Find the unit digit of the numerical expression N = (1! + 2! + 3! + \dots + 50!)^{25} + 8^… — State PSC Exam

Find the unit digit of the numerical expression $N = (1! + 2! + 3! + \dots + 50!)^{25} + 8^{88}$ .

Answer: 9

The unit digit of the given expression is 9.

All factorials from

5!

50!

end in 0, so the sum

(1! + 2! + 3! + \dots + 50!)

ends in

1 + 2 + 6 + 24 = 33

, which has a unit digit of 3. Raising 3 to the power 25 yields a unit digit of 3 because

25 \pmod 4 = 1

. For

8^{88}

, since 88 is divisible by 4 (

88 \pmod 4 = 0

), we take the 4th term of the cyclicity of 8, which is 6. Summing the unit digits yields

3 + 6 = 9

Determine the unit digit of the base sum

(1! + 2! + 3! + \dots + 50!)

Unit digit of base is 3.

Factorials

5!

and above all end in 0 (

5! = 120

), so only

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

contributes to the unit digit.

Calculate the unit digit of

3^{25}

Unit digit is 3.

The cyclicity pattern of 3 has period 4 (3, 9, 7, 1). Exponent

25 \pmod 4 = 1

, giving

3^1 = 3

Calculate the unit digit of

8^{88}

Unit digit is 6.

The cyclicity pattern of 8 has period 4 (8, 4, 2, 6). Exponent

88 \pmod 4 = 0

, so we take the 4th position in the cycle, which is 6.

Add the unit digits of both terms.

Final unit digit is 9.

3 + 6 = 9

, which gives a unit digit of 9.

Cyclicity of numbers and factorial unit digit properties

Question