TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.1

Students can practice TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Exercise 3.1

Question 1.
Which of the following numbers are divisible by 2, by 3 and by 6 ?
(i) 321729
Answer:
The given number is 321729.
Since 9 is in units place, it is not divisible by 2.
Sum of the digits = 3 + 2 + 1 + 7 + 2 + 9 = 24 (a multiple of 3)
So, it is divisible by 3.
∴ The given number is not divisible by 6 as it is not divisible by 2.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.1

(ii) 197232
Answer:
The given number is 197232.
Since 2 is in units place, it is divisible by 2.
Sum of the digits = l+ 9 + 7 + 2 + 3 + 2 = 24 (a multiple of 3)
So, it is divisible by 3.
∴ The given number is divisible by 6 because it is divisible by both 2 and 3.

(iii) 972132
Answer:
The given number is 972132.
Since 2 is in units place, it is divisible by 2.
Sum of the digits = 9 + 7 + 2 + 1 + 3 + 2 = 24 (a multiple of 3)
So, it is divisible by 3.
∴ The given number is divisible also because it is divisible by both 2 and 3.

(iv) 1790184
Answer:
The given number is 1790184.
Since 4 is in units place, it is divisible by 2.
Sum pf the digits = 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30 (a multiple of 3)
So, it is divisible by 3.
∴The given number is divisible by 6 also because it is divisible by both 2 and 3.

(v) 312792
Answer:
The given number is 312792.
Since 2 is in units place, it is divisible by 2.
Sum of the digits = 3 + 1 + 2 + 7 + 9 + 2 = 24 (a multiple of 3) So, it is divisible by 3.
∴ The given number is divisible by 6 also because it is divisible by both 2 and 3.

(vi) 800552
Answer:
The given number is 800552
Since 2 is in units place, it is divisible by 2.
Sum of the digits = 8 + 0 + 0 + 5 + 5 + 2 = 20 (not a multiple of 3)
∴ The given number is not divisible by 3.
∴ The given number is divisible by 2 but not by 3.
So, it is not divisible by 6.

(vii) 4335
Answer:
The given number is 4335.
Since 5 is in units place, it is not divisible by 2.
Sum of the digits = 4 + 3 + 3 + 5 = 15 (a multiple of 3)
So, it is divisible by 3.
∴ The given number is divisible by 3 but not by 2.
So, it is not divisible by 6.

(viii) 726352
Answer:
The given number is 726352.
Since 2 is in units place, it is divisible by 2.
Sum of the digits = 7 + 2 + 6 +3 + 5 + 2 = 25 (not a multiple of 3)
So, it is not divisible by 3.
∴ The given number is divisible by 2 but not by 3.
So, it is not divisible by 6.

Question 2.
Determine which of the following numbers are divisible by 5 and by 10. 25,125,250,1250,10205,70985,45880. Check whether the numbers that are divisible by 10 are also divisible by 2 and 5.
Answer:
Since ‘5’ or ‘0’ is in units place of the above given numbers, they are divisible by 5.
The numbers having ‘0’ in units place (i.e.,)
250, 1250, 45880 are divisible by 10.
250 is divisible by 2 and also by 5. (∵ If ‘0’ is in units place, it is divisible by 2 and by 5).
So, the other two numbers (i.e.,) 1250 and 45880 are also divisible by 2 and by 5.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.1

Question 3.
Fill the table using divisibility test for 3 and 9.

Number Sum of the digits in the number Divisible by
3 9
72 …………………………………………
197 …………………………………………
4689 …………………………………………
79875 …………………………………………
988974 9 + 8 + 8 + 9 + 7 + 4 = 45 Yes Yes

Answer:

Number Sum of the digits in the number Divisible by
3 9
72 7 + 2 = 9 Yes Yes
197 1 + 9 + 7 = 17 No Yes
4689 4 + 6 + 8 + 9 = 27 Yes Yes
79875 7 + 9 + 8 + 7 + 5 = 36 Yes Yes
988974 9 + 8 + 8 + 9 + 7 + 4 = 45 Yes Yes

Question 4.
Make 3 different 3 digit numbers using 1, 9 and 8, where each digit can be used only once. Check which of these numbers are divisible by 9.
Answer:
The 3 different 3 digit numbers using 1, 9 and 8 are 981, 819, 198
Sum of the digits in 981 = 9 + 8 + 1 = 18
18 is divisible by 9. So the number 981 is also divisible by 9.
Sum of the digits in 819 = 8 + 1 + 9 = 18
Sum of the digits in 198 = 1 + 9 + 8 = 18
So these .two numbers (i.e.,) 819 and 198 are also divisible by 9.
(∵ 18 is divisible by 9)

Question 5.
Which numbers among 2, 3, 5, 6, 9 divides 12345 exactly ?
Write 12345 in reverse order and test now which numbers divide it exactly ?
Answer:
(i) 2 does not divide 12345 exactly.
3 divides 12345 exactly.
(∵ Sum of the digits 1 + 2 + 3 + 4 + 5 = 15isa multiple of ‘3’)
5 divides 12345 exactly.
(∵ 5 is in units place)
6 does not divide 12345 exactly
(∵ 12345 is divisible by 3 but not by 6)
9 does not divide 12345 exactly.
(∵ Sum of the digits (i.e.,) 15 is not divisible by 9)

(ii) 2,5, 6, 9 does not divide 54321 exactly.
(iii) 3 divides 54321 exactly.
∴ Sum of the digits is 5 + 4 + 3 + 2 + 1 = 15 is a multiple of 3.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.1

Question 6.
Write different 2 digit numbers using digits 3, 4 and 5. Check whether these numbers are divisible by 2, 3, 5, 6 and 9.
Answer:
The two digit numbers formed with 3, 4 and 5 are 34, 35, 45, 43, 53, 54.
34 is divisible by 2. .
35 is divisible by 5.
45 is divisible by 3, 5 and 9.
43 is not divisible by 2, 3, 5, 6 or 9.
53 is not divisible by 2, 3, 5, 6 or 9.
54 is divisible by 2, 3, 6 and 9.

Question 7.
Write the smallest digit and the great-est possible digit in the blank space of each of the following numbers so that the number formed are divisible by 3.
(i) …………………… 6724
(ii) 4765 ……………………. 2
(iii) 7221 …………………….. 5
Answer:
(i) 2 6724 and 56724
(ii) 476532 and 476592
(iii) 722145 and 722175

Question 8.
Find the smallest number that must be added to 123, so that it becomes exactly divisible by 5.
Answer:
The given number is 123.
The digit in ones place should be 5 so that it becomes exactly divisible by 5.
∴ 2 is to be added to 123.
Then the number becomes
123 + 2 = 125, divisible by 5.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.1

Question 9.
Find the smallest number that has to be subtracted from 256, so that it becomes exactly divisible by 10.
Answer:
The given number is 256.
The digit in units place should be ‘0’ so that it becomes exactly divisible by 10.
6 has to be subtracted from 256. Then the number becomes 256 – 6 = 250 which is exactly divisible by 10.

Leave a Comment