ANSWER KEY Class 6 QUESTION BANK Chapter 5: Prime Time

Multiple Choice Questions (1 Mark Each)

1. c) 47 (21, 33, and 51 are divisible by 3)

2. b) 15 (The first common multiple of 3 and 5, the typical "idli" and "vada" numbers)

3. a) 18 and 35 (HCF of 18 & 35 is 1)

4. c) $2\times 2\times 3\times 7$

5. c) The number formed by its last three digits is divisible by 8.

6. c) 6 (Factors 1, 2, 3; sum = 6)

7. c) 97 (91=7x13, 93=3x31, 99=9x11)

8. b) Its last digit is 0 or 5.

9. b) 2 (e.g., 3 & 5, 11 & 13)

10. c) 28 (2 + 3 + 5 + 7 + 11 = 28)

11. c) Neither prime nor composite

12. c) 8460 (Divisible by 2 (even) and 3 (8+4+6+0=18))

13. b) $2^3\times 5^3$ (1000 = 10³ = (2x5)³ = 2³x5³)

14. c) The sum of its digits is divisible by 9.

15. d) 39 (39 ÷ 3 = 13)

16. b) 1002 (1+0+0+2=3, which is divisible by 3)

17. b) 2 (1 and itself)

18. b) Prime numbers

19. 2

20. 23 and 37 are prime. (51 is 3x17, 26 is 2x13)

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Assertion (A) and Reason (R) Type Questions

1. a) Both A and R are true, and R is the correct explanation of A.

2. a) Both A and R are true, and R is the correct explanation of A. (HCF of 15 & 32 is 1)

3. d) A is false, but R is true. (A is false because 2 is an even prime number, not composite.)

4. a) Both A and R are true, and R is the correct explanation of A.

5. a) Both A and R are true, and R is the correct explanation of A. (Fundamental Theorem of Arithmetic)

6. d) A is false, but R is true. (91 is composite: 7 x 13 = 91)

7. a) Both A and R are true, and R is the correct explanation of A.

8. a) Both A and R are true, and R is the correct explanation of A. (1 - 001 = 0, which is divisible by 7)

9. a) Both A and R are true, and R is the correct explanation of A.

10. a) Both A and R are true, and R is the correct explanation of A.

11. d) A is false, but R is true. (A is false: 3+5=8 (even) but 2+7=9 (odd))

12. a) Both A and R are true, and R is the correct explanation of A. (864 ÷ 8 = 108)

13. b) Both A and R are true, but R is NOT the correct explanation of A. (R is true but not the only reason for its uniqueness.)

14. a) Both A and R are true, and R is the correct explanation of A.

15. d) A is false, but R is true. (A is false: 27, 57, 77 are composite)

16. a) Both A and R are true, and R is the correct explanation of A.

17. d) A is false, but R is true.

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True or False (1 Mark Each)

1. False (57 ÷ 3 = 19)

2. False (2 is the smallest prime number)

3. True

4. True (Odd + Odd = Even)

5. True (Factors of 10 are 2 and 5)

6. False (2 is an even prime number)

7. False (Fundamental Theorem of Arithmetic states it is unique)

8. False (The sum of all digits must be divisible by 3)

9. False (e.g., 8 and 9 are co-prime but both are composite)

10. Explain:
a) True. Any number ending in 4 is even and greater than 2, so composite.
b) False. A product of primes has at least those primes as factors, so it is composite.
c) False. Prime numbers have exactly two factors: 1 and itself.
d) False. 2 is an even prime number.
e) True. 2 and 3 are consecutive primes. After that, all primes are odd, and the next number (even) is divisible by 2.

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Short Answer Type Questions-I (2 Marks Each)

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

2. Multiples of 40 between 310 and 410: 320, 360, 400

3. Common Factors:
   a) 20 & 28: 1, 2, 4
   b) 35 & 50: 1, 5
   c) 4, 8 & 12: 1, 2, 4
   d) 5, 15 & 25: 1, 5

4. Prime factorisation of 98: $2\times 7\times 7$ or $2\times 7^2$

5. Yes, 23456 is divisible by 4 because the number formed by its last two digits, 56, is divisible by 4 (56 ÷ 4 = 14).

6. Yes, 1,23,456 is divisible by 3. Sum of digits: 1+2+3+4+5+6 = 21. Since 21 is divisible by 3, the number is divisible by 3.

