Activity Questions 1.1

Based on the sources, the set of natural numbers is denoted by N. This set includes 0, 1, 2, 3, 4, 5, and so on. The sources explicitly state that whenever they are talking about natural numbers, it always includes a 0, even though some books may not. Natural numbers are primarily used for counting.

The given list of numbers is: 22, -17, 47, -2000, 0, 1, 43, 1729, 6174, -63, 100, 32, -9.

Let’s examine each number in the list to determine if it is a natural number according to the definition provided in the sources:

• 22: This is a non-negative integer greater than 0. It is in the set {0, 1, 2, 3, …}. So, 22 is a natural number.
• -17: This is a negative integer. The set of integers (Z) extends the natural numbers with negative numbers. Natural numbers are non-negative integers starting from 0. So, -17 is not a natural number.
• 47: This is a non-negative integer greater than 0. It is in the set {0, 1, 2, 3, …}. So, 47 is a natural number.
• -2000: This is a negative integer. So, -2000 is not a natural number.
• 0: The sources explicitly state that the set of natural numbers includes 0. So, 0 is a natural number.
• 1: This is a non-negative integer greater than 0. It is in the set {0, 1, 2, 3, …}. So, 1 is a natural number.
• 43: This is a non-negative integer greater than 0. It is in the set {0, 1, 2, 3, …}. So, 43 is a natural number.
• 1729: This is a non-negative integer greater than 0. It is in the set {0, 1, 2, 3, …}. So, 1729 is a natural number.
• 6174: This is a non-negative integer greater than 0. It is in the set {0, 1, 2, 3, …}. So, 6174 is a natural number.
• -63: This is a negative integer. So, -63 is not a natural number.
• 100: This is a non-negative integer greater than 0. It is in the set {0, 1, 2, 3, …}. So, 100 is a natural number.
• 32: This is a non-negative integer greater than 0. It is in the set {0, 1, 2, 3, …}. So, 32 is a natural number.
• -9: This is a negative integer. So, -9 is not a natural number.

The natural numbers in the list are: 22, 47, 0, 1, 43, 1729, 6174, 100, 32.

Counting these numbers, we find there are 9 natural numbers in the given list.

This corresponds to one of the options provided. The number of natural numbers in the list is 9.

Okay, let’s determine the value of M, the number of students in the class, based on the information provided and drawing on the concepts discussed in the sources.

We are given the following facts:

1. There are M students in the class.
2. Each student is distributed 3 pens.
3. After distributing the pens, 4 pens remain.
4. The total number of pens is 70.

The number of students M is a count of objects (students), which aligns with the concept of natural numbers. The sources define the set of natural numbers (N) as {0, 1, 2, 3, 4, 5, …}, explicitly stating that it includes 0. In this context, M must be a positive natural number since there is a class of students. The number of students in a class can also be thought of as the cardinality of the set of students in the class, and for a finite set like a class, the cardinality is a natural number.

The total number of pens distributed can be calculated using multiplication, which the sources describe as repeated addition. If there are M students and each receives 3 pens, the total number of pens distributed is 3 times M, or 3M.

The total number of pens (70) is the sum of the pens distributed (3M) and the pens remaining (4). This involves the addition operation. So, we can write this as an equation:

Total Pens = (Pens per student × Number of students) + Remaining pens 70 = (3 × M) + 4 70 = 3M + 4

To find the value of M, we need to solve this equation. This involves using arithmetic operations. We can isolate the term with M by subtracting 4 from both sides of the equation:

70 - 4 = 3M + 4 - 4 66 = 3M

Next, we need to find M. This involves division, which the sources describe as repeated subtraction. We divide both sides of the equation by 3:

66 / 3 = 3M / 3 22 = M

So, the value of M is 22.

Let’s verify this: If there are 22 students and each gets 3 pens, that’s 22 * 3 = 66 pens distributed. With 4 pens remaining, the total is 66 + 4 = 70 pens, which matches the given total.

The value of M is 22. This is one of the options provided.

