Telangana SCERT Class 6 Maths Chapter 3 Solution – Playing with Numbers. Here in this post we provides Class 6 Maths Playing with Numbers Telangana State Board Solution. Telangana State Board English Class VI Medium Students can download this Solution to Solve out Improve Your Learning Questions and Answers.
Telangana State Board Class 6 Maths Chapter 3 Playing with Numbers Solution:
EXERCISE – 3.1
1.) Which of the following numbers are divisible by 2, by 3 and by 6?
(i) 321729
ANSWER:
We know,
If the One’s digit of a number is 0,2,4,6, or 8, the number is divisible by 2.
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
If a number is divisible by both 2 and 3 then it is also divisible by 6.
Unit digit of number is not even, the number is not divisible by 2.
3+ 2 + 1 + 7 + 2 + 9 = 24 divisible by 3 then the number is divisible by 3.
321729 divisible by 3.
(ii) 197232
ANSWER:
We know,
If the One’s digit of a number is 0,2,4,6, or 8, the number is divisible by 2.
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
If a number is divisible by both 2 and 3 then it is also divisible by 6.
Unit digit of number is even, the number is divisible by 2.
1+ 9 + 7+ 2 + 3+ 2 = 24 divisible by 3 then the number is divisible by 3.
Number is divisible by both 2 and 3 then it is also divisible by 6.
197232 is divisible by 2, by 3 and by 6
(iii) 972132
ANSWER:
We know,
If the One’s digit of a number is 0,2,4,6, or 8, the number is divisible by 2.
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
If a number is divisible by both 2 and 3 then it is also divisible by 6.
Unit digit of number is even, the number is divisible by 2.
9 + 7 + 2 + 1 + 3 + 2 = 24 divisible by 3 then the number is divisible by 3.
Number is divisible by both 2 and 3 then it is also divisible by 6.
972132 is divisible by 2, by 3 and by 6
(iv) 1790184
ANSWER:
We know,
If the One’s digit of a number is 0,2,4,6, or 8, the number is divisible by 2.
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
If a number is divisible by both 2 and 3 then it is also divisible by 6.
Unit digit of number is even, the number is divisible by 2.
1 + 7 + 9 + 0 + 1 + 8 + 4 = 30 divisible by 3 then the number is divisible by 3.
Number is divisible by both 2 and 3 then it is also divisible by 6.
1790184 is divisible by 2, by 3 and by 6
(v) 312792
ANSWER:
We know,
If the One’s digit of a number is 0,2,4,6, or 8, the number is divisible by 2.
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
If a number is divisible by both 2 and 3 then it is also divisible by 6.
Unit digit of number is even, the number is divisible by 2.
3 + 1 + 2 + 7 + 9 + 2 = 24 divisible by 3 then the number is divisible by 3.
Number is divisible by both 2 and 3 then it is also divisible by 6.
312792 is divisible by 2, by 3 and by 6
(vi) 800552
ANSWER:
We know,
If the One’s digit of a number is 0,2,4,6, or 8, the number is divisible by 2.
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
If a number is divisible by both 2 and 3 then it is also divisible by 6.
Unit digit of number is even, the number is divisible by 2.
8 + 0 + 0 + 5 + 5 + 2 = 20 not divisible by 3, hence not divisible by 3.
800552 is divisible by 2
(vii) 4335
ANSWER:
We know,
If the One’s digit of a number is 0,2,4,6, or 8, the number is divisible by 2.
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
If a number is divisible by both 2 and 3 then it is also divisible by 6.
Unit digit of number is not even, the number is not divisible by 2.
4+ 3 + 3 + 5 = 15 divisible by 3 then the number is divisible by 3.
4335 is divisible by 3.
(viii) 726352
ANSWER:
We know,
If the One’s digit of a number is 0,2,4,6, or 8, the number is divisible by 2.
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
If a number is divisible by both 2 and 3 then it is also divisible by 6.
Unit digit of number is even, the number is divisible by 2.
