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What is the remainder when the two-digit,

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What is the remainder when the two-digit, [#permalink]

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What is the remainder when the two-digit, positive integer x is divided by 3?

(1) The sum of the digits of x is 5
(2) The remainder when x is divided by 9 is 5
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What is the remainder when the two-digit, positive integer x is divided by 3?

(1) The sum of the digits of x is 5 --> x can be 14, 41, 23, 32, or 50. Each of this numbers gives the remainder of 2 when divided by 3. Sufficient.

(2) The remainder when x is divided by 9 is 5 --> \(x = 9q + 5 = 9q + 3 + 2 =3(3q+1)+2\) --> the remainder when x is divided by 3 is 2. Sufficient.

Answer: D.
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Re: What is the remainder when the two-digit, [#permalink]

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New post 14 Dec 2013, 10:47
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1. Possible choices - 14,41,23,32,50. All give remainder 2 when divided by 3. So, sufficient.
2. Take for example 14 or any other no. which gives remainder 5 when divided by 9, it will always give remainder 2 when divided by 3. So, sufficient.

So, the correct answer is D.
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Re: What is the remainder when the two-digit, [#permalink]

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A no is divisile by 3 if the sum of the digits is divisible by 3. The remainder any no with 3 can be found by just dividing the sum of digits(of that no) with 3. If it is zero, the no is divisible by 3, otherwise the remainder obtained is same as what we would have got if we divide the no by 3.
1) sum of digits of x is 5. Hence, the remainder obtained by dividing x by 3 is same as remainder obtained by dividing 5 by 3, i.e. 2.
2) x = 9k + 5. => 3*3x + 3 + 2 = 3(3x + 1)+ 2. Now if we divide this by3, the remainder will be 2.

Hence, each statement is sufficient. Ans- D

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Re: What is the remainder when the two-digit, [#permalink]

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New post 30 Jan 2014, 16:51
Bunuel wrote:
What is the remainder when the two-digit, positive integer x is divided by 3?

(1) The sum of the digits of x is 5 --> x can be 14, 41, 23, 32, or 50. Each of this numbers gives the remainder of 2 when divided by 3. Sufficient.

(2) The remainder when x is divided by 9 is 5 --> \(x = 9q + 5 = 9q + 3 + 2 =3(3q+1)+2\) --> the remainder when x is divided by 3 is 2. Sufficient.

Answer: D.


What is the logic in statement 1? How come all numbers that add to 5 give remainder 2? Bunuel, I look forward to your response

PS. Actually, now that I think about it is because they are 2 more than multiples of 3? I guess thats the reason

Cheers
J

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Re: What is the remainder when the two-digit, [#permalink]

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New post 30 Jan 2014, 22:30
jlgdr wrote:
Bunuel wrote:
What is the remainder when the two-digit, positive integer x is divided by 3?

(1) The sum of the digits of x is 5 --> x can be 14, 41, 23, 32, or 50. Each of this numbers gives the remainder of 2 when divided by 3. Sufficient.

(2) The remainder when x is divided by 9 is 5 --> \(x = 9q + 5 = 9q + 3 + 2 =3(3q+1)+2\) --> the remainder when x is divided by 3 is 2. Sufficient.

Answer: D.


What is the logic in statement 1? How come all numbers that add to 5 give remainder 2? Bunuel, I look forward to your response

PS. Actually, now that I think about it is because they are 2 more than multiples of 3? I guess thats the reason

Cheers
J


Yes, the first statement gives the numbers which are 2 more than multiples of 3.
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Re: What is the remainder when the two-digit, [#permalink]

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Re: What is the remainder when the two-digit, [#permalink]

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What is the remainder when the two-digit, positive integer x is divided by 3?

(1) The sum of the digits of x is 5 : rule of divisibility by 3 = sum of the digits must be divisible by 3, since 5 is two more than 3, the remainder is 2
(2) The remainder when x is divided by 9 is 5 : Since 9 is a multiple of 3, and leaves a remainder of 5 then we just know that 3 will enter one more time in 5 and leave a remainder of 2

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Re: What is the remainder when the two-digit, [#permalink]

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New post 09 Jun 2017, 11:53
Imo D
From statement 1
We will get remainder 2 from every number upon division by 3
From statement 2
Again we have remainder​ of 2
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No is divisile by 3 => sum of the digits is divisible by 3.
1) sum of digits of x is 5. Hence, the remainder obtained by dividing x by 3 is same as remainder obtained by dividing 5 by 3, i.e. 2.
2) x = 9k + 5. => 3*3x + 3 + 2 = 3(3x + 1)+ 2. Now if we divide this by 3, the remainder will be 2.

