Prime Time — NCERT Solutions

Using digits 2, 4, 5 each exactly once, the possible three-digit numbers are:
245, 254, 425, 452, 524, 542.

Check each for primality:

• 245 = 5 × 49 = 5 × 7² → composite
• 254 = 2 × 127 → composite (even)
• 425 = 5 × 85 = 5 × 5 × 17 → composite
• 452 = 4 × 113 → composite (even)
• 524 = 4 × 131 → composite (even)
• 542 = 2 × 271 → composite (even)

All even numbers (ending in 2 or 4) are composite. Numbers ending in 5 are divisible by 5, hence composite.

No three-digit prime number can be made using each of 2, 4, and 5 exactly once. The answer is 0.

We need primes $p$ such that $2p + 1$ is also prime.

• $p = 3$: $2(3)+1 = 7$ ✓ (prime) — given
• $p = 5$: $2(5)+1 = 11$ ✓ (prime)
• $p = 7$: $2(7)+1 = 15 = 3 \times 5$ ✗ (not prime)
• $p = 11$: $2(11)+1 = 23$ ✓ (prime)
• $p = 13$: $2(13)+1 = 27 = 3^3$ ✗ (not prime)
• $p = 23$: $2(23)+1 = 47$ ✓ (prime)
• $p = 29$: $2(29)+1 = 59$ ✓ (prime)
• $p = 41$: $2(41)+1 = 83$ ✓ (prime)

Five examples: (3,7), (5,11), (11,23), (23,47), (29,59).

A pair is 'safe' (co-prime) if the only common factor is 1 (i.e., no jump size other than 1 can reach both).

a. 15 and 39:
Factors of 15: 1, 3, 5, 15
Factors of 39: 1, 3, 13, 39
Common factors: 1 and 3. Since they share the factor 3, not safe (Jumpy can use jump size 3).

b. 4 and 15:
Factors of 4: 1, 2, 4
Factors of 15: 1, 3, 5, 15
Common factors: only 1. Safe ✓ (co-prime)

c. 18 and 29:
Factors of 18: 1, 2, 3, 6, 9, 18
Factors of 29: 1, 29 (29 is prime)
Common factors: only 1. Safe ✓ (co-prime)

d. 20 and 55:
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 55: 1, 5, 11, 55
Common factors: 1 and 5. Since they share the factor 5, not safe (Jumpy can use jump size 5).

Jump sizes that reach both 15 and 30 must be common factors of 15 and 30.

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

Common factors of 15 and 30: 1, 3, 5, 15

All jump sizes that can reach both 15 and 30 are: 1, 3, 5, and 15.

Using repeated division by prime numbers:

64:
$64 = 2 \times 32 = 2 \times 2 \times 16 = 2^6$
$$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$$

104:
$104 = 2 \times 52 = 2 \times 2 \times 26 = 2 \times 2 \times 2 \times 13$
$$104 = 2^3 \times 13$$

105:
$105 = 3 \times 35 = 3 \times 5 \times 7$
$$105 = 3 \times 5 \times 7$$

243:
$243 = 3 \times 81 = 3 \times 3 \times 27 = 3 \times 3 \times 3 \times 9 = 3^5$
$$243 = 3^5$$

320:
$320 = 2 \times 160 = 2^2 \times 80 = 2^3 \times 40 = 2^4 \times 20 = 2^5 \times 10 = 2^6 \times 5$
$$320 = 2^6 \times 5$$

141:
$141 = 3 \times 47$ (47 is prime)
$$141 = 3 \times 47$$

1728:
$1728 = 2 \times 864 = 2^2 \times 432 = 2^3 \times 216 = 2^3 \times 8 \times 27 = 2^6 \times 27 = 2^6 \times 3^3$
$$1728 = 2^6 \times 3^3$$

729:
$729 = 3^6$ (since $3^6 = 729$)
$$729 = 3^6$$

1024:
$1024 = 2^{10}$ (since $2^{10} = 1024$)
$$1024 = 2^{10}$$

1331:
$1331 = 11^3$ (since $11^3 = 1331$)
$$1331 = 11^3$$

1000:
$1000 = 8 \times 125 = 2^3 \times 5^3$
$$1000 = 2^3 \times 5^3$$

Given: Prime factorisation has one 2, two 3s, and one 11.

$$\text{Number} = 2^1 \times 3^2 \times 11^1 = 2 \times 9 \times 11 = 198$$

The number is 198.