7 + 2 + 6 + 3 + 5 + 2 = 25 is not divisible by 3
726352 is divisible by 2
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:
We know,
If the One’s digit of a number is 0 or 5, the number is divisible by 5.
If the One’s digit of a number is 0, the number is divisible by 10.
All numbers having One’s digit are 0 and 5 divisible by 5.
Now,
The One’s digit of a number is 0, the number is divisible by 10.
250, 1250, 45880 is divisible by 10.
Now,
10 = 2 x 5
The numbers that are divisible by 10 are also divisible by 2 and 5.
3). Fill the table using divisibility test for 3 and 9
Number | Sum of the digits in the number | Divisible by 3 | Divisible by 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 | Divisible by 9 |
72.
197 4689 79875. 988974 |
7+2.
1 + 9 + 7 4 + 6 + 8 + 9 7 + 9 + 8 + 7 + 5 9 + 8 + 8 + 9 + 7 + 4 = 45 |
Yes Yes
No No. Yes Yes Yes Yes. Yes yes |
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:
We have to make 3 different 3 digit numbers using 1, 9 and 8.
Numbers are 198, 891, and 918
We have to check which of these numbers are divisible by 9
We know,
If sum of digits of a number is divisible by 9 then the number is divisible by 9.
1 + 9 + 8 = 18
8 + 9 + 1 = 18
9 + 1 + 8 = 18
Divisible by 9
198, 891, and 918 all are Divisible by 9
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:
We know,
All the divisibility test rule.
In 12345, the One’s digit of a number is 5, the number is divisible by 5.
1 + 2 + 3 + 4 + 5 = 15 divisible by 3, the number is divisible by 3.
The number 12345 is divisible by 3 and 5.
Now, we reverse the number 12345.
Reverse number is 54321.
5 + 4 + 3 + 2 + 1 = 15 divisible by 3, the number is divisible by 3.
54321 is divisible by 3.
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:
We have to write 2 digit numbers using digits 3, 4 and 5.
34, 35, 43, 45, 53 and 54.
We know,
34 and 54 are even numbers hence divisible by 2.
45 and 54 having sum 9 which are divisible by 3 and 9.
35 and 45 have one’s digit 5, hence divisible by 5.
54 is even number also divisible by 3 hence it is divisible by 6.
7.) Write the smallest digit and the greatest possible digit in the blank space of each of the following numbers so that the numbers formed are divisible by 3.
- __ 6724
ANSWER:
We know,
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
? + 6 + 7 + 2 + 4 =? + 19
The smallest digit = 2
2 + 19 = 21 is divisible by 3
The greatest possible digit = 8
8 + 19 = 27 is divisible by 3
ii.) 4765__ 2
ANSWER:
We know,
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
4 + 7 + 6 + 5 +? + 2 = 24 +?
The smallest digit = 0
0 + 24 = 24 is divisible by 3
The greatest possible digit = 9
9 + 24 = 33 is divisible by 3
iii.) 7221__ 5
ANSWER:
We know,
If sum of digits of a number is divisible by 3 then the number is divisible by 3.
7 + 2 + 2 + 1 +? + 5 = 17 +?
The smallest digit = 1
1 + 17 = 18 is divisible by 3
The greatest possible digit = 7
7 + 17 = 24 is divisible by 3
8.) Find the smallest number that must be added to 123, so that it becomes exactly divisible by 5?
ANSWER:
We have to find the smallest number that must be added to 123, so that it becomes exactly divisible by 5
We know,
If the One’s digit of a number is 0 or 5, the number is divisible by 5.
123 +? = divisible by 5
The smallest number = 2
123 + 2 = 125 is divisible by 5.
9.) Find the smallest number that has to be subtracted from 256, so that it becomes exactly divisible by 10?
ANSWER:
We have to find the smallest number that has to be subtracted from 256, so that it becomes exactly divisible by 10
We know,
If the One’s digit of a number is 0, the number is divisible by 10.
256 – ? = divisible by 10.
The smallest number = 6
256 – 6 = 250 divisible by 10.
EXERCISE – 3.2
1.) Write all the factors of the following numbers.
