Question

Find the last two digits of \[{51^{456}} \times {61^{567}}\]

Answer 1

Hint: It is important that we analyze the repeating pattern of the occurrence of the last two digits of both the numbers. Then we draw a pattern from them and try to reach the last two digits of our required given number.

Complete step by step answer:
We are given to find the last two digits of \[{51^{456}} \times {61^{567}}\] so let us see its pattern,
 \[
  {51^1} = 51 \\
  {51^2} = 2601 = 01 \\
  {51^3} = 132651 = 51 \\
  {51^4} = 6765201 = 01 \\
 \]
So it repeats its last two digits after alternate exponents.
Then for our next number we have,
 \[
  {61^1} = 61 \\
  {61^2} = 3721 = 21 \\
  {61^3} = 226981 = 81 \\
  {61^4} = 13845841 = 41 \\
  {61^5} = 844596301 = 01 \\
  {61^6} = 51520374361 = 61 \\
 \]
This repeats after its every \[5th\] exponent.
So now we have,
 \[
  {51^{456}} = {({51^2})^{228}} = {(01)^{228}} = 01 \\
  {61^{567}} = {61^{565}}{.61^2} = {({61^5})^{113}}.(21) = {(01)^{113}}.21 = (01).21 = 21 \\
 \]
Hence, \[{51^{456}} \times {61^{567}}\] \[ = (01).21 = 21\]
Therefore, the correct answer is \[21\] .

Note:
The concept of finding the units digit of a number to some power of its exponent is comparatively easier than to find the units digit of numbers which are given in multiplied form of their exponents. It is better that we devise a pattern from the cycle of the appearance of their last digit and proceed accordingly to our problem.