If all four of these values are integers that are divisible by 3 ...
... then what statement must be true?
Test some values out if you're unsure. Assume you need to make the most all-encompassing statement possible.
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If a is 72, B is nine, see is three then a is divisible by nine but 8+3 so 11 and that's not divisible
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It doesn't say A can be ANY multiple of 9, but to pick numbers that satisfy this equation, A will always be SOME multiple of 9
at first,
As both B A + C and C are multiples of 3 , then b A 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 × 3 = 9 .................[that works]
Technically need counter example to show the others are not true
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.
Nikita your first "one sentence" solution was the best. You've said
A/B must be in form 3X. (Because if not then (A/B+C) would not be divisible by 3). Thus A=3BX. On the other hand B=3Y. Thus A=9XY.
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)
They are integer values, not necessarily digits.
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Each of the values A,B,C & D where D=((A/B)+C) must be single digit. If each will take multiple values we will get more values and the question (or) answer also need little modification. For example, are you trying to say for the number 81279093, A=81, B=27 and C=09? For the number 81279093 the result is "A,B and C are all divisible by 9" Give stable example for your assumption.
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I don't understand why you're talking about a single long number, instead of 4 distinct integer numbers. "all four of these values are integers" and there are four rectangles with A, B, C and A/B+C. Nowhere does it say that they are digits of a single number. Also we are not trying to find numbers that adhere to the conditions, but the condition that is satisfied for all numbers, so dropping an example like 81,27,9,93 will of course not solve the problem. Ironically, we can actually drop an example (infinite number of examples, in fact),that does solve it (like 27,3,6,15), and we can also prove that the minimum condition is for A to be divisible by 9. But neither do we need to, nor should assume that the numbers in the rectangles are digits.
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@József Inczefi – Thank you. I updated the solution with required modification. By finding digital roots of integer A, B, C and ((A/B)+C) we can generalized the solution.
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@Venkatachalam J – That... actually works :)
If A = 3 i , B = 3 j , C = 3 k , and B A + C = 3 n , then
j i + 3 k = 3 n ⟹ j i = 3 ( n − k ) ( 1 )
Clearly, j i must be divisible by 3 . As a result, A must be divisible by 9 :
A = 9 a ⟹ B A = 3 j 9 a = 3 ( j a )
If B and only B were divisible by 9 , we'd have
B = 9 b ⟹ B A = 9 b 3 i = 3 b i
which is not necessarily an integer.
If A and B were both divisible by 9 , then we'd have
B A = 9 b 9 a = b a
which is not necessarily divisible by 3 .
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 B A + C. Since all of them are are at least divisible by 3, C won't help make B A be divisible by 3, B A 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 B A 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 = 1 8 , B = 6 , c = 2 1 . I used these 3 numbers, however others are also possible. 1 8 / 6 + 2 1 = 2 4 , 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 = B A + C
So we have B A to be an integer to have D to be integer.
Hence B A = 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 = 9
A, B, and C are all multiples of 3. Since C is a multiple of 3 and B A + C is also a multiple of 3, we know that B A is also a multiple of 3. If A and B A 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.
Please note if A/B + C = 3w, C = 3z, then A/B = (A/B + C) - C = 3w - 3z = 3(w - z), therefore 3 | (A/B), this is a fact from number theory, in general if x and y are integers, and d | x, d | y, then d | (ax + by) for any integers a and b. Take d = 3, a = 1, x = A/B + C, b = -1, y = C. Please note that my conclusion is perfectly true.
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that is clear now. if you would have wrote that in the solution you must have got so many upvotes.
need clear explanation.
This is really clear if you are familiar with Number Theory. What part of it is unclear? I just made a change to the last part of the proof.
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A is a multiple of 3, B is a multiple of 3. that doesn't mean b a will be a multiple of 3[suppose, A is 6 & B is 3(they are both multiple of 3) but that doesn't mean b a will be multiple of 3.how could you say that it must be ?
Please if any body else have a question about my solution, let me know. If any of you are number theorist, you see this problem would a trivial example in the first or second week of a first course in Number Theory. What I stated in my comment is actually a trivial theorem, after talking about divisibility of integers.
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.
suppose A is 6[divisible by 3] and B is 3[divisible by 3].then b a is not equal to 3 or divisible by 3. then how could you say that A is divisible by 9?
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Did you miss the part about A must have one more factor of 3 than B? 6 and 3 have one factor of 3 each. Did you even read my post?
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it was not written in the question that a must have at least 1 more factor than B. there can be lots of possibilities.
and you must answer with logic not with shitty anger.
oh! i am sorry if i have hurt you.
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@Mohammad Khaza – You seem to be confusing the proof with the problem statement, and in a way that makes me doubt that you actually read the post.
The problem says that all four are integers and that all four must be divisible by 3 for any choice of A,B,and C that are divisible by 3. Then it asks which of four statements must be true. The statement "A is divisible by 9" is the one, based on the problem given conditions. For A/B to be divisible by 3 which it must be since C is divisible by 3, A must have at least one more factor of 3 than B. (More than that A must be divisible by B to be an integer as required, your 6 and 3 at least do that.) B by hypothesis (fancy word for the problem's given conditions) has at least one, so A must have at least two and is thus divisible by 9. (Now I have repeated the original post's proof for you.) Your comment about 6 and 3 is a non-sequitur since 6/3 + C = 3k + 2 for some integer k and is clearly not divisible by 3 and does not satisfy the given conditions of the problem.
I answered with logic in the original post.
You don't hurt me, you annoy me.
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@Robert DeLisle – that is clear now. and i like to annoy people while i don't understand their solution.
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@Mohammad Khaza – I don't suffer all fools gladly, especially when they propose a combination like 6,3 that does not even satisfy the problem statement that A/B + C is divisible by 3 as a refutation to something I posted. Before you failed understand my solution, you failed to understand the problem.
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@Robert DeLisle – i failed!!!!! do you know that----the thing you see in me is a reflection of you?
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@Mohammad Khaza – I think your remark about 6 and 3 above adequately documents your failure.
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@Robert DeLisle – people come to this world with nothing.so, everything he earns is achievements.you earned indignation.i earned knowledge.
@Robert DeLisle – i like to annoy people while i don't understand the solutions.i am sorry but i will do that
Simple solution:
"Therefore, A is a multiple of 3 x 3 = 9"--A could be a multiple of 12 or other number divisible by 3. how could you know that A must be divisible of 9?
The number could also be a multiple of 12, but the fact that it is of 9 still stands. As both multiples of 3 are multiplied, the resulting A has the same factors as both multiplied which must include 2 lots of 3
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.
if it does not work then what will you try?
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)
if all statements may be true how could you sure that A is a divisible of 9?
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I was referring to the statements given as options to the answers. C might be divisible by 9; there would certainly be answers that show it to be so (9,3,9 for A,B,C for example), but there are also valid solutions where it is not (9,3,6 is one). As A/B must be divisible by three, A must be three times greater than B, and as any multiple of three multiplied by three is a multiple of nine, A must be a multiple of nine.
(It should be noted that A being a multiple of nine does not guarantee that the solution is correct. 9,9,9 is not a valid trio of numbers for the solution. But all trios of numbers where A is not a multiple of nine will be invalid. A must be a multiple of nine, the other numbers may be.)
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We have A = 3 a , B = 3 b , C = 3 c , and B A + C = 3 d , then b a + 3 c = 3 d ⟹ a = 3 b ( d − c ) Thus A = 3 a = 9 b ( d − c ) . A is divisible by 9 .
Counterexample for other choices: Take A = 9 and B = C = 3 .