Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 10 Number Relations Questions and Answers Notes Pdf to clear their doubts.
SCERT Class 6 Maths Chapter 10 Solutions Number Relations
Class 6 Kerala Syllabus Maths Solutions Chapter 10 Number Relations Questions and Answers
Number Relations Class 6 Questions and Answers Kerala Syllabus
Digit Sum (Page Number 154-155)
Question 1.
Complete the table below:
Answer:
Question 2.
The first three digits of a four-digit number, which is a multiple of 9, are 2, 5, 7. What is the number?
Answer:
Here, the first three digits are 2, 5, 7.
The sum is 14.
So the next number should be 4, only after that the sum will be a multiple of 9.
2 + 5 + 7 + 4 = 18
The number is 2574.
Question 3.
(i) Any number, and the number got by changing the order of its digits in any way, both leave the same remainder on division by 9. Why?
(ii) If we change the order of the digits of a multiple of 9 in any way, will we again get a multiple of 9? Why?
Answer:
(i) Because the order is only changing, the sum of the digits are same.
For example, the sum of the digits of the number 3247 is 16.
The remainder when dividing by 9 is 7.
Even if the order has changed, the sum will always be 16.
(ii) Even if the order is changing, the sum is not changing, so it will always be a multiple of 9.
Question 4.
Suppose we take all five-digit numbers made using the digits 1, 2, 3, 4, 5 without repetition.
(i) Which of them are multiples of 3?
(ii) Which of them are multiples of 9?
(iii) What is the remainder on dividing any of these numbers by 9?
Answer:
(i) If we make any numbers with the digits 1, 2, 3, 4, 5 without repetition, the sum will always be a multiple of 3, because the sum of the digits of the number is 1 + 2 + 3 + 4 + 5 = 15, which is a multiple of 3.
(ii) As the sum is 15, none of them will be a multiple of 9.
(iii) When 15 is divided by 9, the remainder is 6.
The sum of the digits of the numbers that are made without repetition will give 6 as a remainder.
Question 5.
Among all the ten-digit numbers using all of the ten digits 0, 1, 2,…, 9, how many are primes? What is the reason?
Answer:
When we add these ten digits, the sum is
0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.
Any ten-digit number formed using all these digits will have a digit sum of 45.
Since 45 is a multiple of both 3 and 9, every such number is divisible by 3 and 9.
Therefore, none of these numbers can be prime.
Question 6.
Which is the least four-digit number without any digit repeating, which is a multiple of 9? What about the least five-digit number?
Answer:
The least four-digit number without any digit repeating is 1023. But it is not a multiple of 9.
If the number should be a multiple of 9, the sum of the digits of that number should also be a multiple of 9.
So the least four-digit number without any digit repeating, which is a multiple of 9 is 1026. And the least five-digit number is 10269.
Playing with Nine (Page Number 157-158)
Question 1.
The difference of any number and the number got by rearranging its digits in any manner is a multiple of 9. Why?
Answer:
When we rearrange the digits of a number, the sum of the digits does not change.
Any number can be written as the sum of its digits plus a multiple of 9.
Therefore, two numbers that have the same digit sum differ only by a multiple of 9.
Hence, the difference between a number and any rearrangement of its digits is always a multiple of 9.
Question 2.
The difference of any three-digit number and the number written in reverse is a multiple of 99. Why? What is the relation between the number by which 99 is multiplied and the digits of the original number?
Answer:
Let’s understand this with an example:
Consider the number 563 and its reverse, 365.
563 – 365 = 198
Write both numbers in expanded form:
563 = 500 + 60 + 3 = (9 × 55 + 5) + (9 × 6 + 6) + 3
365 = 300 + 60 + 5 = (9 × 33 + 3) + (9 × 6 + 6) + 5
Now rewrite more clearly:
563 = 9(55 + 6) + 14
365 = 9(33 + 6) + 14
When we subtract:
Only part 9(55 – 33) = 9 × 22 remains, because all other components are the same and cancel out.
Thus, the difference is always:
9 × 22 = 198 = 99 × 2
Since 22 = 2 × 11, the difference is always a multiple of: 9 × 11 = 99
For any three-digit number, the difference between the number and its reverse is always a multiple of 99.
Take the difference between the first digit and the last digit of the original number.
Original number: 893
Reversed number: 398
(8 – 3) × 99 = 5 × 99 = 495
893 – 398 = 495
Question 3.
What is the relation between the digits of the number got by multiplying 99 by a single-digit number?
Answer:
From the number getting by this, adding the first and last digit, we get the middle number.
Question 4.
Take any three-digit number, write it in reverse, and find the difference of these two numbers. Write the number got as difference in reverse and add to the difference. What number do you get? Do this with the other three-digit numbers. Do you get 1089 always? For what type of numbers do you get a different answer?