Hence, each statement is sufficient. Ans- D

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Re: What is the remainder when the two-digit, [#permalink]

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New post 11 Jul 2017, 08:04
nechets wrote:
What is the remainder when the two-digit, positive integer x is divided by 3?

(1) The sum of the digits of x is 5
(2) The remainder when x is divided by 9 is 5



What is the remainder when the two-digit, positive integer x is divided by 3?

(1) The sum of the digits of x is 5

\(14 - 1 + 4 = 5 = 14/3 =\) Reminder - \(2\)

\(32 - 3 + 2 = 5 - 32/3 =\) Reminder -\(2\)

As we are getting consistent 2 as the reminder this statement is sufficient.

Hence, (1) ===== is SUFFICIENT

(2) The remainder when x is divided by 9 is 5

Possible numbers when divided by 9 to give 5 as the reminder are:

23, 32, 41..... If we divided them by 3 again we get a consistent reminder of 2

Hence, (2) ===== is SUFFICIENT

Hence, Answer is D

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Re: What is the remainder when the two-digit, [#permalink]

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New post 23 Sep 2017, 09:00
Bunuel wrote:
What is the remainder when the two-digit, positive integer x is divided by 3?

......

(2) The remainder when x is divided by 9 is 5 --> \(x = 9q + 5 = 9q + 3 + 2 =3(3q+1)+2\) --> the remainder when x is divided by 3 is 2. Sufficient.

Answer: D.


Hi Bunuel,

What is the rationale behind the manipulation you performed in (2)? Why there has to be a 3q+1? Why can't we just do 3(3q) + 5? (which is wrong..). What did I miss here?

Thanks a lot!

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Re: What is the remainder when the two-digit, [#permalink]

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New post 24 Sep 2017, 01:14
cyy12345 wrote:
Bunuel wrote:
What is the remainder when the two-digit, positive integer x is divided by 3?

......

(2) The remainder when x is divided by 9 is 5 --> \(x = 9q + 5 = 9q + 3 + 2 =3(3q+1)+2\) --> the remainder when x is divided by 3 is 2. Sufficient.

Answer: D.


Hi Bunuel,

What is the rationale behind the manipulation you performed in (2)? Why there has to be a 3q+1? Why can't we just do 3(3q) + 5? (which is wrong..). What did I miss here?

Thanks a lot!


It's not wrong: 3(3q) is divisible by 3 and 5 divided by 3 gives the remainder of 2. In the solution we just re-wrote 9q + 5 so that we directly got (a multiple of 3) + remainder because we separated a number which is less than the divisor (3). Recall that the remainder is always non-negative integer less than divisor \(0\leq{r}<d\), so \(0\leq{r}<3\).

Hope it's clear.
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Re: What is the remainder when the two-digit, [#permalink]

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New post 04 Oct 2017, 12:55
Bunuel wrote:
cyy12345 wrote:
Bunuel wrote:
What is the remainder when the two-digit, positive integer x is divided by 3?

......

(2) The remainder when x is divided by 9 is 5 --> \(x = 9q + 5 = 9q + 3 + 2 =3(3q+1)+2\) --> the remainder when x is divided by 3 is 2. Sufficient.

Answer: D.


Hi Bunuel,

What is the rationale behind the manipulation you performed in (2)? Why there has to be a 3q+1? Why can't we just do 3(3q) + 5? (which is wrong..). What did I miss here?

Thanks a lot!


It's not wrong: 3(3q) is divisible by 3 and 5 divided by 3 gives the remainder of 2. In the solution we just re-wrote 9q + 5 so that we directly got (a multiple of 3) + remainder because we separated a number which is less than the divisor (3). Recall that the remainder is always non-negative integer less than divisor \(0\leq{r}<d\), so \(0\leq{r}<3\).

Hope it's clear.


Thanks a lot, Bunuel!

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Re: What is the remainder when the two-digit,   [#permalink] 04 Oct 2017, 12:55
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