# KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Ex 1.4

Students can Download Maths Chapter 1 Playing with Numbers Ex 1.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 1 Playing with Numbers Ex 1.4

Question 1.
Without actual division using divisibility rules, classify the following numbers as divisible by 3, 4, 5, 11
803, 875, 474, 583, 1067, 350, 657, 684, 2187, 4334, 1905, 2548.
Numbers divisible by 3 means sum of the digits should be divisible by 3 are 474, 657, 684, 2187, 1905
Numbers divisible by 4 are 684, 2548.
Numbers divisible by 5 are 875, 350, 1905
Numbers divisible by 11 are 803, 583, 1067, 4334

Question 2.
How many numbers from 1001 to 2000 are divisible by 4?
A number is divisible by 4 if the last two-digit number is divisible by 4. The numbers between 1001 and 1100 which are divisible by 4 are 1004, 1008, 1012 …. 1100 we get 25 numbers. Similarly from 1101 to 2000, we get numbers from 1201 to 1300 we get 25 numbers and so on. Hence from 1001 to 2000, we get 25 × 10 = 250 numbers which are divisible by 4.

Question 3.
Suppose a 3 digit number abc is divisible by 3. Prove that $$\overline{a b c}+\overline{b c a}+\overline{c a b}$$ is divisible by 9.
A number is divisible by 3
Sum of the digits is divisible by 3.
a + b + c is a multiple of 3.
Hence b + c + a and c + a + b are multiples of 3.
$$\overline{a b c}+\overline{b c a}+\overline{c a b}$$
= a + b + c + b + c + a + c + a + b
= 3a + 3b + 3c
= 3 (a +b + c)
a + b + c is divisible by 3.
Hence it is a multiple of 3.
∴ 3 (a + b + c) is a multiple of 9.
∴ $$\overline{a b c}+\overline{b c a}+\overline{c a b}$$ is divisible by 9.

Question 4.
If $$\overline{4 a 3 b}$$ is divisible by 11 find all possible value of a + b.
The number is divisible by 11 if and only if the difference between the sum of the digits in the odd places and the sum of the digits in the even places is divisible by 11.
(a + b) – (4 + 3) = 0
(a + b) – 7 = 0 or (a + b) – 7 = 11
a + b = 7 or a + b = 11 + 7 = 18
(a + b) – 7 cannot be equal to 22 because it a + b – 7 = 22 then a + b = 29
since a and b are digits their sum cannot be 29.

Question 5.
Prove that a 4 digit palindrome is always divisible by 11.
A 4 digit Palindrome is of the form $$\overline{\text { abba }}$$