- 36
ANSWER:
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
ii.) 23
ANSWER:
Factors of 23 = 1, 23
iii. 96
ANSWER:
Factors of 96 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
iv.) 115
ANSWER:
Factors of 115 = 1, 5, 23, 115
2.) Which of the following pairs are co-prime?
i.) 18 and 35
ANSWER:
We know,
The numbers which have only 1 as the common factor are called co-primes or relatively prime.
In 18 and 35 there is only 1 as the common factor.
18 and 35 pair of co-prime.
ii.) 216 and 215
ANSWER:
We know,
The numbers which have only 1 as the common factor are called co-primes or relatively prime.
216 and 215
In 216 and 215 there is only 1 as the common factor.
216 and 215 pair of co-prime.
iii.) 30 and 415
ANSWER:
We know,
The numbers which have only 1 as the common factor are called co-primes or relatively prime.
In 30 and 415, there are more than 1 common factor.
30 and 415 is not pair of co-prime.
iv.) 17 and 68
ANSWER:
We know,
The numbers which have only 1 as the common factor are called co-primes or relatively prime.
In 17 and 68, there are more than 1 common factor.
17 and 68 is not pair of co-prime.
3.) What is the greatest prime number between 1 and 20?
ANSWER:
We have to find the greatest prime number between 1 and 20.
The greatest prime number between 1 and 20 is 19.
4.) Find the prime and composite numbers between 10 and 30?
ANSWER:
We know,
Numbers whose only factors are 1 and the number itself are called prime numbers.
Numbers having more than two factors are called composite numbers.
Prime numbers between 10 and 30 = 11, 13, 17, 19, 23, 29
Composite numbers between 10 and 30 = 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28
5.) The numbers 17 and 71 are prime numbers. Both these numbers have same digits 1 and 7. Find 2 more such pairs of prime numbers below 100?
ANSWER:
The numbers 17 and 71 are prime numbers.
13-31, 79-97 are also prime numbers. Both these numbers pairs have same digits.
6.) Write three pairs of twin primes below 20?
ANSWER:
We know,
Twin primes are prime numbers that differ from each other by two.
Pairs of twin primes below 20 are (3, 5), (5, 7), (11, 13), (17, 19)
7.) Write two prime numbers whose product is 35?
ANSWER:
We have to write two prime numbers whose product is 35
Two prime numbers are 5 and 7.
5 x 7 = 35
8.) Express 36 as the sum of two odd primes?
ANSWER:
We have to express 36 as the sum of two odd primes.
36 = 13 + 23
Two odd prime numbers are 13 and 23.
9.) Write seven consecutive composite numbers less than 100.
ANSWER:
We have to write seven consecutive composite numbers less than 100.
Seven consecutive composite numbers less than 100 are 90, 91, 92,93,94,95 and 96.
10.) Express 53 as the sum of three primes?
ANSWER:
We have to express53 as the sum of three primes
The three prime numbers are 13 + 17 +23
13 + 17 +23 = 53
11.) Write two prime numbers whose difference is 10?
ANSWER:
We have to write two prime numbers whose difference is 10.
Two prime numbers whose difference is 10 are (3, 13) and (7, 17)
12.) Write three pairs of prime numbers less than 20 whose sum is divisible by 5?
ANSWER:
We have to write three pairs of prime numbers less than 20 whose sum is divisible by 5
Three pairs of prime numbers less than 20 whose sum is divisible by 5 are (2, 3); (3, 7); (7, 13).
EXERCISE – 3.3
1.) Write the missing numbers in the factor tree for 90?
ANSWER:
We have to find missing numbers in the factor tree for 90
(ii)
ANSWER:
2.) Factorise 84 by division method?
ANSWER:
We have to Factorise 84 by division method
84 = 2 × 2 × 3 × 7
3.) Write the greatest 4 digit number and express it in the form of its prime factors?
ANSWER:
We know,
Greatest 4 digit number = 9999
We have to express this in the form of its prime factors
Prime factors of 9999 are 3 × 3 × 11 × 101
4.) I am the smallest number, having four different prime factors. Can you find me?