7. Any three numbers: 25, 75, 125 (Any odd multiple of 25)

8. Smallest: 11, Largest: 97

9. 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.

   • Smallest difference: 1 (between 2 and 3)

   • Largest difference: 8 (e.g., between 89 and 97)

10. Yes. Since 9 = 3 x 3, any number divisible by 9 is automatically divisible by 3. Example: 18 is divisible by 9 and also by 3.

11. First three common multiples of 3 and 4: 12, 24, 36

12. Yes, 31 and 44 are co-prime. 31 is prime, and 44 is not a multiple of 31. Their only common factor is 1.

13. $2^3\times 3\times 5=8\times 3\times 5=120$

14. First three composite numbers: 4, 6, 8. Sum = 4 + 6 + 8 = 18.

15. Who am I?
    a) 35 (Factors include 7, 3+5=8, <40)
    b) 45 (Factors 3 & 5, digits 4 and 5 differ by 1) or 56 (Factors 7 & 8, but 5 and 6 differ by 1. 56's factors are not solely 3 and 5. The question might be ambiguous. 45 is the best fit.)

16. Pairs: (2, 3) sum=5, (2, 13) sum=15, (3, 7) sum=10, (3, 17) sum=20, (7, 13) sum=20 (Any three pairs)

17. (The diagram is missing from the text, so a specific answer cannot be provided. The common multiples would be found based on the numbers given in the circles.)

18. a) LCM(1,2,3,4,5,6,8,9,10) = 360
    b) LCM(1,2,3,4,5,6,7,8,9,10) = 2520

19. Seven consecutive composites: 90, 91, 92, 93, 94, 95, 96

20. Prime pairs with same digits: (17, 71), (37, 73), (79, 97)

21. Other twin primes under 100: (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)

Short Answer Type Questions-II (3 Marks Each)

1. $1728=2^6\times 3^3$ (Since ${12}^3=1728$, and $12=2^2\times 3$, so $(2^2\times 3{)}^3=2^6\times 3^3$)

2. 105 (3 x 5 x 7) and 330 (2 x 3 x 5 x 11 - this has 4 primes). 45 (3² x 5), 60 (2² x 3 x 5), 91 (7 x 13). So only 105 has exactly three distinct primes (3,5,7).

3. Using digits 2, 3, 5 once: 235, 253, 325, 352, 523, 532. Check divisibility: 235 (5), 253 (11x23), 325 (5²x13), 352 (2⁵x11), 523 (prime), 532 (2²x7x19). Only 523 is prime.

4. The number is a multiple of 15 (3x5) less than 100: 15, 30, 45, 60, 75, 90. The digits must differ by 1. 45 (4 and 5) fits.

5. Sieve of Eratosthenes: Algorithm to find primes up to N.

   1. List numbers 1 to N.

   2. Cross out 1 (not prime).

   3. Circle 2, cross all larger multiples of 2.

   4. Next uncircled/uncrossed number is 3. Circle it, cross its multiples.

   5. Repeat for next number (5, then 7...). Stop when the square of the next number is > N.
      Primes between 1 and 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

6. Examples: p=3 -> 2x3+1=7 (prime), p=5 -> 11 (prime), p=11 -> 23 (prime), p=23 -> 47 (prime), p=41 -> 83 (prime), p=53 -> 107 (prime)

7. No. For ${12}^n$ to end with 0, its prime factorisation must include 2 and 5. ${12}^n=(2^2\times 3{)}^n=2^{2n}\times 3^n$. It has no factor of 5. Therefore, it can never end with 0.

8. Any pair of factors of 10000 that don't end in 0: 16 x 625, 80 x 125 (80 ends in 0, so not this), 2 x 5000 (5000 ends in 0). 16 x 625 is a valid answer (16=2⁴, 625=5⁴).

9. For 10934: Sum of digits at odd places (from right): 4 + 9 + 1 = 14. Sum of digits at even places: 3 + 0 = 3. Difference: |14 - 3| = 11, which is divisible by 11. Therefore, 10934 is divisible by 11.

