If p and q are integers, what is the remainder when p^2 + q^2 is divid

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If p and q are positive integers, what is the remainder when p^2 + q^2 is divided by 5?

(1) When p - q is divided by 5, the remainder is 1.
(2) When p + q is divided by 5, the remainder is 2.

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AdiBir · Accepted Answer

Given:  
p and q are positive integers.

To Find:  
Remainder when  p^2 + q^2  is divided by 5.

Pre Process:  
To find the remainder we need to know the last digits of  p^2  and  q^2 . 
Then we can find the last digit of the sum and hence the remainder.

The important thing to note is that p and q are integers.

Now coming to the statements:  
 1)  says When p - q is divided by 5, the remainder is 1.
This can be written as  p - q = 5*a + 1. 
No information about last digits of p and q. We only know the last digit of the difference(1 or 6).
 Not Sufficient

2)  says When p + q is divided by 5, the remainder is 2.
This can be written as  p + q = 5*b + 2. 
No information about last digits of p and q. We only know the last digit of the sum(2 or 7).
 Not Sufficient

Combined  1) and 2): 
 p - q = 5*a + 1  (1)
 p + q = 5*b + 2  (2)
Add (1) and (2) =>  2*p = 5(a+b) + 3 .
Last digit of 2*p is 3 or 8. 
But as p is an integer , so 2*p must be even and last digit of 2*p must be 8.
So last digit of p is 4.

Subtract (2) and (1) =>  2*q = 5(b - a) + 1 
Last digit of 2*q is 1 or 6. 
But as q is an integer , so 2*q must be even and last digit of 2*q must be 8.
So last digit of q is 3.

From this we can easily find the remainder( last digit of  p^2  is 6 and last digit of  q^2  is 9, last digit of   p^2 + q^2  is 5, so remainder is 0.)
  Sufficient  
Option  C  is correct.

Hit Kudos if you like the expanation.

Varunsawhney8 · Answer

Best Guess E

If p and q are positive integers, what is the remainder when p^2 + q^2 is divided by 5?

(1) When p - q is divided by 5, the remainder is 1.
Let 
p-q=5x+1
Insufficient even by sq and solving

(2) When p + q is divided by 5, the remainder is 2.
p+q=5y+2
Insufficient even by sq and solving

By eq. Both as well we will not get any sol.
P=(5X+5Y+3)/2
Q=(5Y-5X+1)/2
So we will not get any answer from above eqn'Insufficient

NiftyNiffler · Answer

If p and q are positive integers, what is the remainder when p^2 + q^2 is divided by 5?

(1) When p - q is divided by 5, the remainder is 1.
(2) When p + q is divided by 5, the remainder is 2.

st1) p^2 + q^2 = (p-q)^2 + 2pq, but since we don't know the remainder of pq, INSUFFICIENT

st2) p^2 + q^2 = (p+q)^2 - 2pq, but since we don't know the remainder of pq, INSUFFICIENT

st1 and st2 taken together, p^2 + q^2 = ((p+q)^2 + (p-q)^2)/2 = rem(2^2 + 1^2) = rem(5) = 0 SUFFICIENT

So, the answer should be  C

exc4libur · Answer

(1) insufic

p-q/5=r1; p-q={1,6,11,16…}
(p,q)=(2,1), p^2+q^2=4+1=5/5=r0
(p,q)=(4,3), p^2+q^2=9+2=11/5=r1

(2) insufic

p+q/5=r2; p+q={2,7,12,17…}
(p,q)=(1,1), p^2+q^2=1+1=2/5=r2
(p,q)=(4,3), p^2+q^2=16+9=25/5=r0

(1/2) sufic

(p+q)^2= ^2= ^2
p^2+q^2+2pq=625m^2+100+500m;

(p-q)^2= ^2= ^2
p^2+q^2-2pq=625n^2+25+250n;

2(p^2+q^2)=625m^2-625n^2+75+500m-250n
p^2+q^2=25(25…-25…+3+20…-20…)/2
p^2+q^2=positive integer, thus:
25(25…-25…+3+20…-20…)/2=positive integer
p^2+q^2=25x=multiple of 5
p^2+q^2/5 has remainder 0

Ans (C)

ashh · Answer

From statement 1 => p - q = 5m + 1 which means that p - q can take values like 1,6,11,16 and so on. Now let us try different values of p,q to check if it holds true for  \frac{p^2 + q^2}{5}  gives remainder as 0.
p = 2, q = 1 then,
 \frac{p^2 + q^2}{5}  =  \frac{2^2 + 1^2}{5} 
 \frac{4 + 1}{5}  =>  \frac{5}{5}  => remainder = 0
p = 5, q = 4 then,
 \frac{p^2 + q^2}{5}  =  \frac{5^2 + 4^2}{5} 
 \frac{25 + 16}{5}  =>  \frac{41}{5}  => remainder = 1

Since we are getting different results in the 2 cases, Statement 1 is not sufficient.

