Calculate the units digit of the following expression:

Question

1!^1 + 2!^2 + 3!^3 +4!^4 +5!^5 ..... + 10!^{10}

A. 1
B. 3
C. 5
D. 7
E. 9

This is  Question 1 of  The e-GMAT Number Properties Marathon

Go to   Question 2 of the Marathon

CHIRANJIV87C · Answer

D) 7

In the expression fifth onward terms will have unit digits 0,
Fourth term's unit digit will be 6,
Third term's unit digit will be 6,
Second term's unit digit will be 4,
First term's unit digit will be 1,
Hence the unit digit of the expression will be given by
0+6+6+4+2+1=17 which has 7 as units digit.

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EgmatQuantExpert · Answer

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  1! = 1  
 o  2! = 2  
 o  3! = 6  
 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.

MartyMurray · Answer

Calculate the units digit of the following expression:

1!^1 + 2!^2 + 3!^3 + 4!^4 +5!^5 ..... + 10!^{10}

This question may appear to require a lot of adding. However, since we're concerned only with the units digits of the  10  terms, we can take advantage of the following.

Any factorial greater than or equal to  5!  will have at least one  5  and one  2  among its factors. As a result, it will be a multiple of  10 , and therefore it will have a units digit of  0 .

Thus, we can ignore the terms above  4!^4  and consider only the units digits of the first four terms.

1!^1 = 1  -> units digit  1

2!^2 = 4  -> units digit  4

3!^3 = 6^3  -> units digit  6

4!^4 = 24^4  -> units digit  6

1 + 4 + 6 + 6 = 17

So, the units digit of the entire expression is  7 .

A. 1
B. 3
C. 5
D. 7
E. 9

Correct answer:   D

Calculate the units digit of the following expression:

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