Question

What is the remainder when ${\left( {5555} \right)^{2222}} + {\left( {2222} \right)^{5555}}$ is divisible by 7?

Answer 1

Hint: Try and reduce 5555 and 2222 into mixed fractions by dividing them with 7 as in case of a mixed fraction of the form $a\dfrac{p}{q}$, p is the remainder part. Then simply by applying the theorem that the number $\left[ {{a^n} + {b^n}} \right]$ is divisible by the number (a + b) we can get the answer.

Complete step-by-step answer:

Given equation is
${\left( {5555} \right)^{2222}} + {\left( {2222} \right)^{5555}}$
We have to find the remainder when this number is divided by 7.
So when we divide 5555 by 7 we get,
$ \Rightarrow \dfrac{{5555}}{7} = 793\dfrac{4}{7}$
And when we divide 2222 by 7 we get,
$ \Rightarrow \dfrac{{2222}}{7} = 317\dfrac{3}{7}$
So the remainder when 5555 and 2222 is divided by 7 are 4 and 3 respectively, because in case of a mixed fraction of the form $a\dfrac{p}{q}$, p is the remainder part.
So, the problem statement is reduces to,
$ \Rightarrow {\left( 4 \right)^{2222}} + {\left( 3 \right)^{5555}}$
Now this is also written as,
$ \Rightarrow {\left( {{4^2}} \right)^{1111}} + {\left( {{3^5}} \right)^{1111}}$
$ \Rightarrow {\left( {16} \right)^{1111}} + {\left( {243} \right)^{1111}}$
Now as we know the theorem which is the number $\left[ {{a^n} + {b^n}} \right]$ is divisible by the number (a + b).
So according to this theorem the number ${\left( {16} \right)^{1111}} + {\left( {243} \right)^{1111}}$ is divisible by (16 + 243) or it is divisible by (259).
And we all know 259 is a multiple of 7.
$ \Rightarrow 259 = 7 \times 37$
Therefore ${\left( {16} \right)^{1111}} + {\left( {243} \right)^{1111}}$ is completely divisible by 7.
Hence ${\left( {5555} \right)^{2222}} + {\left( {2222} \right)^{5555}}$ is completely divisible by 7 leaving remainder zero.
So the remainder when ${\left( {5555} \right)^{2222}} + {\left( {2222} \right)^{5555}}$ is divisible by 7 is zero.
Hence option (A) is correct.

Note: A number which when gets divided by some other number leaves a remainder, if that remainder has the number with which we have divided initially in its factors, then eventually the number is also divisible by that number with which we have divided.