What is the unit digit of 1^(1!) + 2^(2!) + 3^(3!) + ... + 9^(9!) ?

Question

What is the unit digit of  1^{1!} + 2^{2!} + 3^{3!} + ... + 9^{9!}  ?

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

EgmatQuantExpert · Answer

Solution

To find:  
 • The unit digit of the expression  1^1! + 2^2! + 3^3! + … + 9^9!

Approach and Working:   
 • Each and every digit has its own unit digit cycle (the maximum cycle is 4 for any number). Following are the unit digits with their corresponding cycle:
 o For 1: 1 (cycle 1)
o For 2: 2 4 8 6 (cycle 4)
o For 3: 3 9 7 1 (cycle 4)
o For 4: 4 6 (cycle 2)
o For 5: 5 (cycle 1)
o For 6: 6 (cycle 1)
o For 7: 7 9 3 1 (cycle 4)
o For 8: 8 4 2 6 (cycle 4)
o For 9: 9 1 (cycle 2)

Now, 4! Onwards every factorial value is a multiple of 4, or in the form of 4k

Therefore, we can write the expression as:
 •  1^1 + 2^2 + 3^6 + 4^{4k_1} + 5^{4k_2} + 6^{4k_3} + 7^{4k_4} + 8^{4k_5} + 9^{4k_6}  
Considering the unit digits only, we can write
= 1 + 4 + 9 + 6 + 5 + 6 + 1 + 6 + 1 = 9

Hence, the correct answer is option A.

Answer: A

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ScottTargetTestPrep · Answer

This problem tests us on the knowledge of units digit of a number raised to a (positive integer) power. The following is what we need to know:

The units digit of 1, 5, and 6 raised to any power are 1, 5, and 6, respectively.

The units digit pattern of 4 is 4-6. That is, the units digit of 4^odd is 4, and that of 4^even is 6.

The units digit pattern of 9 is 9-1. That is, the units digit of 9^odd is 9, that of 9^even is 1.

The units digit pattern of 2 is 2-4-8-6. The units digit pattern of 3 is 3-9-7-1. The units digit pattern of 7 is 7-9-3-1. The units digit pattern of 8 is 8-4-2-6.

In the cases of 2, 3, 7, and 9, the first number of each pattern is the units digit when the base is raised to the 1st power, the second number is 2nd power, the third number is 3rd power, the fourth number is 4th power, and then the pattern repeats. One thing we have to emphasize is that the last number in each pattern is when the base is raised to the 4th power, but it is also the result when the base is raised to a power that is a multiple of 4. For example, the units digit of 2^24 is 6, that of 3^16 is 1, that of 7^40 is 1, and that of 8^32 is 6 since 24, 16 40 and 32 are all multiples of 4.

Notice that n! is even when n ≥ 2 and n! is a multiple of 4 when n ≥ 4. Therefore, if only the units digit is concerned, we will have the following:

1^1! = 1

2^2! = 2^2 = 4

3^3! = 3^6 = 9

4^4! = 4^even = 6

5^5! = 5

6^6! = 6

7^7! = 7^(multiple of 4) = 1

8^8! = 8^(multiple of 4) = 6

9^9! = 9^even = 1
 
Thus, the sum of the units digits of all the addends is 1 + 4 + 9 + 6 + 5 + 6 + 1 + 6 + 1 = 39, which means the sum of the given expression has a units digit of 9.

Answer: A

RohitSinha · Answer

cycle, number, last digit:
(1),1^1!,1
(2,4,8,6),2^2!,4
(3,6,9,1),3^3!,9
(4,6),4^4!,6
(5),5^5!,5
(6),6^6!(eg; c^4k),6
(7,9,3,1),7^7!,1
(8,4,2,6),8^8!,6
(9,1),9^9!,1
sum of one digits= 9
answer= A

bumpbot · Answer

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What is the unit digit of 1^(1!) + 2^(2!) + 3^(3!) + ... + 9^(9!) ?

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