ANSWER:
We know, smallest prime number is 2.
2 x 3 x 5 x 7 = 210
The smallest number is 210.
EXERCISE – 3.4
1.) Find the HCF of the following numbers by prime factorisation and continued division method?
i.) 18, 27, 36
ANSWER:
We know,
The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors.
We have to find the HCF of the following numbers by prime factorisation and continued division method.
HCF by prime factorisation:
HCF by prime factorisation of 18, 27, 36 = 3 x 3 = 9
HCF by continued division method
18) 27 (1
– 18
Remainder 9) 18 (2
– 18
——————————– 00
HCF of 18 and 27 is 9.
Now, we find HCF of the third number and the HCF of first two numbers.
9) 36 (4
– 36
——————————– 00
HCF of 18, 27 and 36 is 9.
ii.) 106, 159, 265
ANSWER:
We know,
The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors.
We have to find the HCF of the following numbers by prime factorisation and continued division method.
HCF by prime factorisation:
HCF of 106, 159, and 265 is 53.
HCF by continued division method
106) 159 (1
– 106
Remainder 53) 106 (2
– 106
——————————– 00
HCF of 106 and 159 is 53.
Now, we find HCF of the third number and the HCF of first two numbers.
53) 265 (5
– 265
——————————– 00
HCF of 106, 159 and 265 is 53.
iii.) 10, 35, 40
ANSWER:
We know,
The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors.
We have to find the HCF of the following numbers by prime factorisation and continued division method.
HCF by prime factorisation:
HCF of 10, 35, and 40 is 5.
HCF by continued division method
10) 35 (3
– 30
Remainder 5) 30 (6
– 30
——————————– 00
HCF of 10 and 35 is 5.
Now, we find HCF of the third number and the HCF of first two numbers.
5) 40 (8
– 40
——————————– 00
HCF of 10, 35, and 40 is 5.
2.) Find the largest number which is a factor of each of the numbers 504, 792 and 1080?
ANSWER:
We have to find largest number which is a factor of each of the numbers 504, 792 and 1080.
We have to find HCF of the numbers 504, 792 and 1080.
HCF of the numbers 504, 792 and 1080 = 2 x 2 x 2 x 9 = 8 x 9 = 72
3.) The length, breadth and height of a room are 12m, 15m and 18m respectively. Determine the length of longest stick that can measure all the dimensions of the room in exact number of times?
ANSWER:
Given, the length, breadth and height of a room are 12m, 15m and 18m respectively
We have to find, the length of longest stick that can measure all the dimensions of the room
We have to find HCF of 12m, 15m and 18m
HCF of 12m, 15m and 18m is 3 m.
The length of longest stick that can measure all the dimensions of the room is 3 m.
4.) HCF of co-prime numbers 4 and 15 was found as follows by factorisation: 4 = 2 x 2 and 15 = 3 x 5 Since there is no common prime factor, HCF of 4 and 15 is 0.Is the answer correct? If not, what is the correct HCF?
ANSWER:
We know,
When there is no prime factor common the HCF of number is 1.
4 = 2 x 2 and 15 = 3 x 5
The given answer is wrong. The correct HCF of 4 and 15 is 1.
5.) What is the capacity of the largest vessel which can empty the oil from three vesselscontaining 32 litres, 24 litres and 48 litres an exact number of times?
ANSWER:
Given,
Three vessels containing 32 litres, 24 litres and 48 litres.
We have to find capacity of the largest vessel which can empty the oil from three vessels.
We have to find HCF of 32 litres, 24 litres and 48 litres.
HCF of 32 litres, 24 litres and 48 litres = 2 x 2 x 2 = 8 litres.
EXERCISE – 3.5
1.) Find the LCM of the following numbers by prime factorisation method.
i.) 12 and 15
ANSWER:
We have to find LCM
We know,
The least common multiple of two or more given numbers is the lowest (or smallest or least) of their common multiples.
LCM of 12 and 15 = 3 x 4 x 5 = 60
ii.) 15 and 25
ANSWER:
We have to find LCM
We know,
The least common multiple of two or more given numbers is the lowest (or smallest or least) of their common multiples.
