A number when divided by 3 gives a remainder of 1. How many distinct v

Bunuel, chetan2u, yashikaaggarwal, nick1816, IanStewart, GMATinsight, ScottTargetTestPrep, fskilnik.

Hi all,

The cyclicity of the remainder is 1,4,7 when divided by 3.

So can this question be solved by a shortcut using the divisibility properties of 3 and 9 or any of the factors and multiple rules.?

For example, if the cyclicity of the remainder when divided by 3 is 1,4,7 then it will be the same when divided by 9 as well since 9 is the multiple of 3 or 3 is a factor of 9.

Thank you.

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I won’t call 1, 4 and 7 to be cyclicity of remainder but it will surely get you the answer.
The number is 3x+1, and as 3 is a factor of 9, all positive integer values of 3x+1 less than 9 will be the remainder when divided by 9.

I'm not quite sure what you mean by this. When we divide by 3, the only possible remainders are 0, 1 and 2, so it isn't possible for 4 or 7 to be a remainder when we divide by 3. It's true that if a number has a remainder of 1 when divided by 3, that number will give a remainder of 1, 4 or 7 when divided by 9, and that's all you need to observe to answer the intended question.

There's a problem with the wording of the question though:



The question is talking about a specific number. A specific number can only have one remainder when you divide it by 9; we don't need any other information to answer that question. They mean instead to talk about numbers generally that produce a remainder of 1 when divided by 3. Then there are three possible remainders when you divide by 9.

the no. when is divisible by 3 is leaving remainder 1, while dividing by 9 is leaving remainder 1,4,7 on cyclicity ( we did divide those no. with 9 to know the cyclicity)
Instead we can say that, 1) if a no. n is leaving any constant remainder, its square will leave remainder equal to the values for which n is leaving constant remainder till the no. less than n^2 or
2) the square of no. n will leave same distinct values after every n no.
for ex, let say 5 leaves remainder 1 when divided by no. 1,6,11,16,21,26,31................................
its square 25 will leave remainder = 1,6,11,16,21 till the no. less than 25 (21, because 26 is bigger than 25) and then after 21, the cyclicity will continue from 1,6,11,16,21. (after every 5th term) (same as in question where cyclicity is repeating after every 3rd digit)

Your method is correct but using cyclicity word can be dangerous, for instance in highlighted word one can easily interpret that the remainder of a no. when divided by 3 is 1,4,and 7 but its the no. which is leaving remainder 1. the sentence can be ambiguous there, Instead we can use the above stated lines.

Let the number be represented by \(3i + 1\). We can note the remainder when divided by 9 does cycle as \(3i + 1\) can be incremented by multiples of 9's. We can see 9/3 = 3 so every 3 numbers we will return to the original remainder. To use numbers, we can have 4, 7, 10, 13 as possible numbers but when we divide by 9 the remainders can only be 1, 4, 7, thus three possible remainders.

Ans: C

Hi,

I'm extremely sorry. I used the wrong word twice.

What I wanted to say was that the "dividend" has a "space/progression" of 3 between them when divided by 3 with a remainder of 1.

Dividend = Quotient × Divisor + Remainder
1,4,7,10,13... has a space of 3. I misused the word cyclicity for space.

Then I used these number in the above mentioned formula to find the cyclicity of 1,4,7 as remainder for these specific numbers when divided by 9.

So I concluded that 1,4,7 are the 3 distinct remainder as the quotient is the same for these numbers when divided by 9 but the remainder is different.


Sorry for the confusing.

Secondly,I got your methods as well. Thank you dear experts. chetan2u, IanStewart, yashikaaggarwal.

Thank you.

@chetan2u- Dear Chetan Sir, I got till the step that number will be of the form 3q+1.
But what after that? How are we procedding after it? Thanks in advance.

Why 4/9 is not 5? Giving us more remainders.

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