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Difficulty: Very hardUnit Digit and Cyclicity

Determine the unit digit of the finite series and exponent tower expression S=∑k=150(k!)k!+777S = \sum_{k=1}^{50} (k!)^{k!} + 7^{7^7}.

Answer: 0

Answer

The unit digit of the expression is 0.
The sum of the factorial terms from k=1 to k=4 contributes unit digits of 1, 4, 6, and 6, totaling 17 (unit digit 7). Terms for k >= 5 end in 0. The power tower term 7^(7^7) has an exponent 7^7 which leaves a remainder of 3 when divided by 4. Since the cyclicity of 7 is 4, 7^3 gives a unit digit of 3. Adding 7 and 3 results in 10, giving a final unit digit of 0.

Step-by-Step Solution

1
Calculate unit digits of factorial terms for k from 1 to 4
Unit digits are 1, 4, 6, and 6 respectively
1! = 1, 2! = 2, 3! = 6, 4! = 24 (with base ending in 4 raised to an even power 24 giving unit digit 6)
2
Analyze factorial terms for k >= 5
Unit digit is 0 for all k >= 5
Factorials for k >= 5 contain factors 2 and 5, making the trailing digit 0
3
Sum the unit digits of the series sum
Sum of unit digits is 1 + 4 + 6 + 6 = 17, giving unit digit 7
Only terms from k = 1 to 4 contribute to the unit digit of the factorial sum
4
Find the unit digit of the power tower 7^(7^7) using cyclicity modulo 4
7^7 mod 4 = 3, so unit digit is 7^3 mod 10 = 3
The base 7 has a cyclicity of 4, and 7^7 is congruent to 3 modulo 4
5
Add the unit digits of both components
7 + 3 = 10, unit digit is 0
Combining the unit digit of the factorial sum (7) and the tower exponent term (3)

Key Concept

Combining factorial unit digit termination properties with exponent tower cyclicity modulo 4.
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Topics

Unit Digit and Cyclicity

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