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Is a three-digit number PRQ divisible by 11? (1) R=P+Q (2)

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Is a three-digit number PRQ divisible by 11? (1) R=P+Q (2) [#permalink] New post 15 Aug 2003, 02:36
Is a three-digit number PRQ divisible by 11?

(1) R=P+Q
(2) P=R-Q
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Re: simple stuff [#permalink] New post 15 Aug 2003, 06:18
stolyar wrote:
Is a three-digit number PRQ divisible by 11?

(1) R=P+Q
(2) P=R-Q


stolyar, where do you dig out such nice problems? :)

i agree with mistdew. i just plugged in real numbers. you write any numbers from both sides of middle number. you sum those two - you get the middle one. the same applies to choice 2: you subtract from middle number the left one, you'll get the right side number, it is the same with choice 1.

quite interesting!
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 [#permalink] New post 15 Aug 2003, 07:06
It is D, but I did not understand your approach.

As for your ask, I invent questions—this is my hobby.
By the way, can you prove the rule in the forgoing question? It is so simple. No need to plug numbers to make sure that it works.
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Re: simple stuff [#permalink] New post 15 Aug 2003, 07:14
surat wrote:
stolyar wrote:
Is a three-digit number PRQ divisible by 11?

(1) R=P+Q
(2) P=R-Q


stolyar, where do you dig out such nice problems? :)

i agree with mistdew. i just plugged in real numbers. you write any numbers from both sides of middle number. you sum those two - you get the middle one. the same applies to choice 2: you subtract from middle number the left one, you'll get the right side number, it is the same with choice 1.

quite interesting!


First of all, you should quickly see that (2) and (1) are exactly the same. So, it's either D or E.

Plugging in number is helpful, but BE CAREFUL!! Plugging in numbers can only prove something is NOT true all the time, but cannot prove that something IS true all the time. ETS, will trap you by giving you an equation that proves true MOST of the time (sometimes true for every case but one!) -- although this is not the case here.

Here is how you can quickly analyze this type of problem.

The number "PRQ" is equal to:
100P + 10R + Q

We are given P=R+Q so:
100P + 10R + Q
= 100(R+Q) + 10R + Q
= 100R + 10Q + 10R + Q
= 110R + 11Q
= 11(10R + Q) => divisible by 11

Hence, we have now shown that EVERY 3-digit number "PRQ" where P=R+Q is divisible by 11 and can conclusively state that (1) and (2) each are sufficient to answer the question. Answer D.

Q.E.D.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Re: simple stuff [#permalink] New post 15 Aug 2003, 08:18
AkamaiBrah wrote:
surat wrote:
stolyar wrote:
Is a three-digit number PRQ divisible by 11?

(1) R=P+Q
(2) P=R-Q


stolyar, where do you dig out such nice problems? :)

i agree with mistdew. i just plugged in real numbers. you write any numbers from both sides of middle number. you sum those two - you get the middle one. the same applies to choice 2: you subtract from middle number the left one, you'll get the right side number, it is the same with choice 1.

quite interesting!


First of all, you should quickly see that (2) and (1) are exactly the same. So, it's either D or E.

Plugging in number is helpful, but BE CAREFUL!! Plugging in numbers can only prove something is NOT true all the time, but cannot prove that something IS true all the time. ETS, will trap you by giving you an equation that proves true MOST of the time (sometimes true for every case but one!) -- although this is not the case here.

Here is how you can quickly analyze this type of problem.

The number "PRQ" is equal to:
100P + 10R + Q

We are given P=R+Q so:
100P + 10R + Q
= 100(R+Q) + 10R + Q
= 100R + 10Q + 10R + Q
= 110R + 11Q
= 11(10R + Q) => divisible by 11

Hence, we have now shown that EVERY 3-digit number "PRQ" where P=R+Q is divisible by 11 and can conclusively state that (1) and (2) each are sufficient to answer the question. Answer D.

Q.E.D.


Simple, is in it? Akamai, how do you find the foregoing question?
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 [#permalink] New post 16 Aug 2003, 04:27
The number "PRQ" is equal to:
100P + 10R + Q

We are given P=R+Q so:
100P + 10R + Q
= 100(R+Q) + 10R + Q
= 100R + 10Q + 10R + Q
= 110R + 11Q
= 11(10R + Q) => divisible by 11

Hence, we have now shown that EVERY 3-digit number "PRQ" where P=R+Q is divisible by 11 and can conclusively state that (1) and (2) each are sufficient to answer the question. Answer D.

Akamai,
It should be 100R+100Q+10R+Q
= 110R+101Q
But how did u arrive at this solution
= 100(R+Q) + 10R + Q
= 100R + 10Q + 10R + Q
= 110R + 11Q
= 11(10R + Q) => divisible by 11
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 [#permalink] New post 17 Aug 2003, 02:20
Can someone please explain this in more detail i don't see how you get those equations.
  [#permalink] 17 Aug 2003, 02:20
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