Question

Find the unit’s digit of the number ${3^{1001}} \times {7^{1002}} \times {13^{1003}}$ is
A. 1
B. 3
C. 7
D. 9

Answer 1

Hint- First we find the unit digit of the given numbers by raising them to powers of 1,2,3 and so on of the cycle and stop when the unit digits start to form a cycle and then factorize the given powers in the powers of 1,2,3 and so on of the cycle whose unit digits are already known to us and then multiply the obtained unit digits and find the final answer.

Complete step-by-step answer:
Unit’s digit required =(Unit’s digit of ${3^{1001}}$ )$ \times $(Unit’s digit of ${7^{1002}}$)$ \times $ (Unit’s digit of ${13^{1003}}$)
Now we need to analyse unit digit of ${3^n}$, by this the unit digits will form a cycle i.e. repeating after a certain power
$
  {3^1} = 3 \\
  {3^2} = 9 \\
  {3^3} = 7 \\
  {3^4} = 1 \\
  {3^5} = 3 \\
 $
Now we break the given power of 3 into factors to analyse the unit digit according to this cycle
Therefore, we get ${3^{1001}} = \left( {{{\left( {{3^4}} \right)}^{250}} \times {3^1}} \right)$ .
Since unit’s digit of ${3^4} = 1$ and unit’s digit of ${3^{1001}} = \left( {{{\left( 1 \right)}^{250}} \times {3^1}} \right) = 1 \times 3 = 3$
We get the unit’s digit of ${3^{1001}}$ = 3
Now similarly analyse ${7^n}$
$
  {7^1} = 7 \\
  {7^2} = 9 \\
  {7^3} = 3 \\
  {7^4} = 1 \\
  {7^5} = 7 \\
 $
Now, breaking the given power into factors according to the cycle
Therefore, we get ${7^{1002}} = \left( {{{\left( {{7^4}} \right)}^{250}} \times {7^2}} \right)$
Since unit’s digit of ${7^4} = 1$ and unit’s digit of ${7^{1002}} = \left( {{{\left( 1 \right)}^{250}} \times {7^2}} \right) = 9$
We get the unit digit of ${7^{1002}}$ = 9
Now similarly analyse ${13^n}$
$
  {13^1} = 3 \\
  {13^2} = 9 \\
  {13^3} = 7 \\
  {13^4} = 1 \\
  {13^5} = 3 \\
 $
Now, breaking the given power into factors according to the cycle
Therefore, we get ${13^{1003}} = \left( {{{\left( {{{13}^4}} \right)}^{250}} \times {{13}^3}} \right)$
Since unit's digit of ${13^4} = 1$ and unit’s digit of ${13^{1003}} = \left( {{{\left( 1 \right)}^{250}} \times {{13}^3}} \right) = 7$
We get the unit digit of ${13^{1003}}$ = 7
Required unit digits = (Unit’s digit of ${3^{1001}}$ )$ \times $(Unit’s digit of ${7^{1002}}$)$ \times $ (Unit’s digit of ${13^{1003}}$) = $3 \times 9 \times 7$= $189$
Unit’s digit of $189$ is $9$ which is the required answer.
Option D is the right answer.

Note –Make the cycles of the given numbers raising them to the powers carefully and write their unit digit only as answers and stop when the unit digits start to repeat i.e. start to form a cycle. We have to consider only the unit digits which are unique when the number is raised to power 1,2,3 and so on of the cycle. Also, carefully factorize the powers given in question as multiples of 1,2,3 and so on of the cycle.