RS Aggarwal Class 6 Maths Chapter 2 - Factors and Multiples EXERCISE 2B

RS Aggarwal 2021-2022 for Class 6 Maths Chapter 2 - Factors and Multiples 


 EXERCISE 2B


1. Test the divisibility of the following numbers by 2: 

            (i) 2650                                 (ii) 69435     (ii) 59628 

          (iv) 789403         (v) 357986             (vi) 367314 

Solution:-

A number is divisible by 2 if its ones digit is 0, 2, 4, 6 or 8.

(i) The digit in the ones place in 26250 is 0. So,it is divisible by 2

(ii) The digit in the ones place in 69435 is not 0, 2, 4, 6 or 8. So,it is not divisible by 2.

(iii) The digit in the ones place in 59628 is 8, it is divisible by 2.

(iv) The digit in the ones place in 789403 is not 0, 2, 4, 6, or 8. So, it is not divisible by 2.

(v) The digit in the ones place in  357986 is 6. So,it is divisible by 2.

(vi) The digit in the ones place in 367314 is 4. So it is divisible by 2.

2. Test the divisibility of the following numbers by 3:  

            (i) 733 (ii) 10038     (iii) 2070 

            (iv) 524781 (v) 79124     (vi) 872645 

Solution:-

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

(i) 733 is not divisible by 3 because the sum of its digits, 7 + 3 + 3, is 13, which is not divisible     by 3.

(ii) 10038 is divisible by 3 because the sum of its digits, 1 + 0 + 0 + 3 + 8, is 12, which is          divisible by 3.

(iii) 20701 is not divisible by 3 because the sum of its digits, 2 + 0 + 7 + 0 + 1, is 10, which is not divisible by 3.

(iv) 524781 is divisible by 3 because the sum of its digits, 5 + 2 + 4 + 7 + 8 + 1, is 27, which is divisible by 3.

(v) 79124 is not divisible by 3 because the sum of its digits, 7 + 9 + 1 + 2 + 4, is 23, which is not divisible by 3.

(vi) 872645 is not divisible by 3 because the sum of its digits, 8 + 7 + 2 + 6 + 4 + 5, is 32, which is not divisible by 3.

 

 3. Test the divisibility of the following numbers by 4: 

            (i) 618 (ii) 2314             (iii) 63712 

            (iv) 35056         (v) 946126          (vi) 810524 

Solution:-

A number is divisible by 4 if the number formed by the digits in its tens and units place is divisible by 4.

(i)  618 is not divisible by 4 because the number formed by its tens and ones digits is 18, which is not divisible by 4.

(ii) 2314 is not divisible by 4 because the number formed by its tens and ones digits is 14, which is not divisible by 4.

(iii) 63712 is divisible by 4 because the number formed by its tens and ones digits is 12,  which is divisible by 4.

(iv) 35056 is divisible by 4 because the number formed by its tens and ones digits is 56, which is divisible by 4.

(v)  946126 is not divisible by 4 because the number formed by its tens and ones  digits is 26, which is not divisible by 4.

(vi) 810524 is divisible by 4 because the number formed by its tens and ones digits is  24, which is divisible by 4.


4. Test the divisibility of the following numbers by 5: 

            (i) 4965 (ii) 23590     (iii) 35208 

            (iv) 723405 (v) 124684     (vi) 438750

Solution:-

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

(i) 4965 is divisible by 5, because the digit at its ones place is 5.

(ii) 23590 is divisible by 5, because the digit at its ones place is 0.

(iii) 35208 is not divisible by 5, because the digit at its ones place is 8.

(iv) 723405 is divisible by 5, because the digit at its ones place is 5.

(v) 124684 is not divisible by 5, because the digit at its ones place is 4.

(vi) 438750 is divisible by 5, because the digit at its ones place is 0.


5. Test the divisibility of the following numbers by 6: 

            (i) 2070 (ii) 46523     (iii) 71232 

            (iv) 934706 (v) 251780     (vi) 872536 

Solution:-

A number is divisible by 6 if it is divisible by both 2 and 3.

(i) 2070 is divisible by 2 and 3. So, it is divisible by 6.

Checking the divisibility by 2: The number 2070 has 0 in its units place. So, it is divisible by 2.

Checking the divisibility by 3: The sum of the digits of 2070, 2 + 0 + 7 + 0, is 9, which is divisible by 3. So, it is divisible by 3.

(ii) 46523 is not divisible by 2, it is not divisible by 6.

