Prime Time

Given: In the idli-vada game, 'idli' is said for multiples of 3, 'vada' for multiples of 5, and 'idli-vada' for common multiples (multiples of both 3 and 5, i.e., multiples of 15).

Concept: 'Idli-vada' is said at every common multiple of 3 and 5. The LCM of 3 and 5 is 15, so 'idli-vada' is said at every multiple of 15.

The multiples of 15 are: 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, …

Counting: 1st → 15, 2nd → 30, 3rd → 45, …, 10th → 150.

$$10 \times 15 = 150$$

'Idli-vada' is said for the 10th time at the number 150.

Given: Game played from 1 to 90. 'Idli' for multiples of 3, 'vada' for multiples of 5, 'idli-vada' for multiples of both (i.e., multiples of 15).

Part a — Number of times 'idli' is said:
Multiples of 3 up to 90: $\lfloor 90 \div 3 \rfloor = 30$
$$\textbf{Answer: 30 times}$$

Part b — Number of times 'vada' is said:
Multiples of 5 up to 90: $\lfloor 90 \div 5 \rfloor = 18$
$$\textbf{Answer: 18 times}$$

Part c — Number of times 'idli-vada' is said:
Multiples of 15 up to 90: $\lfloor 90 \div 15 \rfloor = 6$
$$\textbf{Answer: 6 times}$$
(These occur at 15, 30, 45, 60, 75, 90.)

Given: Game played from 1 to 900.

Number of times 'idli' is said (multiples of 3 up to 900):
$$\lfloor 900 \div 3 \rfloor = 300$$

Number of times 'vada' is said (multiples of 5 up to 900):
$$\lfloor 900 \div 5 \rfloor = 180$$

Number of times 'idli-vada' is said (multiples of 15 up to 900):
$$\lfloor 900 \div 15 \rfloor = 60$$

Observation: Each answer is exactly 10 times the answer for the game up to 90, because 900 = 10 × 90.

Answers: 'idli' → 300 times; 'vada' → 180 times; 'idli-vada' → 60 times.

Yes, the figure is related to the idli-vada game.

In Fig. 5.1, the numbers are arranged in a grid. The shaded numbers are multiples of 3 (idli), the circled numbers are multiples of 5 (vada), and the numbers that are both shaded and circled are multiples of 15 (idli-vada).

When the game is played till 30, the common multiples (idli-vada) are 15 and 30.
When the game is played till 60, the common multiples are 15, 30, 45, and 60.

To draw the figure for the game up to 60: arrange numbers 1 to 60 in rows of 10. Shade all multiples of 3, circle all multiples of 5, and mark both shading and circle on multiples of 15 (i.e., 15, 30, 45, 60).

The figure visually represents the pattern of multiples of 3 and 5, and their common multiples, exactly as in the idli-vada game.

For each pair, 'idli' = multiples of the smaller number, 'vada' = multiples of the larger number, 'idli-vada' = common multiples (multiples of LCM).

a. 2 and 5 (LCM = 10):
- 'Idli' (multiples of 2 up to 60): 2,4,6,8,10,12,…,60 → 30 times
- 'Vada' (multiples of 5 up to 60): 5,10,15,20,…,60 → 12 times
- 'Idli-vada' (multiples of 10 up to 60): 10,20,30,40,50,60 → 6 times

For the figure: arrange 1–60 in rows of 10; shade even numbers, circle multiples of 5, both shade and circle multiples of 10.

b. 3 and 7 (LCM = 21):
- 'Idli' (multiples of 3 up to 60): 3,6,9,…,60 → 20 times
- 'Vada' (multiples of 7 up to 60): 7,14,21,28,35,42,49,56 → 8 times
- 'Idli-vada' (multiples of 21 up to 60): 21, 42 → 2 times

For the figure: shade multiples of 3, circle multiples of 7, both shade and circle 21 and 42.

c. 4 and 6 (LCM = 12):
- 'Idli' (multiples of 4 up to 60): 4,8,12,…,60 → 15 times
- 'Vada' (multiples of 6 up to 60): 6,12,18,…,60 → 10 times
- 'Idli-vada' (multiples of 12 up to 60): 12,24,36,48,60 → 5 times

For the figure: shade multiples of 4, circle multiples of 6, both shade and circle multiples of 12.

