Individually divisible?

Since $\color{#D61F06}a$ is divisible by $3$ , we can subtract it from the other expressions to get new sums which are still divisible by $3$ . Specifically, the following sums are divisible by $3$ :

${\color{#3D99F6}b}+{\color{#20A900}c}$

${\color{#3D99F6}b}+{\color{#69047E}d}$

${\color{#20A900}c}+{\color{#69047E}d}$

${\color{#3D99F6}b}+{\color{#20A900}c}+{\color{#69047E}d}$

Subtract the first equation from the fourth to get $\color{#69047E}d$ is divisible by $3$ . Subtract the second equation from the fourth to get $\color{#20A900}c$ is divisible by $3$ . Subtract the third equation from the fourth to get $\color{#3D99F6}b$ is divisible by $3$ .

The way in which I reasoned it out is the following:

Given: ${\color{#D61F06}a}$ is divisible by 3.

Therefore from 1st statement: $({\color{#D61F06}a} + {\color{#3D99F6}b} + {\color{#20A900}c}) \% 3 = 0 \implies ({\color{#3D99F6}b}+{\color{#20A900}c}) \% 3 =0$

From statement 4: $({\color{#3D99F6}b}+{\color{#20A900}c}) \% 3 =0 \implies {\color{#69047E}d} \% 3 = 0$

Now, statement 2 implies: ${\color{#3D99F6}b}\%3=0$

and statement 3 implies: ${\color{#20A900}c}\%3=0$

Therefore all the given integers are divisible by 3.

All we need to consider are the remainders of the 4 numbers upon division by 3. For example $13\mod 3 = 1$ .

This means that the numbers can be written anew as either $0$ , $1$ or $2$ .

The case now remains the same, we need the sum of 3 of the four numbers to be divisible by 3.

Due to the Dirichlet principle, since we have 4 numbers to pick and only 3 possible remainders on division by 3, we must have at least 2 of a certain remainder in our set of 4.

Now we know that if there is 2 numbers with remainder $k$ , a third number must also have remainder $k$ , to make the sum of those 3 divisible by 3. We can also apply this rule to the fourth number, meaning that it also has remainder $k$ .

Therefore, for this entire situation to occur, all of the 4 numbers must have the same remainder on division by 3, meaning that if one of them is divisible by 3, all the others must also be divisible by 3.

Since $a + 0 + 0$ is divisible by $3$ , $a$ is divisible by $3$ .

Since $a + c + d$ is divisible by $3$ and $b + c + d$ is divisible by $3$ , we can subtract these two equations and find that $a - b$ is divisible by $3$ as well (since the difference of two values that are divisible by $3$ must also be divisible by $3$ ). And since we already know that $a$ is divisible by $3$ , then b must be divisible $3$ by the same reasoning.

Since $a + b + c$ is divisible by $3$ and we already know that both $a$ and $b$ are divisible by $3$ , $c$ must must be divisible by $3$ .

Finally, since $a + b + d$ is divisible by $3$ and we already know that both $a$ and $b$ are divisible by $3$ , $d$ must must be divisible by $3$ .

Therefore, $b$ , $c$ , and $d$ are guaranteed to be divisible by $3$ .

b+c, c+d and b+d all equal a factor of three, so they all have to be ones too.

Eq5 => a is divisible by 3. So Eq1 => b+c is divisible by 3. And so Eq4 => d is divisible by 3. And so Eq3 => c is divisible by 3. And so again Eq1 => b is divisible by 3.

from $a+0+0$ is divisible by 3, we have
$a=_3 0$ where $=_3$ means the remainder on both sides after dividing by 3 is equal. Since it is divisible by 3, the remainder is 0. Also, if the remainder is 0, then it is divisible by 3. From this, the sums can be written as $a+b+c=_3 0$ $a+b+d=_3 0$ $...$ and this goes for all of those sums since they are divisible by 3. Now let's set two sums $=_3$ to each other: $a+b+c=_3 a+b+d$ The $a+b$ cancels out on both sides just like in a normal equation and we get $c=_3 d$ If you're interested by the fact that we are allowed to apply the normal rules of algebra for " $=$ equality" to " $=_3$ equality" like cancelling out on both sides, then I would recommend learning about modular arithmetic . Now let's substitute $c$ for $d$ in the second sum: $a+d+d=_3 a+0+0=_3 0$ $0+2d=_3 0+0+0$ $2d=_3 0$ $d=_3 0$ If $2d=_3 0$ (it's divisible by 3), then $d$ has to be divisible by 3 since 2 isn't. Now we know $d=_3 c=_3 0$ so let's substitute that in $a+b+d=_3 0$ $0+b+0=_3 0$ $b=_3 0$ now we know that all of the variables are divisible by 3

Given any sum, it would be divisible by 3 if one of its factors is divisible by 3 as well. Assuming a,b are constants and they equal 1,2 respectively, c should be 3 or any of its multiples so that the sum would be divisible by 3 as well. Holding a,b constant, d needs to be a multiple of 3 for the sum to be divisible by 3. Hence the equation in question is b+c+d, and as we agreed that c and are multiples of 3, then this equation is divisible by 3.

