What is the unit digit of the expression S = (1! + 2! + 3! + \dots + 99!) + 3^{99}? — State PSC Exam

What is the unit digit of the expression $S = (1! + 2! + 3! + \dots + 99!) + 3^{99}$ ?

Cevap: 0

The unit digit of the given expression is 0.

The unit digit of a sum is determined by the sum of the unit digits of its individual terms. For the factorial sum

(1! + 2! + 3! + \dots + 99!)

, terms from

5!

onward end in 0 because

5! = 120

. Summing the first four terms yields

1 + 2 + 6 + 24 = 33

, contributing a unit digit of 3. For

3^{99}

, the cyclicity sequence of base 3 is 3, 9, 7, 1 (length 4). Dividing 99 by 4 leaves a remainder of 3, corresponding to

3^3 = 27

, which contributes a unit digit of 7. Summing the two unit digits yields

3 + 7 = 10

, giving a final unit digit of 0.

Evaluate the unit digit contribution of the factorial terms

Unit digit of

(1! + 2! + 3! + \dots + 99!)

is 3

Since

n!

ends in 0 for every

n \ge 5

, only

1! + 2! + 3! + 4! = 33

affects the unit digit.

Find the unit digit of

3^{99}

using pattern cyclicity

Unit digit of

3^{99}

is 7

The cyclicity of base 3 is 4. Since

99 \pmod 4 = 3

, the unit digit corresponds to

3^3 = 27

Combine the unit digits of both parts

Unit digit of

S

is 0

Adding the unit digits gives

3 + 7 = 10

, making the final unit digit 0.

Unit digit evaluation using factorial terminal zeros and exponential cyclicity

Soru