Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Two numbers x and y are such that when divided by 6 they leave remainders 4 and 5 respectively. Find the remainder when \[\left( {{x}^{2}}+{{y}^{2}} \right)\] is divided by 6.

seo-qna
Last updated date: 21st Jul 2024
Total views: 347.7k
Views today: 5.47k
Answer
VerifiedVerified
347.7k+ views
Hint: For solving this question you should know about the general variables addition and making their whole square. In this problem first we will make the whole square as it is provided in the information in the question and then we will find our last expression for this.

Complete step by step answer:
According to our question it is asked to us that if two numbers x and y are such that when divided by 6 they leave remainders 4 and 5 respectively. And we have to find the remainder when divided by 6.
So, as we know that if any question asks us to be any term as something with a remainder also then first we add to the remainder with that and then we find the value of that number or that digit and then if there is asked for more then we find more terms with the help of given statements.
Here, in the question the remainders are 4 and 5 respectively for the numbers x and y when divided by 6. So, if we write our original numbers then we can write it as,
Suppose, \[x=6{{k}_{1}}+4\] and \[y=6{{k}_{2}}+5\]
And \[\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( 6{{k}_{1}}+4 \right)}^{2}}+{{\left( 6{{k}_{2}}+5 \right)}^{2}}\]
If we solve this, then
\[\begin{align}
  & =36k_{1}^{2}+48{{k}_{1}}+16+36k_{2}^{2}+60{{k}_{2}}+25 \\
 & =36k_{1}^{2}+36k_{2}^{2}+48{{k}_{1}}+60{{k}_{2}}+41 \\
\end{align}\]
So, here it is clear that if we divide by 6 the remainder will be 5.

Note: While solving these types of questions you should be careful of the value of the term which is going to evaluate or which is determined according to the question statement and if that will be wrong then the next all calculation will be for different digits and there our question will be wrong.