Based on the sources and our conversation, we can determine the sixteenth prime number.

Firstly, the sources define a prime number as one which has no factors other than 1 and itself. More precisely, a prime number has exactly two factors: 1 and the number itself.

The number 1 is not considered a prime number because it only has one factor, which is 1 itself.

The sources state that the smallest prime number is 2, because its only factors are 1 and 2. This aligns with your starting point, “Consider ‘2’ to be the first prime number”.

After 2, no even numbers can be primes because they are all multiples of 2, meaning 2 divides them. Identifying prime numbers involves checking their factors or using a method like the sieve of Eratosthenes, which involves knocking off multiples of primes. A number a is a factor of b if b mod a is 0, meaning if b is divided by a, there is no remainder.

Let’s list the prime numbers in order, starting from 2, until we reach the sixteenth:

1. 2 (Factors: 1, 2)
2. 3 (Factors: 1, 3)
3. 5 (Factors: 1, 5)
4. 7 (Factors: 1, 7) (Skip 8 - even) (Skip 9 - multiple of 3) (Skip 10 - even)
5. 11 (Factors: 1, 11) (Skip 12 - even)
6. 13 (Factors: 1, 13) (Skip 14 - even) (Skip 15 - multiple of 3 and 5) (Skip 16 - even)
7. 17 (Factors: 1, 17) (Skip 18 - even)
8. 19 (Factors: 1, 19) (Skip 20 - even) (Skip 21 - multiple of 3 and 7) (Skip 22 - even)
9. 23 (Factors: 1, 23) (Skip 24 - even) (Skip 25 - multiple of 5) (Skip 26 - even) (Skip 27 - multiple of 3) (Skip 28 - even)
10. 29 (Factors: 1, 29) (Skip 30 - even)
11. 31 (Factors: 1, 31) (Skip 32 - even) (Skip 33 - multiple of 3 and 11) (Skip 34 - even) (Skip 35 - multiple of 5 and 7) (Skip 36 - even)
12. 37 (Factors: 1, 37) (Skip 38 - even) (Skip 39 - multiple of 3 and 13) (Skip 40 - even)
13. 41 (Factors: 1, 41) (Skip 42 - even)
14. 43 (Factors: 1, 43) (Skip 44 - even) (Skip 45 - multiple of 3 and 5) (Skip 46 - even)
15. 47 (Factors: 1, 47) (Skip 48 - even) (Skip 49 - multiple of 7) (Skip 50 - even) (Skip 51 - multiple of 3 and 17) (Skip 52 - even)
16. 53 (Factors: 1, 53)

Counting through the list, the sixteenth prime number is 53.

Comparing this to the provided options:

• 47 (This is the 15th prime number)
• 61
• 59
• 53 (This is the 16th prime number)

Therefore, the sixteenth prime number is 53.

Based on the sources and our conversation history, we can evaluate which of the given options are possible values for x by using the concept of the modulo operation.

The sources define the modulus as another word for remainder, and it is written as “mod”. The notation a mod b is the remainder when a is divided by b. For instance, 19 mod 5 is 4, because when 19 is divided by 5, the remainder is 4.

A number a is a factor of b if b mod a is 0, meaning there is no remainder when b is divided by a. This indicates that b is a multiple of a.

The conditions given for x are:

1. x mod 3 = 1
2. x mod 7 = 4

This means that when x is divided by 3, the remainder is 1. Also, when x is divided by 7, the remainder is 4.

Let’s check each of the possible values provided:

• 25:

  • Check 25 mod 3: 25 divided by 3 is 8, with a remainder of 1 (25 = 3 × 8 + 1). So, 25 mod 3 = 1. This condition is satisfied.
  • Check 25 mod 7: 25 divided by 7 is 3, with a remainder of 4 (25 = 7 × 3 + 4). So, 25 mod 7 = 4. This condition is also satisfied. Since both conditions are satisfied for 25, 25 is a possible value for x.