Given: Product of three primes (all < 30) = 1955.

Step 1: Check divisibility. $1955$ ends in 5, so divisible by 5.
$$1955 \div 5 = 391$$

Step 2: Factorise 391.
$391 \div 7 = 55.8...$ (no)
$391 \div 11 = 35.5...$ (no)
$391 \div 17 = 23$ ✓
$$391 = 17 \times 23$$

So $1955 = 5 \times 17 \times 23$.

Verify: all three (5, 17, 23) are prime and less than 30. ✓

The three prime numbers are 5, 17, and 23.

Part a: $56 \times 25$
$56 = 2^3 \times 7$
$25 = 5^2$
$$56 \times 25 = 2^3 \times 5^2 \times 7$$

Part b: $108 \times 75$
$108 = 2^2 \times 3^3$
$75 = 3 \times 5^2$
$$108 \times 75 = 2^2 \times 3^3 \times 3 \times 5^2 = 2^2 \times 3^4 \times 5^2$$

Part c: $1000 \times 81$
$1000 = 2^3 \times 5^3$
$81 = 3^4$
$$1000 \times 81 = 2^3 \times 3^4 \times 5^3$$

To get the smallest such number, use the smallest available primes.

Part a: Three different prime numbers
Smallest three primes: 2, 3, 5
$$\text{Smallest number} = 2 \times 3 \times 5 = 30$$

Part b: Four different prime numbers
Smallest four primes: 2, 3, 5, 7
$$\text{Smallest number} = 2 \times 3 \times 5 \times 7 = 210$$

Answers: a. 30, b. 210.

Two numbers are co-prime if they have no common prime factor.

a. 30 and 45:
Guess: Not co-prime (both divisible by 5).
$30 = 2 \times 3 \times 5$
$45 = 3^2 \times 5$
Common prime factors: 3 and 5.
Not co-prime.

b. 57 and 85:
Guess: Not co-prime (both seem divisible by some common factor).
$57 = 3 \times 19$
$85 = 5 \times 17$
No common prime factors.
Co-prime. ✓

c. 121 and 1331:
Guess: Not co-prime ($121 = 11^2$, $1331 = 11^3$).
$121 = 11^2$
$1331 = 11^3$
Common prime factor: 11.
Not co-prime.

d. 343 and 216:
Guess: Co-prime ($343 = 7^3$, $216 = 6^3$).
$343 = 7^3$
$216 = 2^3 \times 3^3$
No common prime factors.
Co-prime. ✓

A number is divisible by another if the prime factorisation of the second is included in (contained within) the prime factorisation of the first.

a. 225 and 27:
$225 = 3^2 \times 5^2$
$27 = 3^3$
$27$ requires $3^3$ but $225$ has only $3^2$. The factorisation of 27 is NOT included in 225.
225 is NOT divisible by 27.

b. 96 and 24:
$96 = 2^5 \times 3$
$24 = 2^3 \times 3$
$24$ requires $2^3$ and $3^1$; $96$ has $2^5$ and $3^1$. The factorisation of 24 IS included in 96.
96 IS divisible by 24. ($96 \div 24 = 4$)

c. 343 and 17:
$343 = 7^3$
$17 = 17$ (prime)
17 is not a factor of 343.
343 is NOT divisible by 17.

d. 999 and 99:
$999 = 3^3 \times 37$
$99 = 3^2 \times 11$
$99$ requires the prime factor 11, but 11 is not in the factorisation of 999.
999 is NOT divisible by 99.

First number: $A = 2 \times 3 \times 7 = 42$
Second number: $B = 3 \times 7 \times 11 = 231$

Are they co-prime?
Common prime factors: 3 and 7.
Since they share prime factors, they are NOT co-prime.

