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If positive integer x is divided by 5, the result is p and [#permalink]

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08 Apr 2012, 02:33

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If positive integer x is divided by 5, the result is p and the remainder 3. If x is divided by 11, the remainder is 3 again, what is the remainder when p is divided by 11?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

I had to plug in numbers, how can you solve this with the remainder formula?

If positive integer x is divided by 5, the result is p and the remainder 3. If x is divided by 11, the remainder is 3 again, what is the remainder when p is divided by 11?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

I had to plug in numbers, how can you solve this with the remainder formula?

If positive integer x is divided by 5, the result is p and the remainder 3: \(x=5p+3\); If positive integer x is divided by 11, the the remainder 3: \(x=11q+3\);

Subtract one from another: \(x-x=(5p+3)-(11q+3)\) --> \(5p=11q\)---> \(\frac{p}{q}=\frac{11}{5}\) --> since both \(p\) and \(q\) are integers then \(p\) must be a multiple of 11, so it yields remainder of zero upon division by 11.

X=5P+3 , x can be 8 13 18 23...58 X=11Q+3, x can be 14,25,....58

To form the equation n=kx+r n=55K+58

Not sure how to proceed.

First of all you don't need to use that approach to solve the problem.

Next, you are making a mistake while deriving a general formula.

Positive integer x is divided by 5, the result is p and the remainder 3: \(x=5p+3\) --> \(x\) can be: 3, 8, 13, ... Notice that the least value of \(x\) for which it gives the remainder of 3 upon division by 5 is 3 itself: 3 divided by 5 yields remainder of 3.

Positive integer x is divided by 11, the the remainder 3: \(x=11q+3\) --> \(x\) can be: 3, 14, 25, ... Th same here the least value of \(x\) is 3: 3 divided by 11 yields remainder of 3.

If positive integer x is divided by 5, the result is p and the remainder 3. If x is divided by 11, the remainder is 3 again, what is the remainder when p is divided by 11?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

I had to plug in numbers, how can you solve this with the remainder formula?

If the remainder is same in both the cases, x = 5p + 3 x = 11q + 3

Re: If positive integer x is divided by 5, the result is p and [#permalink]

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27 Nov 2012, 03:35

ENAFEX wrote:

Bunuel, I tried using this method below as described in

I got stuck. Please help

X=5P+3 , x can be 8 13 18 23...58 X=11Q+3, x can be 14,25,....58

To form the equation n=kx+r n=55K+58

Not sure how to proceed.

Using the same approach, we know that at p=11 the value of X=58, for both the expressions. Hence p is a multiple of 11 so the remainder is 0. Though this is still a more time consuming approach that the ones stated above.

Re: If positive integer x is divided by 5, the result is p and [#permalink]

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07 Mar 2013, 09:23

Hi I have a quick question on this problem. How are you getting to 55 in the combined equation? Why can't X be 3? If you divide 3 by both 5 and 11, the remainder is 3 so I'm not sure what I am missing. Thanks for any help you can give.

Hi I have a quick question on this problem. How are you getting to 55 in the combined equation? Why can't X be 3? If you divide 3 by both 5 and 11, the remainder is 3 so I'm not sure what I am missing. Thanks for any help you can give.

I have discussed the general case there.

Given that: x = 5p + 3 x = 11q + 3

We can say that x = 55a + 3 i.e. when we divide x by 55 (the LCM of 5 and 11), the remainder will be 3 in that case too. To understand this fully, check out the link I gave in my previous post: http://www.veritasprep.com/blog/2011/05 ... emainders/

Sure, the number x can be 3 too. In that case p = 0, q = 0 and a = 0. When you divide p by 11, the remainder will be 0.
_________________

Re: If positive integer x is divided by 5, the result is p and [#permalink]

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31 Jul 2014, 01:15

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BN1989 wrote:

If positive integer x is divided by 5, the result is p and the remainder 3. If x is divided by 11, the remainder is 3 again, what is the remainder when p is divided by 11?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

I had to plug in numbers, how can you solve this with the remainder formula?

When we got 5p = 11k, since 5 and 11 is prime number -> k must be divisible by 5 and p must be divisible by 11 -> A is correct
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......................................................................... +1 Kudos please, if you like my post

Re: If positive integer x is divided by 5, the result is p and [#permalink]

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24 Aug 2015, 03:36

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If positive integer x is divided by 5, the result is p and [#permalink]

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12 Feb 2016, 03:57

I think I took the long road approaching this problem, not sure whether the correct one or not (after looking at the very simple and logical solution given by Bunuel), but this is the way I did it.

x= 5p+3 x=11q+3 p=11c+r

I substituted for p which lead to:

5(11c+r)+3=11q +r --> 5(11c+r)-11q=0 --> 55c+5r-11q=0 --> 11(5c-q)+5r=0. Given that r must be non negative, I concluded that in order for this equation to be 0, r must be equal to 0, as well. Please let me know if this conclusion is faulty.

Re: If positive integer x is divided by 5, the result is p and [#permalink]

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12 Feb 2016, 04:08

Bunuel wrote:

BN1989 wrote:

If positive integer x is divided by 5, the result is p and the remainder 3. If x is divided by 11, the remainder is 3 again, what is the remainder when p is divided by 11?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

I had to plug in numbers, how can you solve this with the remainder formula?

If positive integer x is divided by 5, the result is p and the remainder 3: \(x=5p+3\); If positive integer x is divided by 11, the the remainder 3: \(x=11q+3\);

Subtract one from another: \(x-x=(5p+3)-(11q+3)\) --> \(5p=11q\)---> \(\frac{p}{q}=\frac{11}{5}\) --> since both \(p\) and \(q\) are integers then \(p\) must be a multiple of 11, so it yields remainder of zero upon division by 11.

Answer: A.

Hope it's clear.

I also had the same line of thinking. thanks for all the guidance.

Re: If positive integer x is divided by 5, the result is p and [#permalink]

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27 Feb 2016, 05:03

I have done via number estimation. I have taken list of all multiple values of 11 [ 22,33,44,55,66, ..] and added 3 to each. That results to [25, 36, 47, 58, 69, ...] 58 satisfy with first rule provided in question stem. 58 / 5 = 11 * 5 + 3. Therefore when we divide 55 with 11, remainder is 0.

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