Checking the divisibility by 2: The number 46523 has 3 in its units place, it is not divisible by 2.

(iii) 71232 is divisible by both 2 and 3, it is divisible by 6.

Checking the divisibility by 2: The number has 2 in its units place, it is divisible by 2.

Checking the divisibility by 3: The sum of the digits of the number, 7 + 1 + 2 + 3 + 2, is 15, which  is divisible by 3. So, the number is divisible by 3. 

(iv) 934706 is not divisible by 3, it is not divisible by 6.

Checking the divisibility by 3: Since the sum of the digits of the number, 9 + 3 + 4 + 7 + 0 + 6, is 29, which is not divisible by 3. So, the number is not divisible by 3.

(v)  251780 is not divisible by 3, it is not divisible by 6.

Checking the divisibility by 3: The sum of the digits of the number, 2 + 5 + 1 + 7 + 8 + 0, is 23, which is not divisible by 3. So, the number is not divisible by 3.

(vi)  872536 is not divisible by 3, it is not divisible by 6.

Checking the divisibility by 3: The sum of the digits of the number, 8 + 7 + 2 + 5 + 3 + 6, is 31, which is not divisible by 3. So, the number is not divisible by 3.

6. Test the divisibility of the following numbers by 7: 

            (i) 826 (ii) 117     (iii) 2345 

            (iv) 6021         (v) 14126     (vi) 25368 

Solution:-

To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits. If their difference is a multiple of 7, the number is divisible by 7.

(i) 826 is divisible by 7.

 We have 82 − 2 × 6 = 70, which is a multiple of 7.

(ii) 117 is not divisible by 7.

      We have 11 − 2 × 7 = −3, which is not a multiple of 7.

(iii) 2345 is divisible by 7.

 We have  234 − 2 × 5 = 224, which is a multiple of 7.

(iv) 6021 is divisible by 7. 

 We have 602 − 2 × 1 = 600, which is not a multiple of  7.

(v)  14126 is divisible by 7.

    We have 1412 − 2 × 6 = 1400, which is a multiple of 7.

(vi)  25368 is divisible by 7.

      We have 2536 − 2 × 8 = 2520, which is a multiple of 7.


7. Test the divisibility of the following numbers by 8 

            (i) 9364 (ii) 2138             (iii) 36792 

            (iv) 901674 (v) 136976     (vi) 1790184

Solution:-

A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.

(i) 9364 is not divisible by 8.

  Because the number formed by its hundreds, tens and ones digits, i.e., 364, is not divisible     by 8.

(ii) 2138 is not divisible by 8.

 Because the number formed by its hundreds, tens and ones digits, i.e., 138, is not divisible     by 8.

(iii) 36792 is divisible by 8.

 Because the number formed by its hundreds, tens and ones digits, i.e., 792, is divisible by     8.

(iv) 901674 is not divisible by 8.

 Because the number formed by its hundreds, tens and ones digits, i.e., 674, is not divisible     by 8.

(v) 136976 is divisible by 8.

 Because the number formed by its hundreds, tens and ones digits, i.e., 976, is divisible by     8.

(vi) 1790184 is divisible by 8.

 Because the number formed by its hundreds, tens and ones digits, i.e., 184, is divisible by     8.


8. Test the divisibility of the following numbers by 9: 

            (i) 2358 (ii) 3333             (iii) 98712 

            (iv) 257106 (v) 647514     (vi) 326999

Solution:-

A number is divisible by 9 if the sum of its digits is divisible by 9.

(i)  2358 is divisible by 9, because the sum of its digits, 2 + 3 + 5 + 8, is 18, which is divisible by 9.

(ii)  3333 is not divisible by 9, because the sum of its digits, 3 + 3 + 3 + 3, is 12, which is not divisible by 9.

(iii)  98712 is divisible by 9, because the sum of its digits, 9 + 8 + 7 + 1 + 2, is 27, which is divisible by 9.

(iv)  257106 is not divisible by 9, because the sum of its digits, 2 + 5 + 1 0 + 6, is 21, which is not divisible by 9.

(v)  647514 is divisible by 9, because the sum of its digits, 6 + 4 + 7 + 5 + 1 + 4, is 27, which is divisible by 9.

(vi)  326999 is not divisible by 9, because the sum of its digits, 3 + 2 + 6 + 9 + 9 + 9, is 38, which is not divisible by 9. 