Does one divide the other?
For $A$ to divide $B$: the factorisation of $A$ ($2 \times 3 \times 7$) must be included in $B$ ($3 \times 7 \times 11$). But 2 is in $A$ and not in $B$, so $A$ does not divide $B$.

For $B$ to divide $A$: the factorisation of $B$ ($3 \times 7 \times 11$) must be included in $A$ ($2 \times 3 \times 7$). But 11 is in $B$ and not in $A$, so $B$ does not divide $A$.

Neither number divides the other.

Given: Claim — any two prime numbers are co-prime.

A prime number has only two factors: 1 and itself. If $p$ and $q$ are two different prime numbers, then the only factors of $p$ are 1 and $p$, and the only factors of $q$ are 1 and $q$. Since $p \neq q$, the only common factor is 1.

Therefore, any two distinct prime numbers are co-prime.

(Note: If both primes are the same, e.g., $p = q = 5$, then they are not co-prime since 5 is a common factor. But typically 'two prime numbers' implies two different primes.)

Yes, Guna is right. Any two distinct prime numbers are always co-prime.

Numbers between 120 and 140 divisible by 8:
$8 \times 15 = 120$, $8 \times 16 = 128$, $8 \times 17 = 136$, $8 \times 18 = 144 > 140$.
So: 128 and 136.

Numbers between 1120 and 1140 divisible by 8:
$8 \times 140 = 1120$, $8 \times 141 = 1128$, $8 \times 142 = 1136$, $8 \times 143 = 1144 > 1140$.
So: 1128 and 1136.

Numbers between 3120 and 3140 divisible by 8:
$8 \times 390 = 3120$, $8 \times 391 = 3128$, $8 \times 392 = 3136$, $8 \times 393 = 3144 > 3140$.
So: 3128 and 3136.

Observation: In each range, the numbers divisible by 8 end in the same last three digits (128 and 136). This shows that divisibility by 8 depends only on the last three digits.

Changing last two digits of 8560:
$8560$: last three digits are 560. $560 \div 8 = 70$. So 8560 is already divisible by 8.
To make a different multiple: change last two digits so last three digits form a multiple of 8. For example, change to 856→ 8568 ($568 \div 8 = 71$ ✓) or 8576 ($576 \div 8 = 72$ ✓).

Statement 1: Only the last three digits matter when deciding if a given number is divisible by 8.
Agree. Any number can be written as $1000k + r$ where $r$ is the number formed by the last three digits. Since $1000 = 8 \times 125$, the number $1000k$ is always divisible by 8. Therefore, the original number is divisible by 8 if and only if $r$ (the last three digits) is divisible by 8.

Statement 2: If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8.
Agree. As shown above, original number $= 1000k + r$. If $8 \mid r$ and $8 \mid 1000k$, then $8 \mid (1000k + r)$.

Statement 3: If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8.
Agree. If $8 \mid (1000k + r)$ and $8 \mid 1000k$, then $8 \mid r$.

Part a:
This depends on the student's birth year. For example, if born in 2012:
Leap years from 2012 to 2024: 2012, 2016, 2020, 2024.
(Students should list multiples of 4 from their birth year to the current year, excluding centennial years not divisible by 400.)

Part b: Leap years from 2024 to 2099:
Multiples of 4 from 2024 to 2099:
First: 2024, Last: 2096.
Count: $\frac{2096 - 2024}{4} + 1 = \frac{72}{4} + 1 = 18 + 1 = 19$.

Check for exceptions: 2100 is divisible by 100 but not 400, so it would not be a leap year — but 2100 is outside our range. No year between 2024 and 2099 is divisible by 100 (the only candidate would be 2100).

There are 19 leap years from 2024 to 2099.
(They are: 2024, 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092, 2096.)

A palindrome reads the same forwards and backwards. A 4-digit palindrome has the form $\overline{abba}$ where $a \neq 0$.

For divisibility by 4, only the last two digits matter. The last two digits of $\overline{abba}$ are $\overline{ba}$.