LCM of 15 and 25 = 3 x 5 x 5 = 75
iii.) 14 and 21
ANSWER:
We have to find LCM
We know,
The least common multiple of two or more given numbers is the lowest (or smallest or least) of their common multiples.
LCM of 14 and 21 = 42
iv.) 18 and 27
ANSWER:
We have to find LCM
We know,
The least common multiple of two or more given numbers is the lowest (or smallest or least) of their common multiples.
LCM of 18 and 27 = 2 x 3 x 3 x 3 = 54
v.) 48, 56 and 72
ANSWER:
We have to find LCM
We know,
The least common multiple of two or more given numbers is the lowest (or smallest or least) of their common multiples.
LCM of 48, 56 and 72 = 2 x 2 x 2 x 2 x 3 x 3 x 7
LCM of 48, 56 and 72 = 1008
vi.) 26, 14 and 91.
ANSWER:
We have to find LCM
We know,
The least common multiple of two or more given numbers is the lowest (or smallest or least) of their common multiples.
LCM of 26, 14 and 91 = 13 x 2 x 7 = 182
2.) Find the LCM of the following numbers by division method.
i) 84, 112, 196
ANSWER:
LCM of 84, 112, 196 = 4 x 7 x 3 x 4 x 5
LCM of 84, 112, 196 = 2352
ii.) 102, 119, 153
ANSWER:
LCM of 102, 119, 153 = 17 x 3 x 2 x 3 x 7
LCM of 102, 119, 153 = 2142
iii.) 45, 99, 132, 165
ANSWER:
LCM of 45, 99, 132, 165 = 5 x 11 x 3 x 3 x 4
LCM of 45, 99, 132, 165 =1980
3.) Find the smallest number which when added to 5 is exactly divisible by 12, 14 and 18.
ANSWER:
We have to find the smallest number which when added to 5 is exactly divisible by 12, 14 and 18.
We 1st find LCM of 12, 14 and 18.
LCM of 12, 14 and 18 = 2 x 3 x 2 x 7 x 3
LCM of 12, 14 and 18 = 252
The smallest number which when added to 5 is exactly divisible by 12, 14 and 18 = 252 – 5
The smallest number which when added to 5 is exactly divisible by 12, 14 and 18 = 247
4.) Find the greatest 3 digit number which when divided by 75, 45 and 60 leaves:
i) no remainder
ANSWER:
We have to find LCM of 75, 45 and 60.
LCM of 75, 45 and 60 = 15 x 5 x 3 x 4
LCM of 75, 45 and 60 = 900
ii.) The remainder 4 in each case.
ANSWER:
We have to find LCM of 75, 45 and 60.
LCM of 75, 45 and 60 = 15 x 5 x 3 x 4
LCM of 75, 45 and 60 = 900
The remainder 4 in each case
900 + 4 = 904
5.) Prasad and Raju met in the market on 1st of this month. Prasad goes to the market every 3rd day and Raju goes every 4th day. On what day of the month will they meet again?
ANSWER:
Given, Prasad goes to the market every 3rd day and Raju goes every 4th day.
We have to find on what day of the month they will meet again.
We find LCM of 3rd dayand4th day.
LCM of 3rd day and 4th day = 3 x 4 = 12
On 13th day of the month they will meet again
EXERCISE – 3.6
1.) Find the LCM and HCF of the following numbers?
i) 15, 24
ANSWER:
HCF of 15, 24 = Common factor = 3
LCM of 15, 24 = 3 x 5 x 8 = 120
ii.) 8, 25
ANSWER:
We know, when there is no common factor between numbers then HCF is 1and LCM is multiplication of given number.
HCF of 8, 25 = 1
LCM of 8, 25 = 25 x 8 = 120
iii.) 12, 48
Check their relationship.
ANSWER:
HCF of12, 48 = 4 x 3 = 12
LCM of 12, 48 = 4 x 3 x 4 = 48
2.) If the LCM of two numbers is 216 and their product is 7776, what will be the HCF?
