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

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

(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=[5(5m+2)]^2=[25m+10]^2
p^2+q^2+2pq=625m^2+100+500m;

(p-q)^2=[5(5n+1)]^2=[25n+5]^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)

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.

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
[color=#00a651]sufficient

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.

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?

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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