Question

Find the remainder when $7^{98}$ is divided by 5.

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Answer 1

We try to express 7 in terms of multiples of 5.If we write 7 = 10 - 3,it won't be very easy to expand it.So we will consider the next power of 7,49.

We can write $7^{98}=(49{)}^{49}$.If we write like this, we can write 49 = 50 - 1.It will be easy to expand also as (-1) only changes the sign of terms and the last term will be either 1 or (-1).

$7^{98}=(50-1{)}^{49}$

$(50-1{)}^{49}={}^{49}{C}_0(50{)}^{49}-{}^{49}{C}_1(50{)}^{48}+{}^{49}{C}_2(50{)}^{47}............{}^{49}{C}_{48}(50{)}^1-{}^{49}{C}_{49}$

${}^{49}{C}_0(50{)}^{49}-{}^{49}{C}_1(50{)}^{48}+{}^{49}{C}_2(50{)}^{47}............49\times (50{)}^1$ is multiple of 5.We will replace it by 5k.

$\Rightarrow (50-1{)}^{49}=5k-1$

$\Rightarrow (50-1{)}^{49}=5k-5+4$

$\Rightarrow (50-1{)}^{49}=5(k-1)+4$

$\Rightarrow$ The remainder will be 4.