ANSWER:
Given, LCM of two numbers is 216 and their product is 7776.
We have to find the HCF.
We know,
Product of LCM and HCF of the two numbers = Product of the two numbers.
HCF = Product of the two numbers / LCM
HCF = 7776 / 216
HCF = 36
3.) The product of two numbers is 3276. If their HCF is 6, find their LCM?
ANSWER:
Given, the product of two numbers is 3276. HCF is 6.
We have to find the LCM.
We know,
Product of LCM and HCF of the two numbers = Product of the two numbers.
LCM= Product of the two numbers / HCF
LCM = 3276 / 6
LCM = 546
4.) The HCF of two numbers is 6 and their LCM is 36. If one of the numbers is 12, find the other?
ANSWER:
Given,The HCF of two numbers is 6 and their LCM is 36.one of the numbers is 12.
We have to find other number.
Product of LCM and HCF of the two numbers = Product of the two numbers.
LCM x HCF = 1st number x 2nd number
2nd number = LCM x HCF / 1st number
2nd number = 36 x 6 / 12
2nd number = 18
EXERCISE – 3.7
1.) Which of the following numbers are divisible by 4?
i.) 572
ANSWER:
We know,
A number is divisible by 4, if the number formed by its last two digits (i.e. tens and ones) is divisible by 4.
572 = 72 is divisible by 4 then given number is divisible by 4.
ii.) 21,084
ANSWER:
We know,
A number is divisible by 4, if the number formed by its last two digits (i.e. tens and ones) is divisible by 4.
21,084 = 84 is divisible by 4 then given number is divisible by 4.
iii.) 14,560
ANSWER:
We know,
A number is divisible by 4, if the number formed by its last two digits (i.e. tens and ones) is divisible by 4.
14,560 = 60 is divisible by 4 then given number is divisible by 4.
iv.) 1,700
ANSWER:
We know,
When there is last 2 digit is 0 then the number is divisible by 4.
1,700is divisible by 4.
v.) 2150
ANSWER:
We know,
A number is divisible by 4, if the number formed by its last two digits (i.e. tens and ones) is divisible by 4.
2150 = 50 is not divisible by 4 then given number is not divisible by 4.
2.) Test whether the following numbers are divisible by 8?
i.) 9774
ANSWER:
We know,
A number with 4 or more digits is divisible by 8, if the number formed by its last three digits is divisible by 8.
9774 = last three digits are 774 which is not divisible by 8.
9774is not divisible by 8.
ii.) 5, 31,048
ANSWER:
We know,
A number with 4 or more digits is divisible by 8, if the number formed by its last three digits is divisible by 8.
5, 31,048 = last three digits are048which is divisible by 8.
5, 31,048is divisible by 8.
iii.) 5500
ANSWER:
We know,
A number with 4 or more digits is divisible by 8, if the number formed by its last three digits is divisible by 8.
5500= last three digits are 500 which is not divisible by 8.
5500 is not divisible by 8.
iv.) 6136
ANSWER:
We know,
A number with 4 or more digits is divisible by 8, if the number formed by its last three digits is divisible by 8.
6136 = last three digits are 136 which is divisible by 8.
6136 is divisible by 8.
v.) 4152
ANSWER:
We know,
A number with 4 or more digits is divisible by 8, if the number formed by its last three digits is divisible by 8.
4152 = last three digits are 152 which is divisible by 8.
4152 is divisible by 8.
3.) Check whether the following numbers are divisible by 11?
i.) 859484
ANSWER:
We know,
A given number is divisible by 11, if the difference between the sum of the digits at odd places and the sum of the digits at even places (from the right) is either 0 or divisible by 11.
859484
Sum of the digits at odd places = 4 + 4 + 5 = 13
Sum of the digits at even places = 8 + 9 + 8 = 25
Difference = 25 – 13 = 12
859484is not divisible by 11
ii.) 10824
ANSWER:
We know,
A given number is divisible by 11, if the difference between the sum of the digits at odd places and the sum of the digits at even places (from the right) is either 0 or divisible by 11.
