Question

The last two digits of the number \[{3^{400}}\] are
A). 81
B). 43
C). 29
D). 01

Answer 1

Hint: To solve this question first we split the power into the multiplication of two powers. Then expand the inner part and express that in terms of the sum of two out of then one is multiple of 10. Then expand that power by the binomial theorem and find the last two digits of every term and add all those terms to get the last two digits of the given number.

Complete step-by-step solution:
To find the last two digit of number \[{3^{400}}\]
This can be written as
\[{3^{400}} = {\left( {{3^4}} \right)^{100}}\]
On calculating the power of 3
\[ = {81^{100}}\]
This can be written as
\[ = {\left( {1 + 80} \right)^{100}}\]
Expansion of binomial theorem.
\[{(a + b)^n} = {(a)^n}{\left( b \right)^0}{}^n{c_0} + {(a)^{n - 1}}{\left( b \right)^1}{}^n{c_1} + {(a)^{n - 2}}{\left( b \right)^2}{}^n{c_2} + {(a)^{n - 3}}{\left( b \right)^3}{}^n{c_3} + .......................................... + {(a)^0}{\left( b \right)^n}{}^n{c_0}\]
Put the value of \[a\] , \[b\] and \[n\] . and find the expansion.
On expanding these term by binomial theorem
\[ = {}^{100}{c_0} + {}^{100}{c_1}80 + {}^{100}{c_2}{80^2} + {}^{100}{c_3}{80^3} + ................ + {}^{100}{c_{100}}{80^{100}}\]
Here all the terms are multiples of 10 so that all the last two digit of all the terms are zero except first that is the zero followed by one.
So, all the terms are ended with double zero except the first term that is 1
On adding the last two digits of all the terms we get 01
Final answer:
The last two digits of the number are 01.

Note: Students are unable to form the type of expansion. This question is done by one another method that is looking at the pattern in the last two digits of the number and the power. Students are unable to find that the last two digits of every term are 00. Many of the students don’t know the binomial expansion of higher power.