Note: For drawing, arrange numbers 1–60 in a 6×10 grid and mark accordingly as described above.

(Based on the standard table used in this chapter where multiples of 3 are shaded and multiples of 5 are circled.)

1. Shaded numbers are all multiples of 3. They share the common property of being divisible by 3.

2. Circled numbers are all multiples of 5. They share the common property of being divisible by 5 (their units digit is 0 or 5).

3. Numbers that are both shaded and circled are multiples of both 3 and 5, i.e., multiples of 15. In the range shown (31–70), these are 45 and 60. These numbers are called common multiples of 3 and 5.

Given: Find multiples of 40 between 310 and 410 (not including 310 and 410).

Multiples of 40: …, 280, 320, 360, 400, 440, …

Check:
- $40 \times 8 = 320$ ✓ (310 < 320 < 410)
- $40 \times 9 = 360$ ✓ (310 < 360 < 410)
- $40 \times 10 = 400$ ✓ (310 < 400 < 410)
- $40 \times 11 = 440$ ✗ (greater than 410)

The multiples of 40 between 310 and 410 are: 320, 360, and 400.

Part a:
Given: Number < 40, one factor is 7, sum of digits = 8.

Multiples of 7 less than 40: 7, 14, 21, 28, 35.

Check sum of digits:
- 7 → 7 (not 8)
- 14 → 1+4 = 5 (not 8)
- 21 → 2+1 = 3 (not 8)
- 28 → 2+8 = 10 (not 8)
- 35 → 3+5 = 8 ✓

The number is 35.

Part b:
Given: Number < 100, factors include 3 and 5, one digit is 1 more than the other.

Since 3 and 5 are both factors, the number must be a multiple of LCM(3,5) = 15.

Multiples of 15 less than 100: 15, 30, 45, 60, 75, 90.

Check condition (one digit is 1 more than the other):
- 15 → digits 1 and 5; 5 – 1 = 4 (not 1)
- 30 → digits 3 and 0; 3 – 0 = 3 (not 1)
- 45 → digits 4 and 5; 5 – 4 = 1 ✓
- 60 → digits 6 and 0; 6 – 0 = 6 (not 1)
- 75 → digits 7 and 5; 7 – 5 = 2 (not 1)
- 90 → digits 9 and 0; 9 – 0 = 9 (not 1)

The number is 45.

Given: A perfect number has the sum of all its factors equal to twice the number.

Check numbers between 1 and 10:

- 6: Factors are 1, 2, 3, 6. Sum = 1+2+3+6 = 12 = 2×6 ✓

Verification: $2 \times 6 = 12$ and $1+2+3+6 = 12$. ✓

The perfect number between 1 and 10 is 6.

Part a: 20 and 28
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 28: 1, 2, 4, 7, 14, 28
Common factors: 1, 2, 4

Part b: 35 and 50
Factors of 35: 1, 5, 7, 35
Factors of 50: 1, 2, 5, 10, 25, 50
Common factors: 1, 5

Part c: 4, 8 and 12
Factors of 4: 1, 2, 4
Factors of 8: 1, 2, 4, 8
Factors of 12: 1, 2, 3, 4, 6, 12
Common factors: 1, 2, 4

Part d: 5, 15 and 25
Factors of 5: 1, 5
Factors of 15: 1, 3, 5, 15
Factors of 25: 1, 5, 25
Common factors: 1, 5

A multiple of 25 that is NOT a multiple of 50 must be an odd multiple of 25 (i.e., $25 \times$ an odd number).

- $25 \times 1 = 25$ (not a multiple of 50 ✓)
- $25 \times 3 = 75$ (not a multiple of 50 ✓)
- $25 \times 5 = 125$ (not a multiple of 50 ✓)

Three such numbers are: 25, 75, and 125.

Given: Both numbers are less than 10. The first common multiple (LCM) is greater than 50.

We need two numbers, both less than 10, whose LCM is greater than 50.

Check pairs with LCM > 50:
- 7 and 8: LCM = 56 > 50 ✓ (both < 10)
- 7 and 9: LCM = 63 > 50 ✓ (both < 10)
- 8 and 9: LCM = 72 > 50 ✓ (both < 10)
- 6 and 7: LCM = 42 < 50 ✗
- 7 and 8: LCM = 56 ✓

The problem says the first 'idli-vada' is said after number 50, meaning the LCM > 50.

