 # Important Questions Class 8 Maths Chapter 16 Playing Numbers

Playing with numbers is one of the crucial topics of Maths. It is an important topic as it builds the base for board exams for class 10(CBSE). Apart from studying and practicing problems on numbers in general form from NCERT, students shall also practice these important questions.

Solving these important questions of class 8 maths chapter 16 will help you prepare for CBSE board exams too.

Question 1: Express the given number in generalised form: Given Number – 3458

Ans: 3458 will be written in a generalized way in the following manner,

3458 = 3 x 10 3 + 4 x 102 + 5 x 101 + 8 x 100

Question 2: Express the given number in a normal form – (2 x 1000) + (2 x 10)

Ans: (2 x 1000) + (2 x 10) will be written in a generalised form in the following way:

(2 x 1000) + (2 x 10) = 2000 + 20 = 2020

Question 3: If the following number is divided by 13, find the quotient and remainder.

Solution : dividend = pq + r

1220 = 13 × 93 + 11

Quotient = 93

Remainder = 11

Also check: Whole Numbers , Natural Numbers, Large Numbers.

Question 4: What is the least natural number which is larger than 100 and which leaves the remainder, R = 12 when it is divided by 19.

Ans: 100 = 19 × 5 + 5

As 5 + 7 = 12, and the required number is 100 + 7 = 107

Question 5: What is the smallest number you have to add to 100000 to get a multiple of 1234?

Ans: 100000 = 1234 × 81 + 46 = 1234 – 46 = 1188, so the required number is 1188.

Question 6: How many numbers from 1001 to 2000 are divisible by 4?

Ans: 1004, 1008, 1012, 1016… are the numbers between 1001 and 2000 which are divisible by 4.

Therefore, there will be (2000-1004) 4 + 1 = 250 numbers between 1001 and 2000, and these will be divisible by 4.

Question 7: Show that if a palindrome is 4 digits, it would be divisible by 11?

Ans: Let us assume abba is the four digit palindrome.

So, abba = a x 1000 + cb x 100 + b x 10 + a

= 1001a +110b

= 91 x 11a + 110b

= 11 x (91a + 10b)

Hence Proved.

Question 8: Make a 5 digit number using each of the digits 4,5,6,7,8. Also see that the number made shall be divisible by 132.

Ans: 132 can be written as 11 x 12 = 3 x 4 x 11

As 4 + 5+ 6+ 7+ 8 = 30, the numbers formed by the 4,5,6,7,8 are all divisible by 3.

Let’s check the disability by 11 and 4.

If the number is divisible by 4, if the last 2 digits of the number formed is divisible by 4.

Similarly, if the difference between the sum of digits placed at even and the sum of digits placed at odd places is divisible by 11, then the number is divisible by 11.

Question 9. By which number 345111 is divisible amongst the given options – 15, 12, 3, 9.

Ans: Add all the digits in the number given – 345111

345111 + 3+4+5+1+1+1 = 15, which is divisible by 3. So, 3 is the correct option.

Question 10: Can you tell a 5 digit number which will be divisible by 11 and it should have digits 2,3,4,5,6?

Ans:

So now, 2 – 4 + 3 – 6 + 5 = 0, so 24365 is the least 5 digit number which is also divisible by 11.

Consider the number 24365 formed by the digits 2, 3, 4, 5 and 6

Here, 2 − 4 + 3 − 6 + 5 = 0

Thus, 24365 is the smallest number divisible by 11.