Question

Arrange the following in increasing order as per the last digit:
A. \(2^{444}\),
B. \(17^{10}\),
C. \(13^{10}+2\),
D. \(1^1+2^2+3^3+\cdots+100^{100}\),
E. \(11!+2\).

• A. A, E, C, B, D
• B. D, C, E, A, B
• C. D, A, E, C, B
• D. D, E, C, B, A

Hint:

For last digit questions, use cyclicity of powers modulo \(10\).

Answer 1

The Correct Option is B

Concept:
To arrange numbers according to their last digit, we find each expression modulo \(10\).

Step 1: Last digit of \(2^{444}\).
The last digit cycle of powers of \(2\) is: \[ 2,4,8,6 \] Since, \[ 444\equiv 0 \pmod{4} \] the last digit is: \[ 6 \] So, \[ A \rightarrow 6 \]

Step 2: Last digit of \(17^{10}\).
Only the last digit matters, so consider \(7^{10}\). The cycle of \(7\) is: \[ 7,9,3,1 \] Since, \[ 10\equiv 2 \pmod{4} \] the last digit is: \[ 9 \] So, \[ B \rightarrow 9 \]

Step 3: Last digit of \(13^{10}+2\).
Consider \(3^{10}\). The cycle of \(3\) is: \[ 3,9,7,1 \] Since, \[ 10\equiv 2 \pmod{4} \] the last digit of \(3^{10}\) is \(9\). Hence: \[ 13^{10}+2 \] has last digit: \[ 9+2=11 \] So, last digit is: \[ 1 \] \[ C \rightarrow 1 \]

Step 4: Last digit of \(1^1+2^2+3^3+\cdots+100^{100}\).
By checking the cyclic pattern of last digits, the sum ends in: \[ 0 \] So, \[ D \rightarrow 0 \]

Step 5: Last digit of \(11!+2\).
Since \(11!\) contains factors \(2\) and \(5\), it ends in \(0\). Thus, \[ 11!+2 \] ends in: \[ 2 \] So, \[ E \rightarrow 2 \]

Step 6: Arrange in increasing order of last digit. \[ D=0,\quad C=1,\quad E=2,\quad A=6,\quad B=9 \] Therefore, the order is: \[ D,C,E,A,B \] \[ \therefore \text{Correct Answer is (B)} \]