9. Test the divisibility of the following numbers by 10: 

            (i) 5790 (ii) 63215     (iii) 55555 

Solution:-

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

(i) 5790 is divisible by 10, because its ones digit is 0.

(ii) 63215 is not divisible by 10, because its ones digit is 5, not 0.

(iii) 55555 is not divisible by 10, because its ones digit is 5, not 0.


10. Test the divisibility of the following numbers by 11: 

            (i) 4334 (ii) 83721     (iii) 66311 

            (iv) 137269 (v) 901351     (vi) 8790322 

Solution:-

A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.

(i)  4334 is divisible by 11.

    Sum of the digits at odd places = (4 + 3) = 7

    Sum of the digits at even places = (3 + 4) = 7

    Difference of the two sums = (7 − 7) = 0, which is divisible by 11.

(ii)  83721 is divisible by 11.

    Sum of the digits at odd places = (1 + 7 + 8) = 16

    Sum of the digits at even places = (2 + 3) = 5

    Difference of the two sums = (16 − 5) = 11, which is divisible by 11.

(iii) 66311 is not divisible by 11.

     Sum of the digits at odd places = (1 + 3 + 6) = 10

     Sum of the digits at even places = (1 + 6) = 7

    Difference of the two sums = (10 − 7) = 3, which is not divisible by 11.

(iv) 137269 is divisible by 11.

    Sum of the digits at odd places = (9 + 2 + 3) = 14

    Sum of the digits at even places = (6 + 7 + 1) = 14

    Difference of the two sums = (14 − 14) = 0, which is a divisible by 11.

(v)  901351 is divisible by 11.

    Sum of the digits at odd places = (0 + 3 + 1) = 4

    Sum of the digits at even places = (9 + 1 + 5) = 15

    Difference of the two sums = (4 − 15) = −11, which is divisible by 11.

(vi) 8790322 is not divisible by 11.

    Sum of the digits at odd places = (2 + 3 + 9 + 8) = 22

    Sum of the digits at even places = (2 + 0 + 7) = 9

    Difference of the two sums = (22 − 9) = 13, which is not divisible by 11


11. In each of the following numbers, replace by the smallest number to make it divisible by         3:

             (i) 27*4         (ii) 53*46          (iii) 8 *711

            (iv) 62*35                 (v) 234 *17     (vi) 6 *1054 

Solution:-

(i) Here, 2 + 7 + * + 4 = 13 + * should be a multiple of 3.

     To be divisible by 3, the least value of * should be 2, i.e., 13 + 2 = 15, which is a multiple         of 3.

    ∴ * = 2

(ii) Here, 5 + 3 + * + 4 + 6 = 18 + * should be a multiple of 3.

    As 18 is divisible by 3, the least value of  * should be 0, i.e., 18 + 0 = 18.

     ∴ * = 0

(iii) Here, 8+ * + 7 + 1 + 1 = 17 + * should be a multiple of 3.

    To be divisible by 3, the least value of  * should be 1, i.e., 17 + 1 = 18 , which is a multiple         of 3.

    ∴ * = 1

(iv) Here, 6 + 2 + * + 3 + 5 = 16 + * should be a multiple of 3.

    To be divisible by 3, the least value of  * should be 2, i.e., 16 + 2 = 18, which is a multiple         of 3.

    ∴ * = 2

 (v) Here, 2+ 3 +4 + * + 1 + 7 = 17 + * should be a multiple of 3.

    To be divisible by 3, the least value of  * should be 1, i.e., 17 + 1 = 18, which is a multiple         of 3.

    ∴ * =1

(vi) Here, 6 + * +1 + 0 + 5 + 4 = 16 + * should be a multiple of 3.

    To be divisible by 3, the least value of  * should be 2, i.e., 16 + 2 = 18, which is a multiple         of 3.

    ∴ * =2


12. In each of the following numbers, replace by the smallest number to make it divisible by         9: 

             (i) 65*5         (ii) 2*1335     (iii) 6702* 

            (iv) 91*67                 (v) 6678*1     (vi) 835*86

Solution:-

(i) Here, 6 + 5+ * + 5 = 16 + * should be a multiple of 9.

    To be divisible by 9, the least value of * should be 2, i.e., 16 + 2 = 18, which is a multiple of     9.

    ∴ * =2

(ii) Here, 2 + * + 1 + 3 + 5 = 11 + * should be a multiple of 9.

    To be divisible by 9, the least value of * should be 7, i.e., 11 + 7 = 18, which is a multiple of     9.

