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Re: What is the remainder when the two-digit, positive integer x is divide [#permalink]
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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, positive integer x is divide [#permalink]
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
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Re: What is the remainder when the two-digit, positive integer x is divide [#permalink]
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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, positive integer x is divide [#permalink]
Kuddos chacha (awesome explanation)
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Re: What is the remainder when the two-digit, positive integer x is divide [#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, positive integer x is divide [#permalink]
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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Re: What is the remainder when the two-digit, positive integer x is divide [#permalink]
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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, positive integer x is divide [#permalink]
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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

You liked the answer? Appreciate with a Kudos :good
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Re: What is the remainder when the two-digit, positive integer x is divide [#permalink]
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, positive integer x is divide [#permalink]
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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, positive integer x is divide [#permalink]
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, positive integer x is divide [#permalink]
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.


Hi Bunuel, I still do not understand the conversion to 9q + 3 + 2 =3(3q+1)+2. Could you kindly explain the logic of these two steps? Thanks!
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Re: What is the remainder when the two-digit, positive integer x is divide [#permalink]
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Better do plug in numbers I think.
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Re: What is the remainder when the two-digit, positive integer x is divide [#permalink]
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Re: What is the remainder when the two-digit, positive integer x is divide [#permalink]
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TargetMBA007 wrote:
Bunuel - Tag:Gmatprep


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Added the tag. Thank you!
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Re: What is the remainder when the two-digit, positive integer x is divide [#permalink]
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