Question

The unit digit of the following expression (1!)^99+(2!)^98+(3!)^97+(4!)^96+....+(99!)^1 is

Answer 1

Answer:Hi

Step-by-step explanation:

Answer 2

$\underline{\underline{\Huge{\textbf{Question:}}}}$

The unit digit of the following expression
$\textsf{(1!) ^{99} +(2!) ^{98} +(3!) ^{97} +(4!) ^{96} +....+(99!) ^{1} }$ is

$\underline{\LARGE{\textsf{Concept\: behind:}}}$

Notice the Unit's digit in factorial of the numbers given below:

$\begin{tabular}{|c|l|}\cline{1-2}Number & Value of it's factorial \\\cline{1-2}1&1\\2&2\\3&6\\4&24\\5&120\\6&720\\7&5040\\...&......(0)\\\cline{1-2}\end{tabular}$



From the table, it can be observed that the $\textsf{factorial of the Numbers from and after 5 always has at least a zero at the end}$(i.e., as Unit's digit).

This is due to the fact that factorial contains $\mathbf{2^{m}\times5^{n}}$ which places a zero at the result.

However the number of Zeroes placed at the result depends on the power of 2 or 5 $\textsf{whichever is less}$.

To find the unit digit, the $\textsf{only the first four terms}$ in the expression are considered, as these are the unit's digit contributing numbers.
The Numbers in rest of expression doesn't contribute to Unit's digit.

So, $\textbf{Required\: Unit's\: digit}$ is
$\mathbf{Unit\: digit\: in\: (1)^{99}+(2)^{98}+(6)^{97}+(24)^{96}}$ ———[1]

$\underline{\large{\textbf{Steps\: to\: find\: Unit's\: digit\: in\: (xy...z)^{n} }}}$
Here, z is the last digit of the base.

$\underline{\large{\textsf{Method:}}}$ Divide n by 4 and Note the Remainder

$\underline{\textit{Case-1:}}$ If $\textsf{Remainder = 0}$

Check if z is even/odd

z = $\left \{ {{ \: \: \: Odd; \: \: \: \: Unit's \: digt \: \: \: = 1} \atop {Even; \: \: \: Unit's \: digit \: \: \: = 6}} \right.$

$\underline{\textit{Case-2:}}$ If $\textsf{Remainder = 1}$

Unit's digit is z itself.

$\underline{\textit{Case-3:}}$ If $\textsf{Remainder = 2}$

Unit digit = Unit digit in $z^{2}$

$\underline{\textit{Case-4:}}$ If $\textsf{Remainder = 3}$

Unit digit = Unit digit in $z^{3}$

$\underline{\textit{Note:}}$
If z = 5 ; then last digit in the product is also 5.
There is no need to check any case if z=5.



$\underline{\Large{\textbf{Solution:}}}}$

$\underline{\large{\textit{Step-1:}}}$
Find Unit's digit in $(1)^{99}$

$\implies$ Unit's digit in $(1)^{99}$ = 1 ———[2]

$\underline{\large{\textit{Step-2:}}}$
Find Unit's digit in $(2)^{98}$

Remainder when 98 divided by 4 is 2.

$\implies$ Unit's digit in $(2)^{98} = (2)^{2} = 4$ ———[3]

$\underline{\large{\textit{Step-3:}}}$
Find the Unit's digit in $6^{97}$

Remainder when 97 divided by 4 = 1

$\implies$ Unit's digit in $6^{97}= 6^{1} = 6$ ———[4]

$\underline{\large{\textit{Step-4:}}}$
Find the Unit's digit in $24^{96}$

Remainder when 96 divided by 4 = 0
24 is an Even.

$\implies$ Unit's digit in $24^{96}= 6$ ———[5]

$\underline{\large{\textit{Step-5:}}}$
Substituting [2], [3], [4], [5] in [1]

Required Unit's digit = Unit digit(Sum of Unit's digit in [1])

$\implies$ Required Unit's digit = Unit digit(1 + 4 + 6 + 6 + 0)

$\implies$ Required Unit's digit = Unit's digit (17)

$\implies$ Required Unit's digit = 7.

$\therefore$

$\bigstar\: \textsf{The Unit digit of the Expression:}$
$\mathbf{(1!)^{99}+(2!)^{98}+(3!)^{97}+(4!)^{96}+....+(99!)^{1}\: is\: \underline{\large{7}}}}$