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If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5 --> \(p=8q+5=(8q+4)+1=4(2q+1)+1\) --> so the remainder upon division of p by 4 is 1 (since first term is divisible by 4 and second term yields remainder of 1 upon division by 4). Sufficient.

(2) p is the sum of the squares of two positive integers --> since p is an odd integer then one of the integers must be even and another odd: \(p=(2n)^2+(2m+1)^2=4n^2+4m^2+4m+1=4(n^2+m^2+m)+1\) --> the same way as above: the remainder upon division of p by 4 is 1 (since first term is divisible by 4 and second term yields remainder of 1 upon division by 4). Sufficient.

If p is a positive odd integer, what is the remainder when p [#permalink]

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26 Jul 2012, 08:05

I am not sure if this is discussed, tried to find it but couldn't find easily.

If p is a positive odd integer, what is the remainder when p is divided by 4 ? (1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

I need help, here i thought the answer was A but the official answer given is D. Here's how i did it

(1) P = 8Q + 5, so for all values of Q starting from 1, 4 divided by P results in remainder 1. So sufficient (2) P = X^2+ Y^2, ex X=Y=1, P= 2, remainder 2. ex X=1,Y=2, P=5, remainder 1. Hence Not Sufficient.

I am not sure if this is discussed, tried to find it but couldn't find easily.

If p is a positive odd integer, what is the remainder when p is divided by 4 ? (1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

I need help, here i thought the answer was A but the official answer given is D. Here's how i did it

(1) P = 8Q + 5, so for all values of Q starting from 1, 4 divided by P results in remainder 1. So sufficient (2) P = X^2+ Y^2, ex X=Y=1, P= 2, remainder 2. ex X=1,Y=2, P=5, remainder 1. Hence Not Sufficient.

Confused?

Merging similar topics.

Please hide OA under the spoiler.
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Re: If p is a positive odd integer, what is the remainder when p [#permalink]

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26 Jul 2012, 08:11

Hi Summer101

For St 2: U have to make P an odd Integer. If u select X = Y = 1, Then P become 2 i.e. even. Thats why D is the correct answer.
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Re: If p is a positive odd integer, what is the remainder when p [#permalink]

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25 Jan 2013, 21:43

number plugging is the fastest approach for remainder problems.. because there emerges a definite pattern.. that can make the stem sufficient.
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Re: If p is a positive odd integer, what is the remainder when p [#permalink]

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31 Jan 2013, 05:03

OK.. Let me quote here something..

Let's consider x=1 and y=2... square of two number is 5 (a positive odd integer) and leaves remainder 5 when divided by 8 Again..consider x=1 and y = 6, square of two numbers is 37 (a positive odd integer) and leaves remainder 5 when divided by 8 hence, both the statements are either sufficient to ans this problem..

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

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04 Sep 2013, 11:12

D

1) Is sufficient 5,13 all give 1 as remainder. 2) sum of squares of any two positive integers, but one of them has to be odd and other an even number because p is an odd integer. So consider any pair 3,2 (3^2+2^2) (9+4)/4 1 as remainder. Or 5,2 =>29/4 =>1 as remainder.
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--It's one thing to get defeated, but another to accept it.

The question asks what will be the remainder when P is divided by 4.

Statement 1 says: p=8n+5…..which is when P is divided by 8 we get remainder 5. Now from here on you can do two things. 1. think logically 8n is divided by 4 so we won't have any remainder if we divide 8n by 4 but if we divide 5 by 4 we always get remainder 1. The value of N doesn't really matter because 8N will always be divisible by 4 and 5 will always give remainder 1. 2. plug in number and see what happens. If we put n=1,2,3,4 or so on….we get 13,21,29 respectively. Now in each case we get remainder 1.

So statement 1 is sufficient. Answer should be either A or D. So cross out B C and E.

Statement 2 says: p is sum of square of two integers just plug in numbers and see you will always get remainder 1. So statement 2 is sufficient and Answer is D.

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

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28 Sep 2014, 07:29

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If p is a positive odd integer, what is the remainder when p [#permalink]

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20 Jul 2015, 10:23

Can you please explain why the other number has to be even as p is odd? When we take two odd numbers ie {(1,3),(3,5),(5,7)} etc.. the sum of squares are (10,34,74) and all those give a remainder of 2. Please do explain.

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

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21 Jul 2015, 01:46

1

This post received KUDOS

chandrae wrote:

Can you please explain why the other number has to be even as p is odd? When we take two odd numbers ie {(1,3),(3,5),(5,7)} etc.. the sum of squares are (10,34,74) and all those give a remainder of 2. Please do explain.

Hi Chandrae,

According to the question, p needs to be a positive odd integer.

If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

In statement 2, p can only be odd if one of the squares is odd and the other is even. ( Even + Even = Even, Odd + Odd = Even and Even + Odd = Odd).

Therefore, the cases you mentioned (in which the sum of squares is even) can not be considered as eligible values of p as per the question.

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

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06 Jan 2016, 12:40

1

This post received KUDOS

BANON wrote:

If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

Got it wrong, was happy to see that Statement 1 is sufficient and rushed on Statement 2..)))

(1) p=8k+5: 5, 13, 21, 29 etc. we have a valid pattern here and each time a remainder of 1. Sufficient (2) p is the sum of the squares of two positive integers -> the least possible number is \(1^2+2^2=5\) , 13, 17, 25, 29 and each time we have a remainder of 1 (actually almost the samt thing as above) Sufficient

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

In the original condition, there is 1 variables(p), which should match with the number of equation. So you need 1 equation. For 1) 1euquation, for 2) 1 equation, which is likely to make D the answer. In 1), the remainder from p=8m+5=4(2m+1)+1 is 1, which is unique and sufficient. In 2), from p=odd=a^2+b^2, either a or b should be an even integer and the other should be an odd integer. In case of even^2+odd^2, even^2 is always divided by 4. For odd^2, from 1^2=1, 3^2=9, 5^2=25, 7^2=49, 9^2=81, all of them are divided by 4 and the remainder is 1, which is unique and sufficient. Therefore, the answer is D.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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