56 of 100: Divisible By Three

We have ${\color{#D61F06}A}=3a$ , ${\color{#3D99F6}B}=3b$ , ${\color{#20A900}C}=3c$ , and $\frac{{\color{#D61F06}A}}{{\color{#3D99F6}B}}+{\color{#20A900}C}=3d$ , then $\frac{a}{b}+3c=3d \ \ \ \ \Longrightarrow \ \ \ \ a=3b(d-c)$ Thus ${\color{#D61F06}A}=3a=9b(d-c)$ . ${\color{#D61F06}A}$ is divisible by $9$ .

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Counterexample for other choices: Take ${\color{#D61F06}A}=9$ and ${\color{#3D99F6}B}={\color{#20A900}C}=3$ .

at first,

As both $\frac{A}{B}$ + C and C are multiples of $3$ , then $\frac{A}{b}$ also must be a multiple of 3

secondly,

As, B is a multiple of 3, A divided by a multiple of 3 equals a multiple of 3

so, A must be a multiple of $3 \times 3 = 9$ .................[that works]

A/B must be in form 3X (for some X = 0,1,2,4,...).

(Because if not then (A/B+C) would not be divisible by 3.

We can write every whole number N in the form :

N=3*d + r

Where d=..., - 4, -3, - 2, - 1, 0, 1, 2, 3, 4, ... and r > 0.

If A/B can't be written as 3*X+0,

then (A/B + C) mod 3 =

= ((A/B) mod 3) + (C) mod 3 =

= (3*d + r) mod 3 + 0 =

= r mod 3 > 0)

But A=3Y and B=3Z (Y=0,1,2,4,... Z=1,2,3,4,...) since A and B are both divisible by 3.

Substitution gives us:

3Y/3Z=3X

Y=3XZ

Let XZ=W

Since A=3Y, A is in the form :

A=9W.

By definition, A is divisible by 9.

A = 9 B = 3 C = 6

9/3 = 3; 3 + 6 = 9

9, 3, 6, and 9 are all divisible by 3.

The fraction A/B must come out to be divisible by 3, so that A must be divisible by 9 in order for the A/B fraction to come out as an integer divisible by 3.

Let us use the digital roots of the integer values A, B, C and ((A/B)+C) and generalize the solution. (problem simplified to 4 digits values by checking possible digital root values)

If ${\color{#D61F06}{A}} = 3i$ , ${\color{#3D99F6}{B}} = 3j$ , ${\color{#20A900}{C}} = 3k$ , and $\dfrac{{\color{#D61F06}{A}}}{{\color{#3D99F6}{B}}} + {\color{#20A900}{C}} = 3n$ , then

$\displaystyle \begin{aligned} \frac ij + 3k = 3n \Longrightarrow \frac ij = 3\left(n - k\right) & \tag{1} \end{aligned}$

Clearly, $\dfrac ij$ must be divisible by $3$ . As a result, ${\color{#D61F06}{A}}$ must be divisible by $9$ :

$\displaystyle {\color{#D61F06}{A}} = 9a \Longrightarrow \frac{{\color{#D61F06}{A}}}{{\color{#3D99F6}{B}}} = \frac{9a}{3j} = 3\left(\frac aj\right)$

If ${\color{#3D99F6}{B}}$ and only ${\color{#3D99F6}{B}}$ were divisible by $9$ , we'd have

$\displaystyle {\color{#3D99F6}{B}} = 9b \Longrightarrow \frac{{\color{#D61F06}{A}}}{{\color{#3D99F6}{B}}} = \frac{3i}{9b} = \frac{i}{3b}$

which is not necessarily an integer.

If ${\color{#D61F06}{A}}$ and ${\color{#3D99F6}{B}}$ were both divisible by $9$ , then we'd have

$\displaystyle \frac {{\color{#D61F06}{A}}}{{\color{#3D99F6}{B}}} = \frac{{\color{#D61F06}{9a}}}{{\color{#3D99F6}{9b}}} = \frac ab$

which is not necessarily divisible by $3$ .

${\color{#20A900}{C}}$ need not be divisible by $9$ since the RHS of $(1)$ is already divisible by $3$ .