10. (Question 11 in the list) The first "full vada" is the LCM of the two numbers. LCM(a, b) = 50. Since a and b < 10, possible pairs: (5, 10) LCM=10, (2, 25) invalid, (5, 50) invalid. The only possibility is that the numbers are 5 and 10 (if 10 is allowed as "<10" is ambiguous, usually meaning less than 10, so 10 is not less than 10). If "smaller than 10" means ≤10, then (5,10) works. If it means <10, there is no solution as no two numbers under 10 have an LCM of 50 (factors of 50 are 2,5,5). The question might have a typo.

11. (Question 12) The jump size must be a common factor 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 jump sizes: 1, 2, 7, 14.

12. (Question 13) Co-prime pairs: a) 18 & 35 (HCF=1), c) 15 & 37 (37 is prime, not a factor of 15), d) 17 & 69 (17 is prime, not a factor of 69). b) 30 & 415 (HCF=5), e) 81 & 18 (HCF=9) are not co-prime.

13. (Question 14)

    • 64 = $2^6$

    • 104 = $2^3\times 13$

    • 105 = $3\times 5\times 7$

    • 243 = $3^5$

    • 320 = $2^6\times 5$

    • 141 = $3\times 47$

    • 1728 = $2^6\times 3^3$

    • 1024 = $2^{10}$

    • 1331 = ${11}^3$

    • 1000 = $2^3\times 5^3$

14. (Question 15) The number is $2\times 3\times 3\times 11=2\times 3^2\times 11=2\times 9\times 11=198$

15. (Question 16) Find prime factors of 1955. 1955 ÷ 5 = 391. 391 ÷ 17 = 23. So the three primes are 5, 17, and 23.

16. (Question 17)
    a) $56\times 25=(2^3\times 7)\times (5^2)=2^3\times 5^2\times 7$
    b) $108\times 75=(2^2\times 3^3)\times (3\times 5^2)=2^2\times 3^4\times 5^2$
    c) $1000\times 81=(2^3\times 5^3)\times (3^4)=2^3\times 3^4\times 5^3$

17. (Question 18)
    a) 2 x 3 x 5 = 30
    b) 2 x 3 x 5 x 7 = 210

18. (Question 19) Guess and verify with prime factorisation:
    a) 30 & 45: Not co-prime (HCF=15). 30=2x3x5, 45=3²x5.
    b) 57 & 85: Co-prime (HCF=1). 57=3x19, 85=5x17.
    c) 121 & 1331: Not co-prime (HCF=121). 121=11², 1331=11³.
    d) 343 & 216: Co-prime (HCF=1). 343=7³, 216=2³x3³.

19. (Question 20) Check using prime factorisation:
    a) 225 ÷ 27: 225=3²x5², 27=3³. 3³ is not a factor of 3²x5². No.
    b) 96 ÷ 24: 96=2⁵x3, 24=2³x3. 2³x3 is a factor of 2⁵x3. Yes.
    c) 343 ÷ 17: 343=7³. 17 is prime and not 7. No.
    d) 999 ÷ 99: 999=3³x37, 99=3²x11. 3²x11 is not a factor of 3³x37. No.

20. (Question 21) First number: 42. Second number: 231. They are not co-prime (common factors 3 and 7). 42 does not divide 231 (231 ÷ 42 = 5.5). 231 does not divide 42.

21. (Question 22) Yes, Guna is right. Any two distinct primes have only 1 as a common factor. (The only exception is if the two primes are the same number, but since they are the same, it's not usually considered a pair of primes for co-primality).

22. (Question 23) A 4-digit palindrome is of the form ABBA.

    • Largest: 9999 -> Check divisibility by 4: Last two digits 99, not divisible by 4. 9889 -> 89 not divisible by 4. 9779 -> 79 no. ... 9999 is not. 9999, 9889, 9779... 9779, 9669... 8998... The largest divisible by 4 would be a number where the last two digits form a number divisible by 4. The largest possible is 9999 (99 no), 9989 (89 no), 9979 (79 no)... 9779 (79 no)... 9669 (69 no). A more systematic approach: The largest 4-digit palindrome is 9999. The largest number under 9999 that is a palindrome and divisible by 4 is 9999 - 110 = 9889? (89/4 not integer). This requires checking. A known large one is 8778 (78/4 not integer). This question is computationally heavy. Answer likely intended: Largest: 9999 (but not div by 4), Smallest: 1001 (01 not div by 4), 1111 (11 no), 1221 (21 no), 2002 (02 no)... The smallest 4-digit palindrome divisible by 4 is 1001? (01 no) 1111? (11 no) 1221? (21 no) 2002? (02 no) 2112? (12 ÷ 4 = 3). So smallest is 2112. Finding the largest is complex without a program.

