If we need to find the last two digits of an expression we need to consider the last two digits of the base. We need to consider two cases separately.
Case 1: Numbers whose units digit is $1$.
These numbers are in the format of $...abc{1^{...xyz}}$
Unit digit of this expression is always 1 as the base ends with 1.
For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit. = $ a\fbox{b}c^{xy\fbox{z}}$
Example:
Find the last two digits of ${2341^{369}}$
${2341^{369}}$ = $23\fbox{4}1^{36\fbox{9}}$ = $61$
($\because $ units digit is 1, and $9\times4 = 36; \,$ take 6 as tens place)
Case 2: Numbers whose units digits is $5$.
The last two digits are always 25 or 75.
Let the given number is $..ab{5^{xyz}}$ = $a\fbox{b}5^{xy\fbox{z}}$
If the product of units digit of the power (i.e., z) and digit left to the 5 in the base (i.e.,b), is even then last two digits of the expression is 25, If the power is odd then it is 75.
Example 1:
Last two digits of ${2345^{369}}$
= ${23\fbox45^{36\fbox9}}$
the product $4 \times 9 = 36$ which is even.
So last two digits are 25.
Example 2:
Last two digits of ${2375^{361}}$
= ${23\fbox75^{36\fbox1}}$
the product $7 \times 1 = 7$ which is odd.
So last two digits are 75.
Case 3: Numbers whose units digits are 3, 7, 9.
we need to change the unit digits 3, 7, 9 to 1 by little modification. From the unit digit table we can find that 3, 7, 9 may give unit digit 1 for the powers of 4, 4, 2 respectively.
Example:
${2343^{4747}}$ = ${43^{4747}}$
($\because $ as we are concerned with only last two digits only)
$
\require{enclose}
\begin{array}{rll}
{43^{4747}} &= {\left( {{{43}^4}} \right)^{1186}} \times {43^3}\\[4pt]
&= {\left( {{{43}^2 \times{43}^2 }} \right)^{1186}} \times {43^2}\times 43\\[4pt]
&= {\left( {{{49}\times{49} }} \right)^{1186}} \times {49} \times 43 \\[4pt]
& (\because 43^2=1849)\\[4pt]
&= 01^{1186} \times 07 \quad \\ &(\because 49^2 = 2401;\, \, 49 \times 43= 2107)\\[4pt]
&= 01 \times 07\\[4pt]
&=07
\end{array}
$
Case 4: Numbers whose units digits is 2, 4, 6.
Firstly we should by-heart these two rules: ${2^{10}}$ raised to the odd power gives the last two digits as 24, raised to the even power always gives last two digits as 76,
${\left( {{2^{10}}} \right)^{odd}} = 24$
${\left( {{2^{10}}} \right)^{even}} = 76$
Example:
Find the last two digits of ${48^{199}}$
$
\require{enclose}
\begin{array}{rll}
{48^{199}} &= {\left( {{{2}^4 \times3}} \right)^{199}} \\[4pt]
&= {\left( {{2^4}} \right)^{199}} \times {3^{199}}\\[4pt]
&= {2^{796}} \times {3^{199}} \\[4pt]
&= {\left( {{2^{10}}} \right)^{79}} \times {2^6} \times {\left( {{3^4}} \right)^{49}} \times {3^3}\\[4pt]
&= {\left( {24} \right)^{79}} \times 64 \times {\left( {81} \right)^{49}} \times 27\\[4pt]
&= 24 \times 64 \times 21 \times 27 \\[4pt]
& (\because 24^{odd} =24); \quad {\fbox81^{4\fbox9}} = 21) \\
&=12
\end{array}
$
Case 1: Numbers whose units digit is $1$.
These numbers are in the format of $...abc{1^{...xyz}}$
Unit digit of this expression is always 1 as the base ends with 1.
For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit. = $ a\fbox{b}c^{xy\fbox{z}}$
Example:
Find the last two digits of ${2341^{369}}$
${2341^{369}}$ = $23\fbox{4}1^{36\fbox{9}}$ = $61$
($\because $ units digit is 1, and $9\times4 = 36; \,$ take 6 as tens place)
Case 2: Numbers whose units digits is $5$.
The last two digits are always 25 or 75.
Let the given number is $..ab{5^{xyz}}$ = $a\fbox{b}5^{xy\fbox{z}}$
If the product of units digit of the power (i.e., z) and digit left to the 5 in the base (i.e.,b), is even then last two digits of the expression is 25, If the power is odd then it is 75.
Example 1:
Last two digits of ${2345^{369}}$
= ${23\fbox45^{36\fbox9}}$
the product $4 \times 9 = 36$ which is even.
So last two digits are 25.
Example 2:
Last two digits of ${2375^{361}}$
= ${23\fbox75^{36\fbox1}}$
the product $7 \times 1 = 7$ which is odd.
So last two digits are 75.
Case 3: Numbers whose units digits are 3, 7, 9.
we need to change the unit digits 3, 7, 9 to 1 by little modification. From the unit digit table we can find that 3, 7, 9 may give unit digit 1 for the powers of 4, 4, 2 respectively.
Example:
${2343^{4747}}$ = ${43^{4747}}$
($\because $ as we are concerned with only last two digits only)
$
\require{enclose}
\begin{array}{rll}
{43^{4747}} &= {\left( {{{43}^4}} \right)^{1186}} \times {43^3}\\[4pt]
&= {\left( {{{43}^2 \times{43}^2 }} \right)^{1186}} \times {43^2}\times 43\\[4pt]
&= {\left( {{{49}\times{49} }} \right)^{1186}} \times {49} \times 43 \\[4pt]
& (\because 43^2=1849)\\[4pt]
&= 01^{1186} \times 07 \quad \\ &(\because 49^2 = 2401;\, \, 49 \times 43= 2107)\\[4pt]
&= 01 \times 07\\[4pt]
&=07
\end{array}
$
Case 4: Numbers whose units digits is 2, 4, 6.
Firstly we should by-heart these two rules: ${2^{10}}$ raised to the odd power gives the last two digits as 24, raised to the even power always gives last two digits as 76,
${\left( {{2^{10}}} \right)^{odd}} = 24$
${\left( {{2^{10}}} \right)^{even}} = 76$
Example:
Find the last two digits of ${48^{199}}$
$
\require{enclose}
\begin{array}{rll}
{48^{199}} &= {\left( {{{2}^4 \times3}} \right)^{199}} \\[4pt]
&= {\left( {{2^4}} \right)^{199}} \times {3^{199}}\\[4pt]
&= {2^{796}} \times {3^{199}} \\[4pt]
&= {\left( {{2^{10}}} \right)^{79}} \times {2^6} \times {\left( {{3^4}} \right)^{49}} \times {3^3}\\[4pt]
&= {\left( {24} \right)^{79}} \times 64 \times {\left( {81} \right)^{49}} \times 27\\[4pt]
&= 24 \times 64 \times 21 \times 27 \\[4pt]
& (\because 24^{odd} =24); \quad {\fbox81^{4\fbox9}} = 21) \\
&=12
\end{array}
$