Possible pairs: (7 and 8) with LCM 56, (7 and 9) with LCM 63, or (8 and 9) with LCM 72.

Given: Treasures at 28 and 70. Jump sizes that land on both must be common factors of 28 and 70.

Factors of 28: 1, 2, 4, 7, 14, 28
Factors of 70: 1, 2, 5, 7, 10, 14, 35, 70

Common factors of 28 and 70: 1, 2, 7, 14

Jump sizes that will land on both 28 and 70 are: 1, 2, 7, and 14.

Note: The diagram (Fig. img_3) shows a Venn-diagram style figure with two overlapping circles. The common multiples (in the intersection) are visible but the individual multiples in each circle are erased. Since the figure is not fully visible, we solve based on the standard version of this problem.

In the standard version of this problem in the textbook, the common multiples shown are 12 and 24 (multiples of both 4 and 6, for example).

If the two numbers are 4 and 6 (LCM = 12):
- Multiples of 4 only (not 6): 4, 8, 16, 20, 28, 32, …
- Multiples of 6 only (not 4): 6, 18, 30, …
- Common multiples (multiples of 12): 12, 24, 36, …

Students should identify the two numbers from the given common multiples and then list the remaining multiples in each region accordingly.

Method: Identify the LCM from the common multiples shown, then find the two original numbers, and fill in their individual multiples in the respective regions.

We need the LCM of 1, 2, 3, 4, 5, 6, 8, 9, 10 (excluding 7).

Prime factorisations:
- $1 = 1$
- $2 = 2$
- $3 = 3$
- $4 = 2^2$
- $5 = 5$
- $6 = 2 \times 3$
- $8 = 2^3$
- $9 = 3^2$
- $10 = 2 \times 5$

LCM = highest power of each prime:
$$\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$$

The smallest number that is a multiple of all numbers from 1 to 10 except 7 is 360.

We need the LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Prime factorisations:
- $2 = 2$
- $3 = 3$
- $4 = 2^2$
- $5 = 5$
- $6 = 2 \times 3$
- $7 = 7$
- $8 = 2^3$
- $9 = 3^2$
- $10 = 2 \times 5$

LCM = highest power of each prime present:
$$\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520$$

The smallest number that is a multiple of all numbers from 1 to 10 is 2520.

Given: 2 is a prime and even number.

An even number is divisible by 2. If an even number is greater than 2, it has at least three factors: 1, 2, and itself. Therefore, it cannot be prime.

No, there is no other even prime number. 2 is the only even prime number.

Primes up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Differences between successive primes:
- 3−2=1, 5−3=2, 7−5=2, 11−7=4, 13−11=2, 17−13=4, 19−17=2, 23−19=4, 29−23=6, 31−29=2, 37−31=6, 41−37=4, 43−41=2, 47−43=4, 53−47=6, 59−53=6, 61−59=2, 67−61=6, 71−67=4, 73−71=2, 79−73=6, 83−79=4, 89−83=6, 97−89=8.

Smallest difference = 1 (between 2 and 3).
Largest difference = 8 (between 89 and 97).

The table arranges numbers 1–100 in rows of 10 (decades).

Primes in each decade:
- 1–10: 2, 3, 5, 7 → 4 primes
- 11–20: 11, 13, 17, 19 → 4 primes
- 21–30: 23, 29 → 2 primes
- 31–40: 31, 37 → 2 primes
- 41–50: 41, 43, 47 → 3 primes
- 51–60: 53, 59 → 2 primes
- 61–70: 61, 67 → 2 primes
- 71–80: 71, 73, 79 → 3 primes
- 81–90: 83, 89 → 2 primes
- 91–100: 97 → 1 prime

No, there are not an equal number of primes in every row.

Least primes: 91–100 (only 1 prime).
Most primes: 1–10 and 11–20 (4 primes each).

A prime number has exactly two factors: 1 and itself.