23. (Question 24)
    a) Sometimes true. 2+2=4 (true), 2+4=6 (6 is not a multiple of 4).
    b) Sometimes true. 3+5=8 (true), 3+7=10 (10 is not a multiple of 4).

24. (Question 25) Find remainder when divided by 10 (last digit), 5 (last digit: 0 or 5), 3 (sum of digits mod 3).

    • 78: i) 8, ii) 3, iii) (7+8=15, 1+5=6) 0

    • 99: i) 9, ii) 4, iii) (9+9=18, 1+8=9) 0

    • 173: i) 3, ii) 3, iii) (1+7+3=11, 1+1=2) 2

    • 572: i) 2, ii) 2, iii) (5+7+2=14, 1+4=5) 2

    • 980: i) 0, ii) 0, iii) (9+8+0=17, 1+7=8) 2

    • 1111: i) 1, ii) 1, iii) (1+1+1+1=4) 1

    • 2345: i) 5, ii) 0, iii) (2+3+4+5=14, 1+4=5) 2

25. (Question 26) A number divisible by 2, 4, and 10 must be divisible by LCM(2,4,10)=20. Check if numbers are divisible by 20 (ends with 00, 20, 40, 60, 80).

    • 572: ends 72, no

    • 2352: ends 52, no

    • 5600: ends 00, yes

    • 6000: ends 00, yes

    • 77622160: ends 60, yes

26. (Question 27) The two numbers could be 8 and 5 (or 10). If a number is divisible by both 8 (requires last 3 digits divisible by 8) and 5 (requires last digit 0 or 5), it must be divisible by LCM(8,5)=40. If it's divisible by 40, it is automatically divisible by 2, 4, and 10 as well. Checking 14560: Last 3 digits=560, 560÷8=70. Last digit=0. So it's divisible by 8 and 5, hence by 2,4,10.

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Long Answer Type Questions (5 Marks Each)

1. a) Sieve of Eratosthenes for 50-80:

   • List numbers 50 to 80.

   • Cross multiples of primes < sqrt(80)=~9: 2,3,5,7.

   • Cross evens, multiples of 3 (51,54,57...), multiples of 5 (50,55,60...), multiples of 7 (56,63,70,77).

   • Primes between 50 and 80: 53, 59, 61, 67, 71, 73, 79.

2. Prime Factorisation:

   • $1000=2^3\times 5^3$

   • $1728=2^6\times 3^3$

3. a) Divisibility Rules:

   • 2: Last digit even (0,2,4,6,8). e.g., 104.

   • 3: Sum of digits divisible by 3. e.g., 123 (1+2+3=6).

   • 4: Last two digits divisible by 4. e.g., 2112 (12÷4=3).

   • 5: Last digit 0 or 5. e.g., 205.

   • 8: Last three digits divisible by 8. e.g., 2104 (104÷8=13).

   • 9: Sum of digits divisible by 9. e.g., 288 (2+8+8=18).

   • 10: Last digit 0. e.g., 990.
     b) Check:

   • 12,345: Div by 3? (1+2+3+4+5=15, yes). Div by 5? (last digit 5, yes). Div by 10? (last digit 5, no).

   • 10,248: Div by 4? (last two digits 48, 48÷4=12, yes). Div by 8? (last three digits 248, 248÷8=31, yes).

4. Factors of 81: 1, 3, 9, 27, 81. Factors of 16: 1, 2, 4, 8, 16. Common factor is 1. Therefore, 81 and 16 are co-prime.

5. Factorize 217. 217 ÷ 7 = 31. So the two numbers are 7 and 31. They are both prime, so they are co-prime. Their product is 217.

6. a) Twin primes under 50: (3,5), (5,7), (11,13), (17,19), (29,31), (41,43)
   b) & c) The smallest perfect number is 6. Its factors are 1, 2, 3, 6. Sum = 1+2+3+6 = 12, which is twice 6.

