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Let's solve Question 33 step by step.

The question states: "When a number is divided by 14, 20, or 35, the remainder is always 3. Find the least possible value of the number."

This problem is a case of **congruences**. Let the unknown number be \( N \). The conditions given are:
\[
N \equiv 3 \pmod{14}, \quad N \equiv 3 \pmod{20}, \quad N \equiv 3 \pmod{35}
\]
This means that the number \( N \), when divided by 14, 20, or 35, leaves a remainder of 3. 

To find the least possible value of \( N \), we need to solve for the smallest number that satisfies all three conditions. Notice that the remainder is the same in each case, so we can simplify the problem.

### Step 1: Find the Least Common Multiple (LCM)
We first find the LCM of 14, 20, and 35. The LCM will give us the base multiple that satisfies all the divisibility conditions.

- Prime factorization of 14: \( 14 = 2 \times 7 \)
- Prime factorization of 20: \( 20 = 2^2 \times 5 \)
- Prime factorization of 35: \( 35 = 5 \times 7 \)

The LCM will be the product of the highest powers of all prime factors:
\[
\text{LCM}(14, 20, 35) = 2^2 \times 5 \times 7 = 140
\]

### Step 2: General Solution
Now that we know the LCM is 140, the smallest number \( N \) that satisfies all three conditions must be of the form:
\[
N = 140k + 3
\]
where \( k \) is an integer.

### Step 3: Find the Least Value of \( N \)
To find the least value of \( N \), we start with \( k = 0 \):
\[
N = 140(0) + 3 = 3
\]

But 3 is too small, so we try \( k = 1 \):
\[
N = 140(1) + 3 = 143
\]

Thus, the least possible value of \( N \) is **143**.

---

Would you like more details on any step or have any questions? Here are five related questions you might want to explore:

1. How do you find the least common multiple (LCM) of larger numbers?
2. Can you solve a similar problem with different divisors and remainders?
3. What is the Chinese Remainder Theorem, and how does it apply here?
4. What if the remainder was different for each divisor?
5. How do prime factorizations help in finding the LCM?

**Tip**: When solving congruence problems, check if all conditions have the same remainder. This can greatly simplify the solution!

When a number is divided by 14, 20 or 35, the remainder is always 3. Find the least possible value of the number.

Learn how to solve a congruence problem where a number leaves a remainder of 3 when divided by 14, 20, and 35. This math problem involves key concepts like Least Common Multiple (LCM) and solving for the smallest number using modular arithmetic. Suitable for students in Grades 9-11.

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