3) $~ x = 2 times w$

    4) $~ x = z$

Then

From 2) and 3) we get: $x = w + 4 = 2 times w ~~ implies w = 4$

From there you are done:

$x = 2 times w = 2 times 4 = 8$

$x = z$   therefore   $z=8$

$x + y + z + w = 8 + y + 8 + 4 = 25 implies y = 5$

My solution:

The numbers are $2a,b,2a,a$. The sum is $25$, so $5a+b=25$ means that $b$ is a multiple of $5$. A simple check using condition (2) means $bne0$ as $2a=4$ gives a total of $10$. $b=10$ fails because then $a=3$ and $10-3=7ne4$.

So $b=5, a=4$ and the numbers are $4,5,8,8$.

the first number is 8
the second number is 5
the third number is 8
and the fourth number is 4

What are the four numbers?

• The sum of the whole is 25:

$a + b + c + d = 25$

• Large and small are different by 4:

$x - y = 4$ where $x > y$ and $x$ could be ($a$ or $b$) and y could be ($b$ or $d$)

• The first is 2 times the fourth:

$(a = 2d) ... 2d + b + c + d = 25 => 3d + b + c = 25$

• Third is equal to first:

$(c = a = 2d) ... 2d + b + 2d + d = 25 => 5d + b = 25$

The step 2 is the key to solve the problem. Just replace a, b or d for x and y (excluding c since c = a)...
You will find 2 "integer" solutions:

Solution 1:

when $x = a = 2d$, and $y = d$

$x - y = 4 => 2d - d = 4 => d = 4$

Replace that in the last equation:

$5d + b = 25 => 5*4 + b = 25 => 20 + b = 25 => b = 5$

Solution: $a = 8, b = 5, c = 8, d = 4$

Solution 2: when x = b, and y = a = 2d

b - a = 4 => b = 4 + a => same as => b = 4 + 2d

Replace that in the last equation, to find d:

5d + b = 25 => 5d + (4 + 2d) = 25 => 7d = 25 - 4 => 7d = 21 => d = 3

And then in the case equation, to find b:

b = 4 + 2*3 => b = 10

• a = 2d = 2*3 = 6


• b = 10


• c = 2d = 6


• d = 3
  

I have to remove this 2nd solution since violates the rule #2. 10-3 = 7... in other words, y can't be a since a is not the smallest (a > d), Thanks @Rubio.

All the possible cases are:

* $a - b = 4$, it is possible $a > b$ (no-integer solution)

* $a - d = 4$, we are sure $a > d$ (integer solution)

* $b - d = 4$, it is possible $b > d$ (no-integer solution)

Note these are NOT a valid cases:

* $b - a = 4$, we are sure $a$ is NOT the smallest, $a > d$ (error for solution 2)

* $d - b = 4$, we are sure $d$ is NOT the largest, $a > d$

* $a - c = 4$, impossible $a = c$
* $d - a = 4$, impossible $a > d$