    ∴ * = 7

(iii) Here, 6 + * + 7 + 0 + 2 = 15 + * should be a multiple of 9.

    To be divisible by 9, the least value of * should be 3, i.e., 15 + 3 = 18, which  is a multiple         of 9.

    ∴ * = 3

(iv) Here, 9 + 1 * + 6 + 7 = 23 + * should be a multiple of 9.

    To be divisible by 9, the least value of * should be 4, i.e., 23 + 4 = 27, which is a multiple of     9.

    ∴ * = 4

(v) Here,  6 + 6 + 7 + 8 + * + 1 = 28 + * should be a multiple of 9.

    To be divisible by 9, the least value of * should be 8, i.e., 28 + 8 = 36, which is a multiple         of 9.

    ∴ * = 8

 (vi) Here, 8 + 3 + 5 + * + 8 + 6 = 30 + * should be a multiple of 9.

    To be divisible of 9, the least value of * should be 6, i.e.,  30 + 6 = 36, which is a multiple of     9.

    ∴ * = 6


13. In each of the following numbers, replace * by the smallest number to make it  divisible by     11:

            (i) 26 * 5                 (ii) 39 * 43     (iii) 86 * 72 

            (iv) 467 * 91         (v) 1723 * 4     (vi) 9 * 8071 

Solution:-

(i) 26*5

    Sum of the digits at odd places = 5 + 6 = 11

    Sum of the digits at even places = * + 2

    Difference = sum of odd terms – sum of even terms

    = 11 – (* + 2)  

    = 11 – * – 2

     = 9 – *

    Now, (9 – *) will be divisible by 11 if * = 9.

    i.e., 9 – 9 = 0

    0 is divisible by 11.

    ∴ * = 9

    Hence, the number is 2695.

(ii) 39*43

    Sum of the digits at odd places = 3 + * + 3 = 6 + *

    Sum of the digits at even places = 4 + 9 = 13

    Difference = sum of odd terms – sum of even terms

    = 6 + * – 13  

    = * – 7

    Now, (* – 7) will be divisible by 11 if * = 7.

    i.e., 7 – 7 = 0

    0 is divisible by 11.

    ∴ * = 7

    Hence, the number is 39743.

 (iii) 86*72

    Sum of the digits at odd places  2 + * + 8 = 10 + *

    Sum of the digits at even places  6 + 7 = 13

    Difference = sum of odd terms – sum of even terms

    = 10 + * – 13  

    = * – 3

   Now, (* – 3) will be divisible by 11 if * = 3.

    i.e., 3 – 3 = 0

    0 is divisible by 11.

    ∴ * = 3

     Hence, the number is 86372.


    (iv) 467*91

    Sum of the digits at odd places 1  + * + 6 = 7 + *

    Sum of the digits at even places  9 + 7 + 4 = 20

    Difference = sum of odd terms – sum of even terms

    = (7 + *) − 20  

   = * − 13

  Now, (* −13) will be divisible by 11 if * = 2.

    i.e., 2− 13 = −11

    −11 is divisible by 11.

    ∴ * = 2

     Hence, the number is 467291.

(v) 1723*4

    Sum of the digits at odd places 4+ 3+ 7= 14

    Sum of the digits at even places  *+2+1 = 3 + *

    Difference = sum of odd terms – sum of even terms

    = 14 – (3 + *)  

   = 11 − *

      Now, (11 − *) will be divisible by 11 if * = 0.

    i.e., 11 − 0 = 11

    11 is divisible by 11.

    ∴ * = 0

     Hence, the number is 172304.

(vi) 9*8071

    Sum of the digits at odd places 1+0+*  = 1 + *

    Sum of the digits at even places  7 + 8 + 9 = 24

    Difference = sum of odd terms – sum of even terms

    =1 + *  – 24  

   =   * − 23

      Now, (* − 23) will be divisible by 11 if * = 1.

    i.e., 1 − 23 = −22

    −22 is divisible by 11.

    ∴  * = 1

    Hence, the number is 918071.

    

    14. Test the divisibility of: 

            (i) 10000001 by 11     (ii) 19083625 by 11          (iii) 2134563 by 9 

            (iv) 10001001 by 3      (v) 10203574 by 4         (vi) 12030624 by 8

Solution:-

(i) 10000001 by 11

    10000001 is divisible by 11.

    Sum of digits at odd places = (1 + 0 + 0 + 0) = 1

    Sum of digits at even places = (0 + 0 + 0 + 1) = 1

    Difference of the two sums = (1 − 1) = 0, which is divisible by 11.

