Let's solve each part of the problem by finding the smallest whole number that satisfies each congruence.

Part (a)

We need to find the smallest whole number $x$ such that: $-7 \equiv x \pmod{6}$

To solve this, we need to find the remainder when $-7$ is divided by $6$. We can add $6$ repeatedly until the result is non-negative:

$-7 + 6 = -1$

$-1 + 6 = 5$

So, $-7 \equiv 5 \pmod{6}$. The smallest whole number for $x$ that makes the statement true is 5.

Part (b)

We need to find the smallest whole number $y$ such that: $10 - 17 \equiv y \pmod{9}$

First, compute $10 - 17 = -7$. Now, we need to find the remainder when $-7$ is divided by $9$. Similar to part (a), we add $9$ repeatedly until the result is non-negative:

$-7 + 9 = 2$

So, $-7 \equiv 2 \pmod{9}$. The smallest whole number for $y$ that makes the statement true is 2.

Final Answers:

• (a) $x = 5$
• (b) $y = 2$

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. What is the general method for solving linear congruences?
2. How can we solve a system of congruences using the Chinese Remainder Theorem?
3. What are some real-world applications of modular arithmetic?
4. How does modular arithmetic apply to cryptography?
5. How can we find the multiplicative inverse in modular arithmetic?

Tip: When working with modular arithmetic, always remember that adding or subtracting multiples of the modulus does not change the congruence.