# Understanding Cryptography by Christof Paar and Jan Pelzl - Chapter 1 Solutions - Ex1.9

- 1 min

## Exercise 1.9

Compute $x$ as far as possible without a calculator. Where appropriate, make use of a smart decomposition of the exponent as shown in the example in Sect. 1.4.1:

1. x = 32 mod 13
2. x = 72 mod 13
3. x = 310 mod 13
4. x = 7100 mod 13
5. 7x = 11, mod 13

The last problem is called a discrete logarithm and points to a hard problem which we discuss in Chap. 8. The security of many public-key schemes is based on the hardness of solving the discrete logarithm for large numbers, e.g., with more than 1000 bits.

### Solution

This solution is verified as correct by the official Solutions for Odd-Numbered Questions manual.

These can be performed by performing a smaller exponentiation and reducing:

1.

$3^2\,\mathrm{mod}\,13 \equiv 9\,\mathrm{mod}\,13$

2.

$7^2\,\mathrm{mod}\,13 \equiv 49\,\mathrm{mod}\,13 \equiv 10\,\mathrm{mod}\,13$

3.

$3^{10}\,\mathrm{mod}\,13 \equiv 9^5\,\mathrm{mod}\,13 \equiv 81^2 \times 9\,\mathrm{mod}\,13 \\ \equiv\,3^2 \times 9\,\mathrm{mod}\,13\equiv81\,\mathrm{mod}\,13\equiv3\,\mathrm{mod}\,13$

4.

$7^{100}\,\mathrm{mod}\,13$ $\equiv {7^2}^{50}\,\mathrm{mod}\,13$ $\equiv 49^{50}\,\mathrm{mod}\,13$ $\equiv 10^{50}\,\mathrm{mod}\,13$ $\equiv {10^2}^{25}\,\mathrm{mod}\,13$ $\equiv 100^{25}\,\mathrm{mod}\,13$ $\equiv 9^{25}\,\mathrm{mod}\,13$ $\equiv {9^2}^{12} \times 9\,\mathrm{mod}\,13$ $\equiv 81^{12} \times 9\,\mathrm{mod}\,13$ $\equiv 3^{12} \times 9\,\mathrm{mod}\,13$ $\equiv {3^3}^{4} \times 9\,\mathrm{mod}\,13$ $\equiv 27^{4} \times 9\,\mathrm{mod}\,13$ $\equiv 1^{4} \times 9\,\mathrm{mod}\,13$ $\equiv 9\,\mathrm{mod}\,13$

5. Through trial and error, we can discover the value of $x$:

$7^5\,\mathrm{mod}\,13 \equiv 11\,\mathrm{mod}\,13$