Animal Jumps

Given: 12 is the product.

Working:
$$2 \times 6 = 12$$
$$3 \times 4 = 12$$
$$12 \times 1 = 12$$
$$1 \times 12 = 12$$

Conclusion: Since each of 1, 2, 3, 4, 6, 12 multiplies with another whole number to give 12, all of them are factors of 12, and 12 is a multiple of each of them.

Arrays for 10:
- $1 \times 10$ (1 row of 10)
- $2 \times 5$ (2 rows of 5)
- $5 \times 2$ (5 rows of 2)
- $10 \times 1$ (10 rows of 1)

Factors of 10: 1, 2, 5, 10

Arrays for 14:
- $1 \times 14$
- $2 \times 7$
- $7 \times 2$
- $14 \times 1$

Factors of 14: 1, 2, 7, 14

Arrays for 13:
- $1 \times 13$
- $13 \times 1$

Factors of 13: 1 and 13 only.

Note: 13 can only be arranged in one row or one column — it has exactly two factors (1 and itself). That is why 13 is called a prime number.

Arrays for 20:
- $1 \times 20$
- $2 \times 10$
- $4 \times 5$
- $5 \times 4$
- $10 \times 2$
- $20 \times 1$

Factors of 20: 1, 2, 4, 5, 10, 20

Arrays for 25:
- $1 \times 25$
- $5 \times 5$
- $25 \times 1$

Factors of 25: 1, 5, 25

Arrays for 32:
- $1 \times 32$
- $2 \times 16$
- $4 \times 8$
- $8 \times 4$
- $16 \times 2$
- $32 \times 1$

Factors of 32: 1, 2, 4, 8, 16, 32

Arrays for 37:
- $1 \times 37$
- $37 \times 1$

Factors of 37: 1 and 37 only.

37 is a prime number because it has exactly two factors — 1 and itself.

Arrays for 46:
- $1 \times 46$
- $2 \times 23$
- $23 \times 2$
- $46 \times 1$

Factors of 46: 1, 2, 23, 46

Arrays for 54:
- $1 \times 54$
- $2 \times 27$
- $3 \times 18$
- $6 \times 9$
- $9 \times 6$
- $18 \times 3$
- $27 \times 2$
- $54 \times 1$

Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54

13 and 37 are called prime numbers because each of them has exactly two factors — 1 and the number itself. They cannot be arranged into any rectangular array other than a single row or a single column. In other words, no number other than 1 divides them completely.

Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, …

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, …

Common multiples of 3 and 4: 12, 24, 36, 48, …

Observation: Every common multiple of 3 and 4 is a multiple of $3 \times 4 = 12$. The common multiples are 12, 24, 36, 48, … i.e., multiples of 12.

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, …

Multiples of 6: 6, 12, 18, 24, 30, …

Common multiples of 3 and 6: 6, 12, 18, 24, 30, …

Observation: All multiples of 6 are also multiples of 3. The common multiples are simply the multiples of 6 (the larger number). This is because 6 is itself a multiple of 3.

Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, …

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, …

Common multiples of 4 and 6: 12, 24, 36, 48, 60, 72, …

So beyond 12 and 24, the next common multiples are 36, 48, 60, 72, …

Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, …

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, …

5 Common multiples of 2 and 3: 6, 12, 18, 24, 30

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, …

Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, …

5 Common multiples of 5 and 8: 40, 80, 120, 160, 200

Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, …

Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, …

5 Common multiples of 2 and 4: 4, 8, 12, 16, 20

(All multiples of 4 are common multiples of 2 and 4.)

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, …

Multiples of 9: 9, 18, 27, 36, 45, 54, …

5 Common multiples of 3 and 9: 9, 18, 27, 36, 45

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, …

Multiples of 10: 10, 20, 30, 40, 50, 60, …

5 Common multiples of 5 and 10: 10, 20, 30, 40, 50

Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, …

Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, …

5 Common multiples of 9 and 12: 36, 72, 108, 144, 180

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, …

Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, …

5 Common multiples of 6 and 8: 24, 48, 72, 96, 120

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, …

Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, …

5 Common multiples of 6 and 9: 18, 36, 54, 72, 90

Given: Robby jumps 4 steps at a time; Deeku jumps 6 steps at a time. Both start at 0.

Concept: Both animals can only land on multiples of their jump size.