 (ii) 19083625 by 11

    19083625 is divisible by 11.

    Sum of digits at odd places = (5 + 6 + 8 + 9) = 28

    Sum of digits at even places = (2 + 3 + 0 + 1) = 6

    Difference of the two sums = (28 − 6) = 22, which is divisible by 11.

(iii) 2134563 by 9

    2134563 is not divisible by 9.

    It is because the sum of its digits, 2 + 1 + 3 + 4 + 5 + 6 + 3, is 24, which is not divisible by 9.

(iv) 10001001 by 3

    10001001 is divisible by 3. 

    It is because the sum of its digits, 1 + 0 + 0 + 0 + 1 + 0 + 0 + 1, is 3, which is divisible by 3.

(v) 10203574 by 4                  

    10203574 is not divisible by 4.

    It is because the number formed by its tens and the ones digits is 74, which is not                   divisible by 4.

(vi) 12030624 by 8

     12030624 is divisible by 8.

    It is because the number formed by its hundreds, tens and ones digits is 624, which is            divisible by 8.


15. Which of the following are prime numbers? 

            (i) 103         (ii) 137         (iii) 161         (iv) 179 

           (v) 217                 (vi) 277         (vii) 331         (viii) 397

Solution:-

    A number between 100 and 200 is a prime number if it is not divisible by any prime                number less than 15.

    Similarly, a number between 200 and 300 is a prime number if it is not divisible by any           prime number less than 20.

    (i) 103 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.

    (ii) 137 is a prime number, because it is not divisible by 2, 3, 5, 7 and 11.

    (iii) 161 is a not prime number, because it is divisible by 7.

    (iv) 179 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.

    (v) 217 is a not prime number, because it is divisible by 7.

    (vi) 277 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.

    (vii) 331 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.

    (viii) 397 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.


 16. Give an example of a number 

     (i) which is divisible by 2 but not by 4. 

    (ii) which is divisible by 4 but not by 8. 

  (iii) which is divisible by both 2 and 8 but not by 16.

  (iv) which is divisible by both 3 and 6 but not by 18. 

Solution:-

    (i)  6 is divisible by 2, but not by 4.

    (ii) 20 is divisible by 4, but not by 8.

    (iii) 40 is divisible by both 2 and 8, but not by 16.

    (iv) 42 is divisible by both 3 and 6, but not by 18.


17. Write (T) for true and (F) for false against each of the following statements: 

    (i) If a number is divisible by 4, it must be divisible by 8. 

    (ii) Ifa number is divisible by 8, it must be divisible by 4. 

   (iii) If a number divides the sum of two numbers exactly, it must exactly divide the                            numbers separately. 

  (iv) If a number is divisible by both 9 and 10, it must be divisible by 90 

         Hint. 9 and 10 are co-primes.

  (v) A number is divisible by 18 if it is divisible by both 3 and 6. 

        Hint. 3 and 6 are not co-primes.Consider 186.

Solution:-

    (i) If a number is divisible by 4, it must be divisible by 8. False

        Example: 28 is divisible by 4 but not divisible by 8.

    (ii) If a number is divisible by 8, it must be divisible by 4. True

        Example: 32 is divisible by both 8 and 4.

    (iii) If a number divides the sum of two numbers exactly, it must exactly divide the                     numbers separately. False

        Example: 91 (51 + 40) is exactly divisible by 13. However, 13 does not exactly divide 51             and 40.

    (iv) If a number is divisible by both 9 and 10, it must be divisible by 90. True

        Example: 900 is both divisible by 9 and 10. It is also divisible by 90.

    (v)  A number is divisible by 18 if it is divisible by both 3 and 6. False

        A number has to be divisible by 9 and 2 to be divisible by 18.

        Example: 48 is divisible by 3 and 6, but not by 18.

     (vi) If a number is divisible by 3 and 7, it must be divisible by 21. True

        Example: 42 is divisible by both 3 and 7. It is also divisible by 21.

    (vii) The sum of consecutive odd numbers is always divisible by 4. True

         Example: 11 and 13 are consecutive odd numbers.

        11 + 13 = 24, which is divisible by 4.  

    (viii) If a number divides two numbers exactly, it must divide their sum exactly.  True

        Example: 42 and 56 are exactly divisible by 7.

        42+56 = 98, which is exactly divisible by 7. 







एक टिप्पणी भेजें