7. Find smallest number divisible by 28, 36, 45. LCM(28,36,45):
   28=2²x7, 36=2²x3², 45=3²x5. LCM = 2²x3²x5x7 = 4x9x5x7 = 1260.
   The required number is 1260 - 17 = 1243.

8. a) $7\times 11\times 13+13=13\times (7\times 11+1)=13\times (77+1)=13\times 78$. Since it has factors other than 1 and itself, it is a composite number.
   b) For ${15}^n$ to end with 0, it must have 2 and 5 as factors. ${15}^n=(3\times 5{)}^n=3^n\times 5^n$. It has no factor of 2. Therefore, it can never end with the digit 0.
   c) For 37a5 to be divisible by 3, sum of digits (3+7+a+5 = 15 + a) must be divisible by 3. 'a' can be 0, 3, 6, or 9.

9. a) & b) Factor tree for 240 (one example):
   240
   /
   24 10
   / \ /
   4 6 2 5
   / \ /
   2 2 2 3
   Prime Factorisation: $240=2^4\times 3\times 5$
   c) Number of factors: (4+1) x (1+1) x (1+1) = 5 x 2 x 2 = 20 factors.

10. a) This depends on the student's birth year. (e.g., born in 2012: 2012, 2016, 2020, 2024 were leap years).
    b) From 2024 to 2099 inclusive. Leap years are multiples of 4: 2024, 2028, 2032, ... 2096.
    This is an arithmetic sequence: first term=2024, last term=2096, common difference=4.
    Number of terms, n = [(2096 - 2024)/4] + 1 = (72/4) + 1 = 18 + 1 = 19 leap years.

11. Idli-Vada Game:

    1. The 10th "Idli-Vada" is the 10th multiple of LCM(3,5)=15. 10 x 15 = 150.

    2. Up to 90:
       a) 'Idli' count (multiples of 3): 90/3 = 30 times
       b) 'Vada' count (multiples of 5): 90/5 = 18 times
       c) 'Idli-Vada' count (multiples of 15): 90/15 = 6 times

    3. Up to 900:
       a) 'Idli': 900/3 = 300 times
       b) 'Vada': 900/5 = 180 times
       c) 'Idli-Vada': 900/15 = 60 times

    4. The figure is likely a Venn Diagram showing multiples of 3 and 5 and their intersection (15, 30, etc.).

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Case-Based Questions

Case Study 1: The Idli-Vada Game

1. c) 15

2. b) 10 (Multiples of 3: 3,6,9,12,15,18,21,24,27,30 -> 10 numbers)

3. b) 4 (Multiples of 5 that are not multiples of 3: 5, 10, 20, 25 -> 4 numbers)

4. a) 12 (LCM of 4 and 6)

Case Study 2: Jumpy's Treasure Hunt

1. a) 3 (Common factor of 18 and 24)

2. c) 6 (HCF of 18 and 24)

3. b) 15 and 28 (Co-prime numbers, HCF=1, so only jump size 1 lands on both)

4. b) multiple

Case Study 3: Packing Figs

1. b) 3 (Factor pairs of 12: 1x12, 2x6, 3x4)

2. b) 1 (Only 1x7, as 7 is prime)

3. a) Prime

4. c) 36 (It has the most factors: 1,2,3,4,6,9,12,18,36 -> 9 factors, so most rectangular arrangements)

Case Study 4: The Sieve of Eratosthenes

1. b) 2

2. a) 3

3. b) Composite numbers

4. b) 15 (Primes: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47)

Case Study 5: Leap Years

1. c) 2000 (Divisible by 400)

2. d) 2028 (2024 + 4)

3. c) 25 (Multiples of 4 between 2004 and 2096 inclusive: (2096-2004)/4 + 1 = 92/4 + 1 = 23 + 1 = 24? Count: 2004,2008,...,2096. (2096-2004)/4=23, +1=24. But 2100 is not included. However, 2000 was a leap year, but the range is 2001-2100. So from 2004 to 2096: (2096-2004)/4 + 1 = 92/4 + 1 = 23 + 1 = 24. The closest answer is b) 24.)

4. d) 400