Multiples of 4 (Robby's positions): 4, 8, 12, 16, 20, 24, 28, 32, 36, …

Multiples of 6 (Deeku's positions): 6, 12, 18, 24, 30, 36, …

Common multiples of 4 and 6: 12, 24, 36, 48, …

If the food is at a common multiple (e.g., 12, 24, 36, …), both will reach the food.

Who reaches first?
- Robby reaches 12 in $12 \div 4 = 3$ jumps.
- Deeku reaches 12 in $12 \div 6 = 2$ jumps.

Deeku reaches first because he needs only 2 jumps to reach 12, while Robby needs 3 jumps.

Given: Mowgli jumps 2 steps at a time starting from 0.

Concept: He will land on all multiples of 2, i.e., even numbers: 2, 4, 6, 8, 10, 12, 14, 16, …, 50, …

Friends at even positions (multiples of 2): 4, 12, 14, 50 — these are the ant, frog, bird, and rabbit.

Friends at odd positions (9, 21, 39, 57) cannot be reached by jumps of 2.

Answer: Mowgli will meet the friends at positions 4, 12, 14, and 50. (2 is a common factor of 4, 12, 14, and 50.)

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, …

From the trail, the friends at positions that are multiples of 3 are: 9, 21, 39, 57.

Answer: Mowgli will meet the friends at 9, 21, 39, and 57. 3 is a common factor of 9, 21, 39, and 57.

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, …

From the trail, the position that is a multiple of 5 is: 50.

Answer: 5 is a common factor of the numbers that are multiples of 5 on the trail. Based on the trail described in the chapter, 5 is a common factor of 50 (and any other multiples of 5 shown on the trail).

Multiples of 10: 10, 20, 30, 40, 50, 60, …

From the trail, the position that is a multiple of 10 is: 50.

Answer: 10 is a common factor of the numbers on the trail that are multiples of 10. Based on the trail, 10 is a common factor of 50 (and any other multiples of 10 shown on the trail).

Check: $24 \div 2 = 12$ (no remainder) ✓ and $36 \div 2 = 18$ (no remainder) ✓

Answer: Yes, we can jump by 2 steps to reach both 24 and 36. 2 is a common factor of 24 and 36.

Check: $24 \div 3 = 8$ (no remainder) ✓ and $36 \div 3 = 12$ (no remainder) ✓

Answer: Yes. 3 is a common factor of 24 and 36.

Check: $24 \div 4 = 6$ (no remainder) ✓ and $36 \div 4 = 9$ (no remainder) ✓

Answer: Yes. 4 is a common factor of 24 and 36.

We need numbers that divide both 24 and 36 exactly.

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

Common factors (other than 2, 3, 4 already found): 1, 6, 12

So other possible jumps are 1, 6, and 12.

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

Common factors of 24 and 36: 1, 2, 3, 4, 6, 12

There are 6 common factors in total.

Check: $24 \div 1 = 24$ ✓ and $36 \div 1 = 36$ ✓

Answer: Yes, jumping by 1 step will always reach every number including 24 and 36. 1 is always a common factor of any two numbers.

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 13: 1, 13 (13 is a prime number)

Common factors of 12 and 13: Only 1.

Reason: 13 is a prime number, so its only factors are 1 and 13. Since 13 does not divide 12, the only common factor is 1.

Concept: A number can be reached by jumps of 4 if it is a multiple of 4 (i.e., divisible by 4).

Rule for divisibility by 4: A number is divisible by 4 if its last two digits form a number divisible by 4.

Note: The exact numbers are in an image (img_5) that cannot be seen. However, the method is:
- Check each number: if $\text{number} \div 4$ gives no remainder, it is reachable.
- List all such numbers — 4 is a common factor of those numbers.

*(Apply the above check to each number shown in the image.)*

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 16: 1, 2, 4, 8, 16

Common factors of 12 and 16: 1, 2, 4

Factors of 8: 1, 2, 4, 8

Factors of 12: 1, 2, 3, 4, 6, 12

Common factors of 8 and 12: 1, 2, 4

Factors of 4: 1, 2, 4

Factors of 16: 1, 2, 4, 8, 16

Common factors of 4 and 16: 1, 2, 4

Factors of 2: 1, 2

Factors of 9: 1, 3, 9

Common factors of 2 and 9: 1 only.

