Question

Last two digits in \(9^{50}\) are _ _ _ _ _.

Hint:

For last two digits, work modulo 100. \(9^{10} \equiv 1 \pmod{100}\) is a useful shortcut.

Answer 1

Correct Answer: 1

Concept: To find last two digits, compute \(9^{50} \mod 100\).

Step 1: Find cycle: \[ 9^1 = 09 \] \[ 9^2 = 81 \] \[ 9^3 = 81 \times 9 = 729 \equiv 29 \pmod{100} \] \[ 9^4 = 29 \times 9 = 261 \equiv 61 \pmod{100} \] \[ 9^5 = 61 \times 9 = 549 \equiv 49 \pmod{100} \] \[ 9^6 = 49 \times 9 = 441 \equiv 41 \pmod{100} \] \[ 9^7 = 41 \times 9 = 369 \equiv 69 \pmod{100} \] \[ 9^8 = 69 \times 9 = 621 \equiv 21 \pmod{100} \] \[ 9^9 = 21 \times 9 = 189 \equiv 89 \pmod{100} \] \[ 9^{10} = 89 \times 9 = 801 \equiv 01 \pmod{100} \]

Step 2: Since \(9^{10} \equiv 01 \pmod{100}\): \[ 9^{50} = (9^{10})^5 \equiv (01)^5 = 01 \pmod{100} \]

Step 3: Therefore, last two digits = 01.