We need $\overline{ba}$ (i.e., the two-digit number $10b + a$) to be divisible by 4.

Smallest 4-digit palindrome divisible by 4:
Smallest 4-digit palindromes: 1001, 1111, 1221, 1331, 1441, 1551, 1661, 1771, 1881, 1991, 2002, …

Check last two digits for divisibility by 4:

• 1001: last two digits 01 → 1 ÷ 4 = not divisible
• 1111: last two digits 11 → not divisible
• 1221: last two digits 21 → not divisible
• 1331: last two digits 31 → not divisible
• 1441: last two digits 41 → not divisible
• 1551: last two digits 51 → not divisible
• 1661: last two digits 61 → not divisible
• 1771: last two digits 71 → not divisible
• 1881: last two digits 81 → not divisible
• 1991: last two digits 91 → not divisible
• 2002: last two digits 02 → 2 ÷ 4 = not divisible
• 2112: last two digits 12 → 12 ÷ 4 = 3 ✓

Smallest = 2112

Largest 4-digit palindrome divisible by 4:
Largest 4-digit palindromes (descending): 9999, 9889, 9779, 9669, 9559, 9449, 9339, 9229, 9119, 9009, 8998, 8888, 8778, 8668, …

Check last two digits:

• 9999: 99 ÷ 4 → not divisible
• 9889: 89 ÷ 4 → not divisible
• 9779: 79 ÷ 4 → not divisible
• 9669: 69 ÷ 4 → not divisible
• 9559: 59 ÷ 4 → not divisible
• 9449: 49 ÷ 4 → not divisible
• 9339: 39 ÷ 4 → not divisible
• 9229: 29 ÷ 4 → not divisible
• 9119: 19 ÷ 4 → not divisible
• 9009: 09 ÷ 4 → not divisible
• 8998: 98 ÷ 4 → not divisible (98 = 4×24+2)
• 8888: 88 ÷ 4 = 22 ✓

Largest = 8888

Smallest 4-digit palindrome divisible by 4: 2112; Largest: 8888.

Part a: Sum of two even numbers gives a multiple of 4.
Sometimes true.

Examples:

• $2 + 2 = 4$ ✓ (multiple of 4)
• $2 + 4 = 6$ ✗ (not a multiple of 4)
• $4 + 8 = 12$ ✓ (multiple of 4)
• $2 + 6 = 8$ ✓ (multiple of 4)
• $2 + 8 = 10$ ✗ (not a multiple of 4)

The sum of two even numbers is always even, but not always a multiple of 4. Sometimes true.

Part b: Sum of two odd numbers gives a multiple of 4.
Sometimes true.

Examples:

• $1 + 3 = 4$ ✓ (multiple of 4)
• $1 + 5 = 6$ ✗ (not a multiple of 4)
• $3 + 5 = 8$ ✓ (multiple of 4)
• $1 + 7 = 8$ ✓ (multiple of 4)
• $1 + 9 = 10$ ✗ (not a multiple of 4)

The sum of two odd numbers is always even, but not always a multiple of 4. Sometimes true.

Key rules:

• Remainder when divided by 10 = units digit
• Remainder when divided by 5: if units digit is 0 or 5, remainder = 0; if units digit is 1 or 6, remainder = 1; if 2 or 7, remainder = 2; if 3 or 8, remainder = 3; if 4 or 9, remainder = 4.
• Remainder when divided by 2: 0 if even, 1 if odd.

Given: 14560. We need to find two numbers from {2, 4, 5, 8, 10} whose divisibility implies divisibility by all others.

Note that:

• $10 = 2 \times 5$, so divisibility by 10 implies divisibility by both 2 and 5.
• $8 = 2^3$, so divisibility by 8 implies divisibility by 4 and 2.
• If a number is divisible by both 8 and 10, then it is divisible by 8 (hence by 4 and 2) and by 10 (hence by 5 and 2). So divisibility by all five is guaranteed.

Check: $14560 \div 8 = 1820$ ✓ and $14560 \div 10 = 1456$ ✓.

