Question

The remainder on dividing $5^{99}$ by 11 is

Hint:

When dealing with large powers and modular arithmetic, break down the exponents using the properties of divisibility to simplify calculations and find remainders efficiently.

Answer 1

Correct Answer: 9

The correct answer is 9.
$5^{99}=5^4.5^{95}$
$=625{\left[5^5\right]}^{19}$
$=625[3125{]}^{19}$
$=625[3124+1{]}^{19}$
$=625[11k\times 19+1]$
$=625\times 11k\times 19+625$
$=11k_1+616+9$
$=11\left(k_2\right)+9$
Remainder $=9$

Answer 2

\[ 5^{99} = 5^{4 \cdot 25} \cdot 5^{19} \] \[ = 625 \cdot 5^{19} \] \[ = 625 \cdot (3125)^{19} \] \[ = 625 \cdot (3124 + 1)^{19} \] Now, using the binomial expansion of \( (3124 + 1)^{19} \), we can write the expression as: \[ = 625 \cdot \left( \sum_{n=0}^{19} \binom{19}{n} 3124^{19-n} 1^n \right) \] Simplifying further: \[ = 625 \cdot \left( 3124^{19} + 19 \cdot 3124^{18} + \dots + 1 \right) \] \[ = 625 \cdot 3124^{19} + 625 \cdot 19 \cdot 3124^{18} + \dots + 625 \] The remainder when dividing by some divisor \(k\) is given as: \[ \text{Remainder} = 9 \]