Playing With Numbers
Exercise 2.5 (Tests for Divisibility of Numbers)
Question 1 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
Test of Divisibility by 2 A number is divisible by 2, if its unit’s digit is 0, 2, 4, 6 or 8. i. 6520 The unit’s digit of 8652 is 2, so it is divisible by 2.
ii. 984325 The unit’s digit of 984325 is 5, so it is not divisible by 2.
iii. 367314 The unit’s digit of 367314 is 4, so it is divisible by 2. 
Question 2 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
Test of Divisibility by 3 A number is divisible by 3 if the sum of its digit is divisible by 3. i. 70335 Sum of the digits = 7+0+3+3+5 = 18, which is divisible by 3. Hence, the number 70335 is divisible by 3.
ii. 607439 Sum of the digits = 6+0+7+4+3+9 = 29, which is not divisible by 3. Hence, the number 607439 is not divisible by 3.
iii. 9082746 Sum of the digits = 9+0+8+2+7+4+6 = 36, which is divisible by 3. Hence, the number 9082746 is divisible by 3.

Question 3 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
Test of Divisibility by 6 A number is divisible by 6, if it is divisible by both 2 and 3. i. 7020 The unit’s digit of 7020 is 0, so it is divisible by 2. Sum of the digits = 7+0+2+0= 9, which is divisible by 3. Hence, the number 7020 is divisible by 3. Therefore, the number 7020 is divisible by 6.
ii. 56423 The unit’s digit of 56423 is 3, so it is not divisible by 2. Sum of the digits = 5+6+4+2+3= 20, which is not divisible by 3. Hence, the number 5623 is not divisible by 3. Therefore, the number 56423 is not divisible by 6.
iii. 732510 The unit’s digit of 732510 is 0, so it is divisible by 2. Sum of the digits = 7+3+2+5+1+0= 18, which is divisible by 3. Hence, the number 732510 is divisible by 3. Therefore, the number 732510 is divisible by 6. 
Question 4 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
Test of Divisibility by 4 A number is divisible by 4, if the number formed by its digits in ten’s and unit’s places is divisible by 4. i. 786532 The number formed by ten’s and unit’s digits is 32, which is divisible by 4. Hence, the number 786532 is divisible by 4.
ii. 1020531 The number formed by ten’s and unit’s digits is 31, which is not divisible by 4. Hence, the number 1020531 is not divisible by 4.
iii. 9801523 The number formed by ten’s and unit’s digits is 23, which is not divisible by 4. Hence, the number 9801523 is not divisible by 4.

Question 5 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
Test of Divisibility by 8 A number is divisible by 8, if the number formed by its digits in hundred’s, ten’s and unit’s places is divisible by 8. i. 8364 The number formed by hundred’s, ten’s and unit’s digits is 364, is not divisible by 8. Hence, the number 8364 is not divisible by 8.
ii. 7314 The number formed by hundred’s, ten’s and unit’s digits is 314, is not divisible by 8. Hence, the number 7314 is not divisible by 8.
iii. 36712 The number formed by hundred’s, ten’s and unit’s digits is 712, is divisible by 8. Hence, the number 36712 is divisible by 8. 
Question 6 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
Test of Divisibility by 9 A number is divisible by 9, if the sum of its digits is divisible by 9. i. 187245 Sum of its digits = 1+8+7+2+4+5 = 27, which is divisible by 9. Hence, the given number is divisible by 9.
ii. 3478 Sum of its digits = 3+4+7+8 = 22, which is not divisible by 9. Hence, the given number is not divisible by 9.
iii. 547218 Sum of its digits = 5+4+7+2+1+8 = 27, which is divisible by 9. Hence, the given number not divisible by 9.

Question 7 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
Test of Divisibility by 11 A number is divisible by 11, if the difference of the sum of its digits in odd places and the sum of its digits in even places (starting from unit’s place) is either 0 or multiple of 11. i. 5335 Sum of its digits in odd places = 5+3 = 8 Sum of its digits in even places = 3+5 = 8 Difference of the two sums = 88=0, which is divisible by 11. Hence, the number 5335 is divisible by 11.
ii. 70169803 Sum of its digits in odd places = 3+8+6+0 = 17 Sum of its digits in even places = 0+9+1+7 = 17 Difference of the two sums = 1717=0, which is divisible by 11. Hence, the number 70169803 is divisible by 11.
iii. 10000001 Sum of its digits in odd places = 1+0+0+0 = 1 Sum of its digits in even places = 0+0+0+1 = 1 Difference of the two sums = 11=0, which is divisible by 11. Hence, the number 10000001 is divisible by 11. 
Question 8 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
A number is divisible by 3 if the sum of its digit is divisible by 3. i. 75*5 Sum of the digits = 7+5+*+5= 17+* Nearest number divisible by 3 is the number 18. Therefore, value of *= 1817 = 1. Hence, the number 7515 is divisible by 3.
ii. 35*64 Sum of the digits = 3+5+*+6+4= 18+* Nearest number divisible by 3 is the number 18. Therefore, value of *= 1818 = 0. Hence, the number 35064 is divisible by 3.
iii. 18*71 Sum of the digits = 1+8+*+7+1 = 17+* Nearest number divisible by 3 is the number 18. Therefore, value of *= 1817 = 1. Hence, the number 18171 is divisible by 3.

Question 9 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
A number is divisible by 9, if the sum of its digits is divisible by 9. i. 67*19 Sum of the digits = 6+7+*+1+9 = 23+* Nearest number divisible by 9 is the number 27. Therefore, value of *= 2723 = 4. Hence, the number 67419 is divisible by 9.
ii. 66784* Sum of the digits = 6+6+7+8+4+* = 31+* Nearest number divisible by 9 is the number 36. Therefore, value of *= 3631 = 5. Hence, the number 667845 is divisible by 9.
iii. 538*8 Sum of the digits = 5+3+8+*+8 = 24+* Nearest number divisible by 9 is the number 27. Therefore, value of *= 2724 = 3. Hence, the number 53838 is divisible by 9. 
Question 10 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
A number is divisible by 11, if the difference of the sum of its digits in odd places and the sum of its digits in even places (starting from unit’s place) is either 0 or multiple of 11. i. 86*72 Sum of its digits in odd places = 2+*+8 = 10+* Sum of its digits in even places = 7+6 = 13 Difference of the two sums should either be 0 or multiple of 11 * = 3 [As 13(10+3) = 0, is divisible by 11] Therefore, the missing value is 3. Hence, the number 86372 is divisible by 11.
ii. 467*91 Sum of its digits in odd places = 1+*+6 = 7+* Sum of its digits in even places = 9+7+4 = 20 Difference of the two sums should either be 0 or multiple of 11 Value of * = 2 [as 20(7+2)=11] Hence, the number 467291 is divisible by 11.
iii. 9*8071 Sum of its digits in odd places = 1+0+* = 1+* Sum of its digits in even places = 7+8+9 = 24 Difference of the two sums should either be 0 or multiple of 11 Value of * = 1 [as 24(1+1)=22] Hence, the number 918071 is divisible by 11.

Question 11 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
i. 10 is the number which is divisible by 2 but not by 4. ii. 15 is the number which is divisible by 3 but not by 6. iii. 28 is the number which is divisible by 4 but not by 8. iv. 48 is the number which is divisible by both 4 and 8 but not by 32. 
Question 12 (Refer Book  Mathematics Class VI R.D. Sharma) 
Solution: 
i. False ii. True iii. False iv. True v. False vi. True vii. False viii. True ix. False x. True
