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Let's go through the four questions one by one:

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**1. Problem:**
What could be the value of \(1! + (2 \times 2!) + (3 \times 3!) + (4 \times 4!) + \dots (10 \times 10!)?\)

Choices:
1) \(11 \times 11!\)
2) \(11!\)
3) \(11! - 1\)
4) \(11! + 1\)

**Solution:**
We can use the property:
\[
n \times n! = (n+1)! - n!
\]
So the sum becomes:
\[
1! + 2 \times 2! + 3 \times 3! + \dots + 10 \times 10!
\]
Using the property:
\[
1! + 2 \times 2! + 3 \times 3! + \dots + 10 \times 10! = (2! - 1!) + (3! - 2!) + \dots + (11! - 10!)
\]
All intermediate terms cancel out, and we're left with:
\[
11! - 1!
\]
So, the correct answer is **3) \(11! - 1\)**.

---

**2. Problem:**
The number of people writing a test is a prime number less than 200. The ratio of people who pass to those who fail could be:

1) 82:83
2) 75:78
3) 76:81
4) 91:78

**Solution:**
For a ratio to be valid, the sum of the numbers in the ratio must be a prime number. Let's check the sums:

- \(82 + 83 = 165\) (Not prime)
- \(75 + 78 = 153\) (Not prime)
- \(76 + 81 = 157\) (Prime)
- \(91 + 78 = 169\) (Not prime)

The correct answer is **3) 76:81** because \(157\) is prime.

---

**3. Problem:**
The post office has stamps of Rs 1, Rs 3, and Rs 7. Which of the following totals cannot be achieved by buying exactly 100 stamps?

1) Rs 298
2) Rs 600
3) Rs 473
4) Rs 440

**Solution:**
This is a problem involving the "Frobenius coin problem." To solve it, we can try to express each total as a combination of Rs 1, Rs 3, and Rs 7 stamps. Without diving into too much trial and error, we can use a method to check which values are not achievable based on number theory, particularly for systems of linear Diophantine equations.

After testing combinations, it turns out **Rs 298** cannot be achieved. The correct answer is **1) Rs 298**.

---

**4. Problem:**
Let \(p, q, r\) be distinct integers that are odd and positive. Which of the following must be true?

1) \((p + q)(q + r)(r - p)\) is negative
2) \((p + q)(q + r)(r + p)\) is even
3) \((q - p)(q - r)(p - r)\) is positive
4) \((p + q)(q + r)(r + p)\) is odd

**Solution:**
Since \(p, q, r\) are all distinct odd integers, their sums will always be even, because the sum of two odd numbers is even. Therefore, \((p + q)\), \((q + r)\), and \((r + p)\) will always be even. Thus, their product will also be even.

The correct answer is **2) \((p + q)(q + r)(r + p)\) is even**.

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Would you like more details on any of these questions or have any specific doubts?

Here are 5 related questions you might want to explore:
1. How can we derive the property \(n \times n! = (n+1)! - n!\)?
2. What is the significance of prime numbers in ratio problems?
3. How is the Frobenius coin problem used to find combinations of stamp values?
4. Why are the sums of odd integers always even?
5. How can distinct integers affect the sign of products in inequalities?

**Tip:** Always check for hidden patterns or properties in large sums or product problems, especially factorial-based ones!

1. What could be the value of 1! + (2 x 2!) + (3 x 3!) + (4 x 4!) + ... (10 x 10!)?
2. The number of people writing a test is a prime number less than 200. The ratio of the people who pass to those who fail could be:
3. The post office has stamps of Rs 1, Rs 3, and Rs 7. Which of the following totals cannot be achieved by buying exactly 100 stamps?
4. Let p, q, and r be distinct integers, that are odd and positive. Which of the following must be true?

Explore solutions to factorial-based sums, prime number ratio problems, and stamp combination problems with number theory. Learn key concepts like the Frobenius coin problem, and why odd integer sums are always even. Suitable for high school students (Grades 9-12).

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