Question

The unit’s digit in the product
${7}^{35} \times {3}^{71} \times {11}^{55}$
is:
a.1
b.3
c.7
d.9

​

Answer 1

Consider $7^{35}.$

$\begin{aligned}&7\equiv 7\pmod{10}\\ \\ &7^2\equiv 49\pmod{10}\\ \\ \Longrightarrow\ \ &7^2\equiv 9\pmod{10}\ \ \ \ \ [\because\ 49\equiv 9\pmod{10}]\\ \\ &(7^2)^2\equiv 9^2\pmod{10}\\ \\ \Longrightarrow\ \ &7^4\equiv 81\pmod{10}\\ \\ \Longrightarrow\ \ &7^4\equiv 1\pmod{10}\ \ \ \ \ [\because\ 81\equiv 1\pmod{10}]\end{aligned}$

$\begin{aligned}&(7^4)^8\equiv 1^8\pmod{10}\\ \\ \Longrightarrow\ \ &7^{32}\equiv 1\pmod{10}\\ \\ &7^{32}\times 7^3\equiv 1\times 3\pmod{10}\ \ \ \ \ [\because\ 7^3=343\equiv3\pmod{10}]\\ \\ \Longrightarrow\ \ &7^{35}\equiv 3\pmod{10}\end{aligned}$

Hence the ones digit of $7^{35}$ is 3.

Consider $3^{71}.$

$\begin{aligned}&3\equiv 3\pmod{10}\\ \\ &3^4\equiv 81\pmod{10}\\ \\ \Longrightarrow\ \ &3^4\equiv 1\pmod{10}\ \ \ \ \ [\because\ 81\equiv 1\pmod{10}]\\ \\ &(3^4)^{18}\equiv 1^{18}\pmod{10}\\ \\ \Longrightarrow\ \ &3^{72}\equiv 1\pmod{10}\\ \\ \Longrightarrow\ \ &3^{72}\equiv 21\pmod{10}\ \ \ \ \ [\because\ 21\equiv 1\pmod{10}]\\ \\ &\dfrac{3^{72}}{3}\equiv\dfrac{21}{3}\pmod{10}\\ \\ \Longrightarrow\ \ &3^{71}\equiv 7\pmod{10}\end{aligned}$

Hence the ones digit of $3^{71}$ is 7.

The ones digit of $11^{n}$ is always 1, for any whole number n. Hence so is the ones digit of $11^{55}.$

Thus,

$\begin{aligned}&7^{35}\times 3^{71}\times 11^{55}\equiv 3 \times 7\times 1\pmod{10}\\ \\ \Longrightarrow\ \ &7^{35}\times 3^{71}\times 11^{55}\equiv 21\pmod{10}\\ \\ \Longrightarrow\ \ &7^{35}\times 3^{71}\times 11^{55}\equiv\mathbf{1}\pmod{10}\ \ \ \ \ [\because\ 21\equiv 1\pmod{10}]\end{aligned}$

Hence the answer is (a) 1.