From statement 2 => p + q = 5k + 2 which means that p + q can take values like 2,7,12,17 and so on. Now let us try different values of p,q to check if it holds true for  \frac{p^2 + q^2}{5}  gives remainder as 0.
p = 1, q = 1 then,
 \frac{p^2 + q^2}{5}  =  \frac{1^2 + 1^2}{5} 
 \frac{1 + 1}{5}  =>  \frac{2}{5}  => remainder = 2
p = 4, q = 3 then,
 \frac{p^2 + q^2}{5}  =  \frac{4^2 + 3^2}{5} 
 \frac{16 + 9}{5}  =>  \frac{25}{5}  => remainder = 0

Since we are getting different results in the 2 cases, Statement 2 is not sufficient.

Now combining both the statements 1 & 2 we get  p - q = 5m + 1  and  p + q = 5k + 2 . we need to check if  \frac{p^2 + q^2}{5}  gives remainder as 0 so to get that we square both the results and add them.

p^2 + q^2 - 2pq = 25m^2 + 1 + 10m  +  p^2 + q^2 + 2pq = 25k^2 + 4 + 10k

2(p^2 + q^2) = 25m^2 + 25k^2 + 10m + 10k + 5

2(p^2 + q^2) = 5(5m^2 + 5k^2 + 2m + 2k + 1)

=>  \frac{p^2 + q^2}{5}  gives remainder as 0.

So option C.

HoneyLemon · Answer

Is it C ? 
If p and q are positive integers, what is the remainder when p^2 + q^2 is divided by 5?

(1) When p - q is divided by 5, the remainder is 1.
 the unit digit of P -q can be 1 or 6 ... P ,q can be 4 and 3 so ... p^2 +q^2 = 25 .. divisible by 5 -- rem 0 .. so p , q can be 10 and 4 ., p^2 +q^2 =116 .. rem -- 1 --  not sufficient  
 
(2) When p + q is divided by 5, the remainder is 2.

the unit digit of P +q can be 2 or 7 ... P ,q can be 4 and 3 so ... p^2 +q^2 = 25 .. divisible by 5 -- rem 0 .. so p , q can be 12 or 10 ., p^2 +q^2 =244.. rem -- 4 --  not sufficient

Combining 1 and 2 . 
the unit digit of P -q can be 1 and the unit digit of P +q is 7
 so ... p^2 +q^2 = 25 .. divisible by 5 -- rem 0
  sufficient

minustark · Answer

If p and q are positive integers, what is the remainder when p^2 + q^2 is divided by 5?

(1) When p - q is divided by 5, the remainder is 1.
(2) When p + q is divided by 5, the remainder is 2.

1) When p = 3, q = 2, then p^2 + q^2 = 9 + 4 = 13 (when divided by 5, remainder is 3) . Again when, p = 2, q = 1, p^2 +q^2 = 5 (remainder is 0). Not sufficient.

2) p = q = 1, then p^2 + q^2 = 2 ( remainder is 2). When, p = 5, q = 2, p^2 + q^2 = 29 (remainder is 4). Not sufficient.

Together, p- q = 5k + 1.....(i)
and, p + q = 5k + 2.....(ii)
Now, p = 4, q =3 will meet both the equations, p^2 + q^2 = 25 (remainder is 0). Again, when p = 9, q =8, p^2 + q^2 = 81 + 64 = 145 (divisible by 5). So, we can infer that p and q will be consecutive integers and p > q. When p is even , we can write the sum as (2k)^2 +(2k -1)^2 = 4k^2 + 4k^2 - 4k + 1 = 8k^2 - 4k + 1. For k = 1, it will be 5. For k =2, it will be 32 - 8 + 1= 33 - 8 = 25, always divisible by 5.
When p is odd,p^2 + q^2 = (2k+1)^2 + (2k)^2 = 4k^2 + 4k + 1 + 4k^2 = 8k^2 + 4k +1. Now, as p + q is at least 2, and p - q is at least 1, so k can assume values as integers that is 1 less than multiples of 5 (e.g 4, 9, 14, 19 etc ). When, p = 19, the sum will be ....1 + ...4 = an integer ending with 5, so the remainder in all cases will be 0.

C is the answer.

MikeAlphaZulu · Answer

Hi Ash,

Thanks for posting your solution to this problem.
I was following everything up until the last two lines.

It's pretty clear that  2(p^2 + q^2)  is divisible by 5, but how did you deduce that  (p^2 + q^2)  is also divisible by 5?

MBAB123 · Answer

Hey if twice of a number is a divisible by 5, then that number is definitely divisible by 5 as multiplying by 2 does not affect the number's divisibility by 5.

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If p and q are integers, what is the remainder when p^2 + q^2 is divid

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