https://file.aiquickdraw.com/mathbot/file/1728119670863s36peubi.jpg

Let's solve the problems visible in your image.

### Problem 33:
The problem asks to find the smallest number that when divided by 14, 20, or 35 leaves a remainder of 3.

This is a problem involving **modular arithmetic**. We want to find the least number \( x \) such that:
\[
x \equiv 3 \pmod{14}, \quad x \equiv 3 \pmod{20}, \quad x \equiv 3 \pmod{35}
\]

This is the same as finding the least common multiple (LCM) of 14, 20, and 35, and then adding 3 to it. So:
1. First, find the LCM of 14, 20, and 35.
   - Prime factorizations:
     - \( 14 = 2 \times 7 \)
     - \( 20 = 2^2 \times 5 \)
     - \( 35 = 5 \times 7 \)
   - The LCM will include the highest powers of all primes involved: \( 2^2 \times 5 \times 7 = 140 \).
   
2. The smallest number divisible by 14, 20, and 35 is 140. Therefore, the least number that satisfies all these conditions is \( 140 + 3 = 143 \).

So, the smallest possible value of the number is **143**.

---

### Problem 34:
Tommy has:
- 102 blue pens,
- 144 red pens,
- 210 green pens.

He packs them in such a way that each packet contains the same number of pens, and the number of pens in each packet is the same across all colors.

#### (a) At most how many pens are there in each packet?
To find the maximum number of pens in each packet, we need to determine the greatest common divisor (GCD) of 102, 144, and 210.

- Prime factorizations:
  - \( 102 = 2 \times 3 \times 17 \)
  - \( 144 = 2^4 \times 3^2 \)
  - \( 210 = 2 \times 3 \times 5 \times 7 \)

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

Thus, the greatest number of pens that can be 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?
To find the number of packets for each color, we divide the total number of pens by the number of pens in each packet:

- For blue pens: \( 102 \div 6 = 17 \) packets,
- For red pens: \( 144 \div 6 = 24 \) packets,
- For green pens: \( 210 \div 6 = 35 \) packets.

So, Tommy will have:
- 17 packets of blue pens,
- 24 packets of red pens,
- 35 packets of green pens.

Let me know if you'd like further clarification or more details on any part!

---

Here are 5 related questions for practice:
1. What is the least number that leaves a remainder of 5 when divided by 12, 18, and 24?
2. If you have 80 pencils, 120 pens, and 160 erasers, what is the maximum number of packets you can divide them into so that each packet contains an equal number of items?
3. What is the smallest number that leaves a remainder of 2 when divided by 9, 15, and 25?
4. Find the least common multiple of 16, 20, and 24.
5. What is the greatest common divisor of 156, 252, and 312?

### Tip:
When solving problems involving the least common multiple (LCM) or greatest common divisor (GCD), always break the numbers down into their prime factors—it simplifies the calculations!

Problem 33: Find the least possible number that when divided by 14, 20, or 35 leaves a remainder of 3. 
Problem 34: Tommy has 102 blue pens, 144 red pens, and 210 green pens. He packs the pens of the same color in packets so that each packet has the same number of pens. 
(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 modular arithmetic problems and apply LCM and GCD concepts. Learn how to find the least number divisible by 14, 20, and 35 with a remainder of 3, and how to divide pens equally using the GCD of the quantities.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation