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Zorluk: ZorUnit 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})?

Cevap: 8

Cevap

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 20262(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.

Adım Adım Çözüm

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 k5k \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 S202632026S^{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=3i=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.

Anahtar Kavram

Unit Digit and Cyclicity of Factorial and Exponential Expressions
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