We observe that: amod3 = 0 => a = 3n Let a+b+c = x, a+b+d = y, a+c+d = z, b+c+d=w We know that (x,y,z,w)mod3 = 0 => x=3k, y=3l, z=3m, w=3u So our equations can be rewritten as: a = 3n, 3n+b+c = 3k, 3n+b+d = 3l, 3n+c+d = 3m, b+c+d = 3u => a = 3n, b+c = 3(k-n), b+d = 3(l-n), c+d = 3(m-n), b+c+d = 3u If from the 5th equation we subtract the 4th we get: b = 3[u-(m-n)] ,if from the 5th equation we subtract the 3rd we get: c = 3[u-(l-n)] ,if from the 5th equation we subtract the 2nd we get: d = 3[u-(k-n)] Therefore b,c,d are divisible by 3. (I don't know how to format text here yet)

We know that any multiple of three subtracted from another multiple of three is equal to a multiple of three. Using this principle, we can determine the value of a, b, and c, in terms of wether or not they are divisable by three. Since a+0+0 is divisable by three, a is divisable by three. We can now remove a from all the remaining equations and know the result will be divisable by three. We see that b+c is a multiple of three and can therefore be removed with the guarantee that the result will be divisable be there. This shows that c is divisable by three. We can then use this in the equations c+b and c+d to determine that b and d are divisable by three. Therefore, a, b, c, d and all possible sums of these values are all divisable by three.

Add first three lines: 3(a+b+c+d), (a+b+c+d)-(a+b+c)=d, (a+b+c+d)-(a+c+d)=b... etc

Considering eqns of form a+b+c=3k , a+b+d=3k ;a+c+d=3k : b+c+d=3k and a=3k ; where k can be any integer just to represent the divisibility by 3. Now subtract the eqn 4 by eqn 1 which gives a-d =3k => 3k-d=3k => d=3k therefore d is divisible by 3. Similarly all can be proved.

Every numbers are divisible by 3. Even when its division results in a fraction. The question never said that results must be integer.

Using modular $3$ arithmetic, the fifth equation implies that $a \equiv 0$ ; from this, the first equation then implies that $b + c \equiv 0$ . So, the fourth equation implies that $d \equiv 0$ , and hence the second and third equation implies that $b \equiv c \equiv 0$ .

There are just sums, so if every number its divisible by 3, the sum will be divisible by 3. Ex a=b=c=d=3

Let X,Y,Z,H,Q be random integers;
a+0+0 is divisible by 3 => a=3X ; ......(1)
then
a+b+c => 3X+b+c=3Y;......(2)
a+b+d => 3X+b+d=3Z;......(3)
a+c+d => 3X+c+d=3H;......(4)
- b+c+d =3Q =>b+c=3Q-d.......(5)


;
Let (5) in (2): => 3X+3Q-d=3Y=> d=3(X+Q-Y) =>It means d is divisibel by 3.
The proves for b and c can be easily derived from above method.

Because a + 0 + 0 is divisible by 3, a must be divisible by 3. Knowing this, we can know that all the other sums involving a will also be divisible by 3 when a is subtracted. So b+c, c+d, and b+d are all divisible by 3. Therefore in b+c+d, any 2 of these letters combine to form an integer divisible by 3, so to add the 3rd letter and still form a number divisible by 3, it must also be divisible 3.

a is divisible by 3 (fifth expression). Watching the first expression either

1) a, b and c is divisible by 3, thus (by the second expression) d is divisible by 3, or

2) the remainders of b/3 and c/3 are 1 and 2, in not particular order (two cases).

• 2/a) if the remainder of b/3 is 1, using the second expression comes that the remainder of d/3 is 2. From the third expression comes that the remainder of c/3 is 1, which contradicts 2).
• 2/b) the problem is symmetric for b and c, we don't need to examine the other case, where the remainder of c/3 is 1. If you insist, we can come to a contradiction the same way as in 2/a).

As 2) leads to contradiction, 1) is true, all numbers must be divisible by 3.

Side note: We didn't need the fourth expression. You can remove one of the first four expressions, to be exact. Doesn't matter, which one. We need the fifth, without that we could just conclude that all of the numbers give the same remainder when divided by 3.

$3|a \implies 3|(b + d) \implies 3|c \implies 3|d \implies 3|b$ .

Just add all of the expressions:

$4a + 3b + 3c + 3d$

= $4a + 3(b + c + d)$

Once can clearly see that b, c and d are now clearly divisible by 3.