10824
Sum of the digits at odd places = 4 + 8 + 1 = 13
Sum of the digits at even places = 2 + 0 = 2
Difference = 13 – 2 = 11
10824 is divisible by 11.
iii.) 20801
ANSWER:
We know,
A given number is divisible by 11, if the difference between the sum of the digits at odd places and the sum of the digits at even places (from the right) is either 0 or divisible by 11.
20801
Sum of the digits at odd places = 1 + 8 + 2 = 11
Sum of the digits at even places = 0 + 0 = 0
Difference = 11 – 0 = 11
20801is divisible by 11.
4.) Verify whether the following numbers are divisible by 4 and by 8?
i.) 2104
ANSWER:
We know,
A number is divisible by 4, if the number formed by its last two digits (i.e. tens and ones) is divisible by 4.
A number with 4 or more digits is divisible by 8, if the number formed by its last three digits is divisible by 8.
2104 = 04 divisible by 4
2104 = 104 divisible by 8
2104divisible by 4 and by 8
ii). 726352
ANSWER:
We know,
A number is divisible by 4, if the number formed by its last two digits (i.e. tens and ones) is divisible by 4.
A number with 4 or more digits is divisible by 8, if the number formed by its last three digits is divisible by 8.
726352= 52 divisible by 4
726352= 352 divisible by 8
726352divisible by 4 and by 8
iii.) 1800
ANSWER:
We know,
A number is divisible by 4, if the number formed by its last two digits (i.e. tens and ones) is divisible by 4.
A number with 4 or more digits is divisible by 8, if the number formed by its last three digits is divisible by 8.
1800 = 00 divisible by 4
1800= 800 divisible by 8
1800divisible by 4 and by 8
5.) Find the smallest number that must be added to 289279, so that it is divisible by 8?
ANSWER:
We have to find the smallest number that must be added to 289279, so that it is divisible by 8.
The smallest number be x.
x + 289279 = divisible by 8
We know,
A number with 4 or more digits is divisible by 8, if the number formed by its last three digits is divisible by 8.
289279 = 279 / 8 = remainder is 7
To get divisible by 8 we add 1 in remainder 7.
1 + 289279 = divisible by 8.
The smallest number is 1.
6.) Find the smallest number that can be subtracted from 1965, so that it becomes divisible by 4?
ANSWER:
We have to find thesmallest number that can be subtracted from 1965, so that it becomes divisible by 4.
We know,
A number is divisible by 4, if the number formed by its last two digits (i.e. tens and ones) is divisible by 4.
The smallest number be x.
1965 – x = divisible by 4
65 / 4 = remainder is 1.
To get divisible by 4 we subtract 1 in remainder 1.
1965 – 1 = divisible by 4
The smallest number is 1.
7.) Write all the possible numbers between 1000 and 1100 that are divisible by 11?
ANSWER:
We have to find all the possible numbers between 1000 and 1100 that are divisible by 11.
Numbers between 1000 and 1100 that are divisible by 11 are 1001, 1012, 1023, 1034, 1045, 1056, 1067, 1078, and 1089.
8.) Write the nearest number to 1240 which is divisible by 11?
ANSWER:
We have to find nearest number to 1240 which is divisible by 11
We know,
A given number is divisible by 11, if the difference between the sum of the digits at odd places and the sum of the digits at even places (from the right) is either 0 or divisible by 11.
1240
Sum of the digits at odd places = 0 + 2 = 2
Sum of the digits at even places = 4 + 1 = 5
Difference = 5 – 2= 3
But we require difference 0.
We add 3 in 1240
Nearest number is 1243.
9.) Write the nearest number to 105 which is divisible by 4?
ANSWER:
We have to find nearest number to 105 which is divisible by 4.
We know,
A number is divisible by 4, if the number formed by its last two digits (i.e. tens and ones) is divisible by 4.
105 = last 2 digits are 05
05/ 4 = remainder 1
We subtract 1 from 105 for nearest number to 105 which is divisible by 4.
Nearest number to 105 which is divisible by 4 = 104