Guna checked divisibility by 8 and 10. Divisibility by 8 ensures divisibility by 4 and 2; divisibility by 10 ensures divisibility by 5 and 2. Together, all five are covered.

For a number to be divisible by all of 2, 4, 5, 8, and 10, it must be divisible by LCM(2,4,5,8,10) = 40.

A number is divisible by 40 if it is divisible by both 8 and 5 (i.e., divisible by 8 and ends in 0).

• 572: Does not end in 0 → not divisible by 5 or 10. No.
• 2352: Does not end in 0 → not divisible by 5 or 10. No.
• 5600: Ends in 0 ✓. Last three digits: 600. $600 \div 8 = 75$ ✓. Yes.
• 6000: Ends in 0 ✓. Last three digits: 000. $0 \div 8 = 0$ ✓. Yes.
• 77622160: Ends in 0 ✓. Last three digits: 160. $160 \div 8 = 20$ ✓. Yes.

Numbers divisible by all of 2, 4, 5, 8 and 10: 5600, 6000, and 77622160.

$10000 = 2^4 \times 5^4$

We need to split this into two factors, neither ending in 0. A number ends in 0 only if it has both 2 and 5 as factors. So we must give all the 2s to one number and all the 5s to the other.

$$2^4 = 16 \quad \text{and} \quad 5^4 = 625$$

Check: $16 \times 625 = 10000$ ✓
Units digit of 16 is 6 (not 0) ✓
Units digit of 625 is 5 (not 0) ✓

The two numbers are 16 and 625.

Box 1: 5, 7, 12, 35

• 5 is special: it is the only single-digit number; also the only multiple of 5 that is prime.
• 7 is special: it is the only prime number that is not a factor of 35 among the primes listed; actually 7 is prime and a factor of 35.
• 12 is special: it is the only even number; it is the only composite number not involving 5 or 7; it is the only multiple of 4.
• 35 is special: it is the only two-digit composite number; it is the only multiple of both 5 and 7.

Box 2: 3, 8, 11, 24

• 3 is special: it is the only single-digit odd prime.
• 8 is special: it is the only power of 2 (and the only even number that is not a multiple of 3).
• 11 is special: it is the only two-digit prime number.
• 24 is special: it is the only composite number divisible by both 2 and 3; it is the only multiple of 8.

Box 3: 27, 3, 123, 31

• 3 is special: it is the only single-digit number; it is the only prime.
• 27 is special: it is the only perfect cube ($3^3$).
• 31 is special: it is the only two-digit prime.
• 123 is special: it is the only three-digit number; it is the only multiple of 41.

Box 4: 17, 27, 44, 65

• 17 is special: it is the only prime number.
• 27 is special: it is the only odd number that is a perfect cube ($3^3$).
• 44 is special: it is the only even number; it is the only multiple of 4.
• 65 is special: it is the only multiple of 5 and 13; it is the only number divisible by 5.

We need to fill a 3×3 grid with prime numbers such that:

• Row 1 product = 63, Row 2 product = 27, Row 3 product = 190
• Column 1 product = 45, Column 2 product = 42, Column 3 product = 171

Prime factorisations:

• $63 = 3^2 \times 7 = 3 \times 3 \times 7$
• $27 = 3^3 = 3 \times 3 \times 3$
• $190 = 2 \times 5 \times 19$
• $45 = 3^2 \times 5 = 3 \times 3 \times 5$
• $42 = 2 \times 3 \times 7$
• $171 = 3^2 \times 19 = 3 \times 3 \times 19$

Let the grid be:
$$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$$

Row 1: $a \times b \times c = 63 = 3 \times 3 \times 7$
Row 2: $d \times e \times f = 27 = 3 \times 3 \times 3$
Row 3: $g \times h \times i = 190 = 2 \times 5 \times 19$

Column 1: $a \times d \times g = 45 = 3 \times 3 \times 5$
Column 2: $b \times e \times h = 42 = 2 \times 3 \times 7$
Column 3: $c \times f \times i = 171 = 3 \times 3 \times 19$

From Row 3: $g = 2, h = 5, i = 19$ (or permutations).
From Column 1: $a \times d \times 2 = 45$ → $a \times d = 22.5$ — not integer with $g=2$.