- 23: Check divisibility by primes up to $\sqrt{23} \approx 4.8$, i.e., 2 and 3. 23 is odd; 2+3=5, not divisible by 3. 23 is prime. ✓
- 51: 5+1=6, divisible by 3. $51 = 3 \times 17$. 51 is not prime.
- 37: Check primes up to $\sqrt{37} \approx 6.1$, i.e., 2, 3, 5. 37 is odd; 3+7=10, not divisible by 3; does not end in 0 or 5. 37 is prime. ✓
- 26: Even number, $26 = 2 \times 13$. 26 is not prime.

The prime numbers are 23 and 37.

Primes less than 20: 2, 3, 5, 7, 11, 13, 17, 19.

We need pairs whose sum is divisible by 5:

- $2 + 3 = 5$ ✓ (multiple of 5) → Pair: (2, 3)
- $2 + 13 = 15$ ✓ (multiple of 5) → Pair: (2, 13)
- $7 + 13 = 20$ ✓ (multiple of 5) → Pair: (7, 13)
- (Other valid pairs: (3,7)=10 ✓, (2,3)=5 ✓, (11,19)=30 ✓)

Three pairs: (2, 3), (2, 13), and (7, 13). [Other valid answers are also acceptable.]

We need pairs of prime numbers up to 100 that use the same digits (just in different order — i.e., reversals of each other).

Checking all two-digit primes and their reversal:
- 13 and 31: both prime ✓
- 17 and 71: both prime ✓
- 37 and 73: both prime ✓
- 79 and 97: both prime ✓
- 11 and 11: same number, not a pair
- 12 reversed is 21 = 3×7, not prime
- 14 reversed is 41 (prime) but 14 is not prime

Pairs of prime numbers up to 100 with the same digits:
$$(13, 31),\ (17, 71),\ (37, 73),\ (79, 97)$$

We need 7 consecutive numbers, all composite (none prime).

Consider numbers 90 to 96:
- 90 = 2×45 (composite)
- 91 = 7×13 (composite)
- 92 = 4×23 (composite)
- 93 = 3×31 (composite)
- 94 = 2×47 (composite)
- 95 = 5×19 (composite)
- 96 = 2×48 (composite)

All seven numbers 90, 91, 92, 93, 94, 95, 96 are composite.

Seven consecutive composite numbers between 1 and 100: 90, 91, 92, 93, 94, 95, 96.

Twin primes are pairs of primes differing by 2.

Primes up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Checking pairs with difference 2:
- (3, 5) ✓ — given
- (5, 7) ✓
- (11, 13) ✓
- (17, 19) ✓ — given
- (29, 31) ✓
- (41, 43) ✓
- (59, 61) ✓
- (71, 73) ✓

Other twin prime pairs between 1 and 100 (besides (3,5) and (17,19)):
$$(5,7),\ (11,13),\ (29,31),\ (41,43),\ (59,61),\ (71,73)$$

a. There is no prime number whose units digit is 4.
TRUE. Any number ending in 4 is even (divisible by 2). If it is greater than 2, it has at least three factors (1, 2, and itself), so it cannot be prime.

b. A product of primes can also be prime.
FALSE. A product of two or more primes has at least four factors (1, each prime, and the product itself), so it is composite. For example, $2 \times 3 = 6$, which is not prime.

c. Prime numbers do not have any factors.
FALSE. Every prime number has exactly two factors: 1 and itself. For example, 7 has factors 1 and 7.

d. All even numbers are composite numbers.
FALSE. The number 2 is even but it is prime (its only factors are 1 and 2).

e. 2 is a prime and so is the next number, 3. For every other prime, the next number is composite.
TRUE. Every prime greater than 3 is odd. The number after any odd number is even (and greater than 2), hence divisible by 2, hence composite. So for every prime $p > 3$, the number $p+1$ is even and greater than 2, making it composite.

We find the prime factorisation of each:

- $45 = 3 \times 3 \times 5 = 3^2 \times 5$ → only 2 distinct primes (3 and 5)
- $60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5$ → 3 distinct primes, but not a product of exactly three (has repeated factor)
- $91 = 7 \times 13$ → only 2 distinct primes
- $105 = 3 \times 5 \times 7$ → exactly 3 distinct primes, each appearing once ✓
- $330 = 2 \times 3 \times 5 \times 11$ → 4 distinct primes

The number that is the product of exactly three distinct prime numbers is 105 $(= 3 \times 5 \times 7)$.