Question

Difficulty: HardUnit Digit and Cyclicity

What is the unit digit of the composite expression E=(1!+2!+3!+⋯+20!)2026+(31×32×33×⋯×320)E = (1! + 2! + 3! + \dots + 20!)^{2026} + (3^1 \times 3^2 \times 3^3 \times \dots \times 3^{20})?

Answer: 8

Answer

The unit digit of the expression is 8.
Evaluating the expression requires breaking it down into two components. First, for the factorial sum 1!+2!+3!+⋯+20!1! + 2! + 3! + \dots + 20!, every term from 5!5! onward contains factors of both 2 and 5, so its unit digit is 0. The unit digit of the sum is determined solely by 1!+2!+3!+4!=331! + 2! + 3! + 4! = 33, which has a unit digit of 3. Raising 3 to the power 2026 gives 320263^{2026}. Since the unit digits of powers of 3 repeat in cycles of 4 (3, 9, 7, 1) and 2026≡2(mod4)2026 \equiv 2 \pmod 4, the unit digit of 320263^{2026} is 32=93^2 = 9.

Second, the product 31×32×⋯×3203^1 \times 3^2 \times \dots \times 3^{20} simplifies using the exponent addition rule to 31+2+⋯+20=32103^{1+2+\dots+20} = 3^{210}. Dividing 210 by 4 leaves a remainder of 2, so 32103^{210} also has a unit digit of 32=93^2 = 9.

Adding the unit digits of both terms gives 9+9=189 + 9 = 18, resulting in a final unit digit of 8.

Step-by-Step Solution

1
Find the unit digit of the inner factorial sum S=1!+2!+3!+⋯+20!S = 1! + 2! + 3! + \dots + 20!.
The unit digit of SS is 3.
For all k≥5k \ge 5, k!k! is divisible by 10 and ends in 0. Thus, only the sum of the first four terms 1!+2!+3!+4!=1+2+6+24=331! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 determines the unit digit.
2
Calculate the unit digit of the first term S2026≡32026S^{2026} \equiv 3^{2026}.
The unit digit of the first term is 9.
The unit digits of powers of 3 repeat in a cycle of 4 (3, 9, 7, 1). Dividing the exponent 2026 by 4 gives a remainder of 2 (2026=4×506+22026 = 4 \times 506 + 2). Therefore, the unit digit is 32=93^2 = 9.
3
Simplify the exponential product P=31×32×33×⋯×320P = 3^1 \times 3^2 \times 3^3 \times \dots \times 3^{20}.
The product simplifies to 32103^{210}.
By exponent multiplication rules, 31×32×⋯×320=3∑i=120i3^1 \times 3^2 \times \dots \times 3^{20} = 3^{\sum_{i=1}^{20} i}. The sum of the first 20 positive integers is 20×212=210\frac{20 \times 21}{2} = 210.
4
Calculate the unit digit of 32103^{210}.
The unit digit of the second term is 9.
Dividing the exponent 210 by 4 gives a remainder of 2 (210=4×52+2210 = 4 \times 52 + 2). Therefore, the unit digit is 32=93^2 = 9.
5
Combine the unit digits of the two terms.
The unit digit of EE is 8.
Adding the unit digits gives 9+9=189 + 9 = 18. The unit digit of 18 is 8.

Key Concept

Unit Digit and Cyclicity of Factorial and Exponential Expressions
Estimated Time:2m 0s
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