If p is a positive odd integer, what is the remainder when p is divide

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.

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

Answer D

Also see the post from Ron(MGMAT) it's really good:
https://www.manhattanprep.com/gmat/foru ... t3557.html

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.

GIven: P is a positive odd integer.
Required: Remainder of p/4

Statement 1:Remainder (p/8) = 5
Hence p = 8k + 5 = 8k + 4 + 1
Hence Remainder (p/5) = 1
SUFFICIENT

Statement 2: p = x^2 + y^2
We are given that p = odd
Hence p = (2k)^2 + (2n+1)^2
p = 4(k^2 + n^2) + 4n + 1
Remainder of (p/4) = 1
SUFFICIENT

Correct Option: D

Alternative way to solve the first statement is-
As 4 is the factor of 8, so if the number leaves a remainder of 5 when divided by 8, it will leave the remainder 5-4 =1, when divided by 4. For ex if n left a remainder of 3 (or any positive integer less than 4), then remainder 3 will remain same when n is divided by 4. Detailed theory below-


If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.

If the value of ‘r’ is greater than the value of the factor, then we have to take the remainder of ‘r’ divided by the factor to get the remainder.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 3 (which is a factor of 21) will be remainder of 5/3, which is 2.

Question stem analysis:
From question stem we know that R is less than divisor i.e. R is less than 4, as the dividend is odd it can be either 3 or 1.

Statement I is pretty straightforward. Dividend can be 5, 5+8, 5+16....everytime we will get 1 as remainder.

Statement II the smallest non-negative will be "2", hence when we raise it to power of 2, we will always have 4*Someinteger form. Hence 2/4/6 raised to power of 2 will be divisible by 4. however for odd number from 1/3/5/ raised to power of two, we will always get remainder as 1.

Even^2/4 + Odd^2/4 has remainder as 1. sufficient.

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