• 34:

  • Check 34 mod 3: 34 divided by 3 is 11, with a remainder of 1 (34 = 3 × 11 + 1). So, 34 mod 3 = 1. This condition is satisfied.
  • Check 34 mod 7: 34 divided by 7 is 4, with a remainder of 6 (34 = 7 × 4 + 6). So, 34 mod 7 = 6. This condition is not satisfied (it must be 4). Since the second condition is not satisfied, 34 is not a possible value for x.

• 58:

  • Check 58 mod 3: 58 divided by 3 is 19, with a remainder of 1 (58 = 3 × 19 + 1). So, 58 mod 3 = 1. This condition is satisfied.
  • Check 58 mod 7: 58 divided by 7 is 8, with a remainder of 2 (58 = 7 × 8 + 2). So, 58 mod 7 = 2. This condition is not satisfied (it must be 4). Since the second condition is not satisfied, 58 is not a possible value for x.

• 67:

  • Check 67 mod 3: 67 divided by 3 is 22, with a remainder of 1 (67 = 3 × 22 + 1). So, 67 mod 3 = 1. This condition is satisfied.
  • Check 67 mod 7: 67 divided by 7 is 9, with a remainder of 4 (67 = 7 × 9 + 4). So, 67 mod 7 = 4. This condition is also satisfied. Since both conditions are satisfied for 67, 67 is a possible value for x.

Therefore, the possible values for x among the given options are 25 and 67.

Based on the information from the sources, let’s determine the value of P - 2Q.

The problem states that P and Q are perfect squares and represent days of the month. A perfect square is a number that is the square of an integer. The sources explain that perfect squares must be non-negative, as squaring a positive or a negative number results in a positive number, and 0 squared is 0. They give examples like 1, 4, 9, 16, and 25. A perfect square is an integer whose square root is also an integer.

Since P is the day of the month for Sheetal’s birthday in October, P must be an integer between 1 and 31, inclusive. The perfect squares in this range are:

• 1² = 1
• 2² = 4
• 3² = 9
• 4² = 16
• 5² = 25
• 6² = 36 (too large for a day in October)

So, P must be one of the values in the set {1, 4, 9, 16, 25}.

Similarly, Q is the day of the month for Karthik’s birthday in November. Q must be an integer between 1 and 30, inclusive (November has 30 days). The perfect squares in this range are:

• 1² = 1
• 2² = 4
• 3² = 9
• 4² = 16
• 5² = 25
• 6² = 36 (too large for a day in November)

So, Q must be one of the values in the set {1, 4, 9, 16, 25}.

We are told that Sheetal was born 10 days before Karthik. Sheetal’s birthday is on day P in October. Karthik’s birthday is on day Q in November.

The number of days remaining in October after Sheetal’s birthday is (Number of days in October) - P. October has 31 days. Days remaining in October = 31 - P.

Karthik’s birthday is Q days into November. The total number of days between Sheetal’s birthday and Karthik’s birthday is the sum of the remaining days in October and the days in November until Karthik’s birthday.

(31 - P) + Q = 10

We need to find the values of P and Q from the possible perfect squares {1, 4, 9, 16, 25} that satisfy this equation. Let’s rearrange the equation: Q - P = 10 - 31 Q - P = -21 or P - Q = 21

Now we check the possible perfect square values for P and Q:

• If P = 1, then 1 - Q = 21 => Q = 1 - 21 = -20. -20 is not a valid day.
• If P = 4, then 4 - Q = 21 => Q = 4 - 21 = -17. -17 is not a valid day.
• If P = 9, then 9 - Q = 21 => Q = 9 - 21 = -12. -12 is not a valid day.
• If P = 16, then 16 - Q = 21 => Q = 16 - 21 = -5. -5 is not a valid day.
• If P = 25, then 25 - Q = 21 => Q = 25 - 21 = 4. 4 is a perfect square (2²) and a valid day in November.

So, the values that satisfy the conditions are P = 25 and Q = 4.

Finally, we need to find the value of P - 2Q. P - 2Q = 25 - 2(4) = 25 - 8 = 17.

The value of P - 2Q is 17.