Try $g = 5, h = 2, i = 19$:
Column 1: $a \times d \times 5 = 45$ → $a \times d = 9 = 3 \times 3$
Column 2: $b \times e \times 2 = 42$ → $b \times e = 21 = 3 \times 7$
Column 3: $c \times f \times 19 = 171$ → $c \times f = 9 = 3 \times 3$

Row 1: $a \times b \times c = 63$; Row 2: $d \times e \times f = 27$.

Let $a = 3, d = 3$ (from $a \times d = 9$).
Let $b = 3, e = 7$ (from $b \times e = 21$).
Let $c = 3, f = 3$ (from $c \times f = 9$).

Check Row 1: $3 \times 3 \times 3 = 27 \neq 63$. ✗

Try $b = 7, e = 3$:
Row 1: $a \times 7 \times c = 63$ → $a \times c = 9 = 3 \times 3$, so $a = 3, c = 3$.
Row 2: $d \times 3 \times f = 27$ → $d \times f = 9 = 3 \times 3$, so $d = 3, f = 3$.
Check Column 1: $3 \times 3 \times 5 = 45$ ✓
Check Column 3: $3 \times 3 \times 19 = 171$ ✓

$$\boxed{\begin{pmatrix} 3 & 7 & 3 \\ 3 & 3 & 3 \\ 5 & 2 & 19 \end{pmatrix}}$$

Verification:

• Row 1: $3 \times 7 \times 3 = 63$ ✓
• Row 2: $3 \times 3 \times 3 = 27$ ✓
• Row 3: $5 \times 2 \times 19 = 190$ ✓
• Col 1: $3 \times 3 \times 5 = 45$ ✓
• Col 2: $7 \times 3 \times 2 = 42$ ✓
• Col 3: $3 \times 3 \times 19 = 171$ ✓

Prime factorisations:

• $343 = 7^3 = 7 \times 7 \times 7$
• $66 = 2 \times 3 \times 11$
• $44 = 2^2 \times 11 = 2 \times 2 \times 11$
• $28 = 2^2 \times 7 = 2 \times 2 \times 7$
• $154 = 2 \times 7 \times 11$
• $231 = 3 \times 7 \times 11$

Let the grid be:
$$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$$

Row 1: $a \times b \times c = 343 = 7 \times 7 \times 7$ → $a = b = c = 7$

Column 1: $7 \times d \times g = 28$ → $d \times g = 4 = 2 \times 2$, so $d = 2, g = 2$.
Column 2: $7 \times e \times h = 154$ → $e \times h = 22 = 2 \times 11$.
Column 3: $7 \times f \times i = 231$ → $f \times i = 33 = 3 \times 11$.

Row 2: $d \times e \times f = 66 = 2 \times 3 \times 11$. With $d = 2$: $e \times f = 33 = 3 \times 11$.
Row 3: $g \times h \times i = 44 = 2 \times 2 \times 11$. With $g = 2$: $h \times i = 22 = 2 \times 11$.

From $e \times f = 33$ and $e \times h = 22$ and $f \times i = 33$ and $h \times i = 22$:

Let $e = 3, f = 11$: then $h = 22/3$ — not integer.
Let $e = 11, f = 3$: then $h = 22/11 = 2$ and $i = 33/3 = 11$.
Check Row 3: $2 \times 2 \times 11 = 44$ ✓

$$\boxed{\begin{pmatrix} 7 & 7 & 7 \\ 2 & 11 & 3 \\ 2 & 2 & 11 \end{pmatrix}}$$

Verification:

• Row 1: $7 \times 7 \times 7 = 343$ ✓
• Row 2: $2 \times 11 \times 3 = 66$ ✓
• Row 3: $2 \times 2 \times 11 = 44$ ✓
• Col 1: $7 \times 2 \times 2 = 28$ ✓
• Col 2: $7 \times 11 \times 2 = 154$ ✓
• Col 3: $7 \times 3 \times 11 = 231$ ✓