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Let's solve the two questions from the image you uploaded.

### Question 33:
*When a number is divided by 14, 20, or 35, the remainder is always 3. Find the least possible value of the number.*

We need to find the smallest number \( x \) such that:
\[
x \equiv 3 \pmod{14}, \quad x \equiv 3 \pmod{20}, \quad x \equiv 3 \pmod{35}
\]
This means the number is 3 more than a multiple of the least common multiple (LCM) of 14, 20, and 35.

1. The prime factorization of the numbers:
   - \( 14 = 2 \times 7 \)
   - \( 20 = 2^2 \times 5 \)
   - \( 35 = 5 \times 7 \)

2. The LCM must contain the highest powers of all primes involved:
   \[
   \text{LCM}(14, 20, 35) = 2^2 \times 5 \times 7 = 140
   \]

Thus, the least number \( x \) is:
\[
x = 140k + 3
\]
For the smallest value, set \( k = 0 \), so the least number is:
\[
x = 140 \times 0 + 3 = 3
\]

However, since \( x = 3 \) is too small to be divided by 14, 20, and 35, we move to the next \( x \) by setting \( k = 1 \):
\[
x = 140 \times 1 + 3 = 143
\]

The smallest number that satisfies all conditions is **143**.

---

### Question 34:
*Tommy has 102 blue pens, 144 red pens, and 210 green pens. He packs all the pens of the same color in packets such that the numbers of pens in all packets are the same.*

#### (a) At most how many pens are there in each packet?
This is asking for the greatest common divisor (GCD) of the numbers 102, 144, and 210.

- The prime factorization of the numbers:
  - \( 102 = 2 \times 3 \times 17 \)
  - \( 144 = 2^4 \times 3^2 \)
  - \( 210 = 2 \times 3 \times 5 \times 7 \)

The common factors are \( 2 \times 3 = 6 \).

Thus, the most number of pens in each packet is **6**.

#### (b) If the number of pens in each packet is equal to the number found in (a), how many packets of pens are there?

Now, we divide the total number of pens of each color by 6:
\[
\text{Blue packets} = \frac{102}{6} = 17, \quad \text{Red packets} = \frac{144}{6} = 24, \quad \text{Green packets} = \frac{210}{6} = 35
\]

The total number of packets is:
\[
17 + 24 + 35 = 76
\]

So, Tommy packs **76** packets of pens.

---

Let me know if you need more details or have questions. Here are five related questions to think about:

1. How do you compute the LCM and GCD of larger numbers efficiently?
2. What happens if the remainder changes in question 33 (e.g., remainder = 5)?
3. Can we use another method (like modular arithmetic) to solve question 33?
4. What if Tommy had 300 pens of each color—how would the answer change?
5. How do prime factorizations help in finding the GCD?

**Tip:** Always factorize numbers into primes first to simplify GCD and LCM calculations.

33. When a number is divided by 14, 20, or 35, the remainder is always 3. Find the least possible value of the number.
34. Tommy has 102 blue pens, 144 red pens, and 210 green pens. He packs all the pens of the same colour in packets such that the numbers of pens in all packets are the same. (a) At most how many pens are there in each packet? (b) If the number of pens in each packet is equal to the number found in (a), how many packets of pens are there?

Solve a two-part math problem involving LCM, GCD, and modular arithmetic. In the first part, find the smallest number that leaves a remainder of 3 when divided by 14, 20, and 35. In the second part, calculate the maximum number of pens per packet and the total number of packets for 102 blue, 144 red, and 210 green pens. Suitable for Grades 8-10.

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