# NCERT Solutions For Class 6 Maths Chapter 3 Exercise 3.3

## Ncert Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.3:

**Ncert Solutions for Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.3:**In this Chapter playing with numbers We learn the basics of the Numbers System In class 6 maths ncert Exercise 3.3 Solutions. This Chapter Playing with Numbers is very important for Class 6 maths Students to get a High Score in their Exam. And we help the Class 6 Students to Achieve Their Dream By providing Class 6 Maths Ncert Solutions Chapter 3 Playing with Numbers Exercise 3.3 With Free Pdf Download. And We also Provide Video solutions Of Playing with Numbers Class 6 maths Ncert Solutions.

### Ncert Solutions for Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.3 pdf:-

**Exercise 3.3**Class 6 maths NCERT solutions Chapter 3 Playing with Numbers Numbers Pdf download:-

### Ncert Solution for Class 6 Maths Chapter 3 Playing Numbers Exercise 3.3 Tips:-

**Tests for Divisibility of Numbers**

Is the number 38 divisible by 2? by 4? by 5?

By actually dividing 38 by these numbers we find that it is divisible by 2 but

not by 4 and by 5.

Let us see whether we can find a pattern that can tell us whether a number is

divisible by 2, 3, 4, 5, 6, 8, 9, 10 or 11. Do you think such patterns can be easily

seen?

**Divisibility by 10:**Charu was looking at the multiples of

10. The multiples are 10, 20, 30, 40, 50, 60, … . She found

something common in these numbers. Can you tell what?

Each of these numbers has 0 in the one’s place.

She thought of some more numbers with 0 at one’s place

like 100, 1000, 3200, 7010. She also found that all such numbers are divisible

by 10.

She finds that if a number has 0 in the one’s place then it is divisible by 10.

Can you find out the divisibility rule for 100?

**Divisibility by 5:**Mani found some interesting pattern in the numbers 5, 10,

15, 20, 25, 30, 35, … Can you tell the pattern? Look at the units place. All these

numbers have either 0 or 5 in their one’s place. We know that these numbers are

divisible by 5.

Mani took up some more numbers that are divisible by 5, like 105, 215,

6205, 3500. Again these numbers have either 0 or 5 in their one’s places.

He tried to divide the numbers 23, 56, 97 by 5. Will he be able to do that?

Check it. He observes that a number which has either 0 or 5 in its ones

the place is divisible by 5, other numbers leave a remainder.

Is 1750125 divisible 5?

**Divisibility by 2:**Charu observes a few multiples of 2 to be 10, 12, 14, 16…

and also numbers like 2410, 4356, 1358, 2972, 5974. She finds some pattern

in one’s place of these numbers. Can you tell that? These numbers have only

the digits 0, 2, 4, 6, 8 in the one’s place.

She divides these numbers by 2 and gets remainder of 0.

She also finds that the numbers 2467, 4829 are not divisible by 2. These

numbers do not have 0, 2, 4, 6 or 8 in their one’s place.

Looking at these observations she concludes that a number is divisible

by 2 if it has any of the digits 0, 2, 4, 6 or 8 in its one’s place.

**Divisibility by 3:**Are the numbers 21, 27, 36, 54, 219 divisible by 3? Yes,

they are.

Are the numbers 25, 37, 260 divisible by 3? No.

Can you see any pattern in the one’s place? We cannot, because numbers

with the same digit in the one’s places can be divisible by 3, like 27, or may

not be divisible by 3 like 17, 37. Let us now try to add the digits of 21, 36, 54

and 219. Do you observe anything special? 2+1=3, 3+6=9, 5+4=9, 2+1+9=12.

All these additions are divisible by 3.

Add the digits in 25, 37, 260. We get 2+5=7, 3+7=10, 2+6+0 = 8.

These are not divisible by 3.

We say that if the sum of the digits is a multiple of 3, then the number

is divisible by 3.

Is 7221 divisible by 3?

**Divisibility by 6:**Can you identify a number which is divisible

by both 2 and 3? One such number is 18. Will 18 be divisible by

2×3=6? Yes, it is.

Find some more numbers like 18 and check if they are divisible

by 6 also.

Can you quickly think of a number which is divisible by 2 but

not by 3?

Now for a number divisible by 3 but not by 2, one example is

27. Is 27 divisible by 6? No. Try to find numbers like 27.

From these observations, we conclude that if a number is

divisible by 2 and 3 both then it is divisible by 6 also.

**Divisibility by 4:**Can you quickly give five 3-digit numbers divisible by

4? One such number is 212. Think of such 4-digit numbers. One example is

1936.

Observe the number formed by the ones and tens places of 212. It is 12;

which is divisible by 4. For 1936 it is 36, again divisible by 4.

Try the exercise with other such numbers, for example with 4612;

3516; 9532.

Is the number 286 divisible by 4? No. Is 86 divisible by 4? No.

So, we see that a number with 3 or more digits is divisible by 4 if the

the number formed by its last two digits (i.e. ones and tens) is divisible by 4.

Check this rule by taking ten more examples.

Divisibility for 1 or 2 digit numbers by 4 has to be checked by actual division.

**Divisibility by 8:**Are the numbers 1000, 2104, 1416 divisible by 8?

You can check that they are divisible by 8. Let us try to see the pattern.

Look at the digits at ones, tens and hundreds place of these numbers. These

are 000, 104 and 416 respectively. These two are divisible by 8. Find some more

numbers in which the number formed by the digits at units, tens and hundreds

place (i.e. last 3 digits) is divisible by 8. For example, 9216, 8216, 7216, 10216,

9995216 etc. You will find that the numbers themselves are divisible by 8.

We find that a number with 4 or more digits is divisible by 8 if the

the number formed by the last three digits is divisible by 8.

Is 73512 divisible by 8?

The divisibility for numbers with 1, 2 or 3 digits by 8 has to be checked by

actual division.

**Divisibility by 9:**The multiples of 9 are 9, 18, 27, 36, 45, 54,… There are

other numbers like 4608, 5283 that are also divisible by 9.

Do you find any pattern when the digits of these numbers are added?

1 + 8 = 9, 2 + 7 = 9, 3 + 6 = 9, 4 + 5 = 9

4 + 6 + 0 + 8 = 18, 5 + 2 + 8 + 3 = 18

All these sums are also divisible by 9.

Is the number 758 divisible by 9?

No. The sum of its digits 7 + 5 + 8 = 20 is also not divisible by 9.

These observations lead us to say that if the sum of the digits of a number

is divisible by 9, then the number itself is divisible by 9.