Question

1. (i) find the sum of the prime numbers that are between 12 and 42.
2. (ii) which of the following numbers are divisible by:

(a) 2? 1020
(b) 3? 125 130
(c) 5? 1332 185
(d) 10? 132

Explanation:

Part (5): Find the sum of the prime numbers between 12 and 42.

Step 1: Identify prime numbers between 12 and 42

Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. Let's list numbers between 12 and 42 and check for primality:

• 13: Prime (divisors 1, 13)
• 14: Not prime (divisible by 2, 7)
• 15: Not prime (divisible by 3, 5)
• 16: Not prime (divisible by 2)
• 17: Prime (divisors 1, 17)
• 18: Not prime (divisible by 2, 3)
• 19: Prime (divisors 1, 19)
• 20: Not prime (divisible by 2, 5)
• 21: Not prime (divisible by 3, 7)
• 22: Not prime (divisible by 2, 11)
• 23: Prime (divisors 1, 23)
• 24: Not prime (divisible by 2, 3)
• 25: Not prime (divisible by 5)
• 26: Not prime (divisible by 2, 13)
• 27: Not prime (divisible by 3)
• 28: Not prime (divisible by 2, 7)
• 29: Prime (divisors 1, 29)
• 30: Not prime (divisible by 2, 3, 5)
• 31: Prime (divisors 1, 31)
• 32: Not prime (divisible by 2)
• 33: Not prime (divisible by 3, 11)
• 34: Not prime (divisible by 2, 17)
• 35: Not prime (divisible by 5, 7)
• 36: Not prime (divisible by 2, 3)
• 37: Prime (divisors 1, 37)
• 38: Not prime (divisible by 2, 19)
• 39: Not prime (divisible by 3, 13)
• 40: Not prime (divisible by 2, 5)
• 41: Prime (divisors 1, 41)

So the prime numbers between 12 and 42 are: 13, 17, 19, 23, 29, 31, 37, 41.

Step 2: Sum these prime numbers

Now, sum them up:

$$ 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 $$

Let's calculate step by step:

• \(13 + 17 = 30\)
• \(30 + 19 = 49\)
• \(49 + 23 = 72\)
• \(72 + 29 = 101\)
• \(101 + 31 = 132\)
• \(132 + 37 = 169\)
• \(169 + 41 = 210\)

Answer:

210

Part (6): Which of the following numbers are divisible by:

(a) Divisible by 2?

A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). Let's check each number:

• 1020: Last digit 0 (even) → Divisible by 2.
• 125: Last digit 5 (odd) → Not divisible by 2.
• 1332: Last digit 2 (even) → Divisible by 2.
• 185: Last digit 5 (odd) → Not divisible by 2.
• 132: Last digit 2 (even) → Divisible by 2.
• 10?: Assuming it's 10 (last digit 0, even) or 10x (if x is even). But from the list, 1020, 1332, 132 are divisible by 2.

(b) Divisible by 3?

A number is divisible by 3 if the sum of its digits is divisible by 3.

• 1020: \(1 + 0 + 2 + 0 = 3\) (3 is divisible by 3) → Divisible by 3.
• 125: \(1 + 2 + 5 = 8\) (8 not divisible by 3) → Not divisible by 3.
• 1332: \(1 + 3 + 3 + 2 = 9\) (9 divisible by 3) → Divisible by 3.
• 185: \(1 + 8 + 5 = 14\) (14 not divisible by 3) → Not divisible by 3.
• 132: \(1 + 3 + 2 = 6\) (6 divisible by 3) → Divisible by 3.
• 10?: If 10, \(1 + 0 = 1\) (not divisible by 3); if 105, etc. But from the list, 1020, 1332, 132 are divisible by 3.

(c) Divisible by 5?

A number is divisible by 5 if its last digit is 0 or 5.

• 1020: Last digit 0 → Divisible by 5.
• 125: Last digit 5 → Divisible by 5.
• 1332: Last digit 2 → Not divisible by 5.
• 185: Last digit 5 → Divisible by 5.
• 132: Last digit 2 → Not divisible by 5.
• 10?: If last digit 0 or 5, e.g., 10 (0) or 15 (5). From the list, 1020, 125, 185 are divisible by 5.

(d) Divisible by 10?

A number is divisible by 10 if its last digit is 0.

• 1020: Last digit 0 → Divisible by 10.
• 125: Last digit 5 → Not divisible by 10.
• 1332: Last digit 2 → Not divisible by 10.
• 185: Last digit 5 → Not divisible by 10.
• 132: Last digit 2 → Not divisible by 10.
• 10?: If last digit 0 (e.g., 10) → Divisible by 10. From the list, 1020 is divisible by 10.

Summarized Answers for Part (6):

• (a) Divisible by 2: 1020, 1332, 132 (and 10 if 10 is considered)
• (b) Divisible by 3: 1020, 1332, 132
• (c) Divisible by 5: 1020, 125, 185
• (d) Divisible by 10: 1020

(Note: If "10?" is a typo and should be "10", then 10 is also divisible by 2, 5, 10, and 10's digit sum is 1 (not divisible by 3), so adjust accordingly.)