Solution

• The units digit of the expression will depend on the units digits of the factorials and on the powers of each term

• To simplify things, let us temporarily focus on the units digit of the factorials and then we will move on to their exponents.

• While simplifying the factorials of the numbers we will observe the following:

o \(4! = 24\): Units digit =\(4\)

o \(5! = 120\): Units digit =\(0\)

o We do not need to calculate any further since after \(4!\), the value of the next terms will end with zeroes.

Example: \(6! = 5!*6 = 120*6 = 720\) : Units digit: \(0\)

This can be logically deduced because we have a \(5\) and \(2\) in each factorial starting from \(5!\), which gives a zero at the end every time.

• Since zero added to any number yield the same number, we can ignore it and in turn ignore all terms from \(5! ^5\) to \(10! ^{10}\).

• This method helped us eliminate all the terms after \(4!\) and we can re-write the whole expression as shown below:

\(1! ^1 + 2!^ 2 + 3! ^3 + 4! ^4\).

Now we can calculate the units digit of each of the factorials and find out the units digit of our main expression.

• Units digit of \(1! ^1\):

o We know that the cyclicity of \(1\) is \(1\).

o Thus, units digit is \(1\)

• Units digit of \(2! ^2\):

o We know \(2^2 = 4\), thus the units digit is \(4\)

• Units digit of \(3! ^3\) or \(6^3\):

o Cyclicity of \(6\) is \(1\)

o Thus, units digit of \(6^3\) = \(6\)

• Units digit of \(4!^ 4\) or \(24^4\):

o Cyclicity of \(4\) is \(2\).

o The given number is of the form \(4^{2k}\), where “k” is any natural number. Hence the units digit =\(6\)

The units digit of the expression will be the sum of all the individual unit digits calculated above.

• Units digit of \(1!^1 + 2!^2 + 3!^3 + 4!^4\) = \(1 + 4 + 6 + 6 = 17\), which implies the units digit is \(7\).

Since, the units digit of the given expression is 7, and the correct answer is Option D.

_________________

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