Factors of 3: 1, 3

Factors of 5: 1, 5

Common factors of 3 and 5: 1 only.

Factors of 20: 1, 2, 4, 5, 10, 20

Factors of 5: 1, 5

Common factors of 20 and 5: 1, 5

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 15: 1, 3, 5, 15

Common factors of 12 and 15: 1, 3

Factors of 9: 1, 3, 9

Factors of 21: 1, 3, 7, 21

Common factors of 9 and 21: 1, 3

Factors of 6: 1, 2, 3, 6

Factors of 27: 1, 3, 9, 27

Common factors of 6 and 27: 1, 3

Answer: False (F)

Reason: Even numbers can have odd factors. For example, factors of 12 (an even number) include 1, 3, and 6 — here 1 and 3 are odd. So factors of even numbers need not be even.

Answer: False (F)

Reason: Multiples of odd numbers can be even. For example, $3 \times 2 = 6$ (even), $5 \times 4 = 20$ (even). So multiples of odd numbers can indeed be even.

Answer: True (T)

Reason: If an even number were a factor of an odd number, then the odd number would be divisible by 2, making it even — a contradiction. Therefore, factors of odd numbers are always odd.

Answer: True (T)

Reason: For any two numbers $a$ and $b$, their product $a \times b$ is always a multiple of both $a$ and $b$. For example, consecutive numbers 4 and 5: $4 \times 5 = 20$, which is a multiple of both 4 and 5. So their product is always a common multiple.

Answer: True (T)

Reason: Consecutive numbers differ by 1. If a number d > 1 divided both $n$ and $n+1$, it would also divide their difference $(n+1) - n = 1$, which is impossible for d > 1. Hence, the only common factor of two consecutive numbers is 1.

Answer: True (T)

Reason: A factor of a number must divide it exactly (without remainder). Division by 0 is undefined in mathematics. Therefore, 0 cannot be a factor of any number.

Given: Sher Khan hunts every 3rd day; Bagheera hunts every 5th day. Both start on Day 0 (same day).

Concept: They hunt together on common multiples of 3 and 5.

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, …

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, …

Common multiples of 3 and 5: 15, 30, 45, 60, …

Answer: They will hunt together on Day 15, Day 30, Day 45, Day 60, and so on — i.e., every 15th day.

Given: Sher Khan is at 25; Baloo is at 30. Mowgli must reach 30 but NOT land on 25.

Concept: Mowgli lands on multiples of his jump size. He needs a jump size that:
- Is a factor of 30 (to reach Baloo), AND
- Is NOT a factor of 25 (to avoid Sher Khan).

Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30

Factors of 25: 1, 5, 25

Factors of 30 that are NOT factors of 25: 2, 3, 6, 10, 15, 30

Answer: Mowgli could jump 2, 3, 6, 10, 15, or 30 steps at a time. These jumps will take him to 30 (Baloo's house) without landing on 25 (Sher Khan's house).

Given: Kaa is at 21; Akela is at 35. Mowgli must reach both 21 and 35.

Concept: He needs a jump size that is a common factor of 21 and 35.

Factors of 21: 1, 3, 7, 21

Factors of 35: 1, 5, 7, 35

Common factors of 21 and 35: 1 and 7

Answer: Mowgli should jump 7 steps at a time (or 1 step, but 7 is the more interesting answer). With jumps of 7: he lands on 7, 14, 21, 28, 35 — meeting both Kaa and Akela.

Concept/Rules:
- Divisible by 2: Last digit is 0, 2, 4, 6, or 8.
- Divisible by 5: Last digit is 0 or 5.
- Divisible by 10: Last digit is 0.
- Divisible by 2, 5, and 10: Last digit is 0 (divisible by 10 implies divisible by both 2 and 5).

Sorting method:
- If a number ends in 0 → divisible by 2, 5, and 10 → Category (d).
- If a number ends in 5 → divisible by 5 only (not by 2 or 10) → Category (b).
- If a number ends in 2, 4, 6, or 8 → divisible by 2 only (not by 5 or 10) → Category (a).
- No number can be divisible by 10 only (since divisibility by 10 always implies divisibility by both 2 and 5).

Note: The exact numbers are in images that cannot be read. Apply the above rules to each number shown in the images to sort them into the correct categories.