The first 3 numbers divisible by 3 are 3, 6, and 9. Then, A/B must be a whole number so it could be 9/3 or 6/3. However, 6/3 can't work because the sum would be 11, which is not divisible by 9. Therefore, A = 9, B = 3, C = 6. By: 3rd grade students (Jia, Ashley, Yeseo, and Aaryav)

Adding any two integers divisible by three results in a number divisible by three. This can be proven by dividing two multiples of three by three (which must result in integers), adding them, and multiplying the by three again (which also must result in an integer). Therefore, A/B and C are both divisible by three. If A, and B are divisble by three, then A/B results in an integer. Since A/B must also be divisble by three, A must be divisble by 3*3, which equals 9. This proves that:

A must be divisble by 9. From here, test out any values for B and C that are not multiples of 9, and you'll still get a valid statement, proving that:

A, and only A, must be divisble by 9.

The numbers equal, a=9, b=3 and c=12. Focus on the last equation. If 9/3+12=15 15/3=5 Only a is divisible by 9. This results in the following equations: a(9)/3=3 b(6)/3=2 and c(12)/3=4

Answer: Only a is divisible by 9. Because a is 9.

So we can automatically start working with $\frac{A}{B}$ + C. Since all of them are are at least divisible by 3, C won't help make $\frac{A}{B}$ be divisible by 3, $\frac{A}{B}$ has to be divisible by 3 on it's own. So, We have 2 cases (minimum) here

Case 1: B is divisible by 3, Simply if A was also only divisible by 3 and not 3^x (x is greater than or equal to 2) then B's 3 would "cancel" out A's 3 leaving $\frac{A}{B}$ not divisible by 3.

Case 2: B is divisible by 3^x (x is greater than or equal to 2) A would at least have to be 3 times B, so this one uses the same logic as the previous.

$A=18, B=6, c=21$ . I used these 3 numbers, however others are also possible. $18/6+21=24$ , of which 24 is divisible by 3. So, the only variable here that is divisible by $9$ is $**A**$

If A,B and C all are multiples if 3 and A/B+C is also divisible by 3, then A/B is divisible by three. This is possible in only one of the following two cases:

1) If A's prime Factorizations is A = 3•p, p is the product of the other prime factors of A. And B is not divisible by 3 (otherwise A/B is not divisble by 3, which it is by asumption).

2) if A's prime factorization is A = 3^n • p, p is the product of the other prime factors of A, for any positive integer n, And B's prime factorization is B=3^{n-1}•q, q is product of other prime factors of B, for any positive integer n.

Obviously 1 is false by assumption, so 2 is true. A is divisible for any n and p, b is not.

A, B and C are integer numbers.

The last one is a composed one

$D = \frac{A}{B} + C$

So we have $\frac{A}{B}$ to be an integer to have D to be integer.

Hence $\frac{A}{B} = K$ with K integer.

As A and B share the same one 3 factor to be divisible by 3, to have D divisible by 3 the number A need to have at least one other 3 as its factors to be sure that D is divisible by 3...

So at least A is divisible by $3*3=\boxed{9}$

A, B, and C are all multiples of 3. Since C is a multiple of 3 and $\frac{A}{B}+C$ is also a multiple of 3, we know that $\frac{A}{B}$ is also a multiple of 3. If A and $\frac{A}{B}$ are both multiples of 3, then A must be a multiple of 9.

Since 3 | (A/B + C) and 3 | C, so by definition of divisibility, A/B + C = 3x, C = 3y and therefore A/B = (A/B + C) - C = 3x - 3y = 3(x - y) = 3k, where x, y, and k are integers. Since 3 | B, B = 3t (here t is an integer), therefore A = (A/B)B = 3k(3t) = 9kt therefore 9 | A.

Since C is divisible by 3, A/B must be divisible by 3 in order that A/B + C is divisible by 3.

A must have at least one more factor of 3 than B so that A/B is divisible by 3. In addition A must be divisible by B in order that all the given numbers are integers as stated.

B has at least one factor of 3. A/B must be divisible by 3. Therefore A must have at least two factors of 3, that is, divisible by 9.

Simple solution:

• As both $\frac{A}{B}$ + C and C are multiples of three, then $\frac{A}{B}$ must also be a multiple of 3
• As B is a multiple of 3, A divided by a multiple of 3 equals a multiple of 3
• Therefore, A is a multiple of 3 x 3 = 9

Every time I see these kinds of problem, I like to try out the first 3 numbers. Therefore I tried 3,6 and 9 and rearranged them until they work.

Given that C is divisible by three, for A/B+C to be divisible by three then A/B must also be divisible by three. Given that B is divisible by three, A must therefore be divisible by nine, making the first statement the correct answer (note that all statements may be true, but only the first one must be true)