Answer:
638 – 836 = 198
198 + 891 = 1089
692 + 296 = 396
396 + 693 = 1089
603 – 306 = 297
297 + 792 = 1089
If the difference between the digits in the ones and hundreds place is 1, those numbers won’t get these answers.
392 – 293 = 99
99 + 99 = 198
544 – 445 = 99
99 + 99 = 198
Class 6 Maths Chapter 10 Kerala Syllabus Number Relations Questions and Answers
Class 6 Maths Number Relations Questions and Answers
Question 1.
Write the following numbers as the multiple of 9 and the sum of the digits.
(a) 527
(b) 4325
(c) 38
(d) 6428
Answer:
(a) 527 = 500 + 20 + 7
= (9 × 55 + 5) + (9 × 2 + 2) + 7
= (9 × 57) + 14
(b) 4325 = 4000 + 300 + 20 + 5
= (9 × 444 + 4) + (9 × 33 + 3) + (9 × 2 + 2) + 5
= (9 × 479) + 14
(c) 38 = 30 + 8
= (9 × 3 + 3) + 8
= (9 × 3) + 11
(d) 6428 = 6000 + 400 + 20 + 8
= (9 × 666 + 6) + (9 × 44 + 4) + (9 × 2 + 2) + 8
= (9 × 712) + 20
Question 2.
From the numbers given below, which of them are multiples of 9?
(a) 329
(b) 1484
(c) 3299
(d) 5238
(e) 3483
Answer:
5238 and 3483
(the number those which the sum of their digits are multiples of 9)
Question 3.
From the numbers given below, which of them are multiples of 3?
(a) 486
(b) 394
(c) 6288
(d) 5430
(e) 3841
Answer:
486, 6288, and 5430
(the number those which the sum of their digits are multiples of 3)
Question 4.
From the numbers given below, which of them are multiples of both 3 and 9?
(a) 492
(b) 648
(c) 5264
(d) 3897
(e) 5486
Answer:
648 and 3897
(multiples of 9 and also multiples of 3)
Question 5.
Find the remainder of the following numbers when divided by 9?
(a) 4238
(b) 5269
(c) 4845
(d) 9890
(e) 4241
Answer:
(a) 8
(b) 4
(c) 3
(d) 8
(e) 2
Question 6.
Find the remainder of the following numbers when divided by 3?
(a) 394
(b) 1574
(c) 3248
(d) 1375
(e) 2489
Answer:
(a) 1
(b) 2
(c) 2
(d) 1
(e) 2
Question 7.
Suppose we take all four-digit numbers made using the digits 5, 9, 4, 3 without repetition, which of them are multiples of 9? Are there any multiples of 3?
Answer:
Since the sum of the digits of the numbers is not a multiple of 9, there will be no numbers with multiples of 9.
Since the sum of the digits of the numbers are the multiple of 3, all numbers formed will be multiples of 3.
Question 8.
3284_ is a five-digit number. If this is a multiple of 9, what digit will be in the place of ones?
Answer:
The digit in the ones place is 1.
(3 + 2 + 8 + 4 + 1 = 18)
Question 9.
From the following pairs of numbers, which of them have a difference that is a multiple of 9?
(a) 327,426
(b) 1624,1382
(c) 5248,3268
Answer:
327, 426, and 5248, 3268
Class 6 Maths Chapter 10 Notes Kerala Syllabus Number Relations
→ Any natural number is equal to the sum of a multiple of nine and the sum of the digits of the number.
→ The remainder on dividing any number by 9 is the remainder on dividing the sum of the digits of the number by 9.
→ If the sum of the digits of a number is a multiple of 9, then the number itself is a multiple of 9.
→ The remainder on dividing any number by 3 is the remainder on dividing the sum of the digits of the number by 3.
→ If the sum of the digits of a number is a multiple of 3, then the number itself is a multiple of 3.
In this chapter, we discuss the special properties of the factors of 9 and 3. From this, we learn how to determine the remainders when a number is divided by 9 or 3.
Digit Sum
If we take any two-digit number, it will be a multiple of 10, or the sum of a digit less than ten is added to the multiple of 10.
30 = 3 × 10
35 = 3 × 10 + 5
In detail, we can say that a two-digit number is the sum of multiples of the first digit, with 10 added to the second digit.
74 = 7 × 10 + 4
Let us now check whether the two-digit number has any relation with multiples of 9.
If we take any two-digit number, it is the sum of the first digit, which is multiplied by 9, and the sum of those two digits.
14 = 1 × 9 + 5
23 = 2 × 9 + 5
44 = 4 × 9 + 8
65 = 6 × 9 + 11
Try to write the following numbers as multiples of nine and the sum of the digits of the number.
(1) 16
(2) 24
(3) 35
(4) 63
(5) 87
Answer:
(1) 16 = 1 × 9 + 7
(2) 24 = 2 × 9 + 6
(3) 35 = 3 × 9 + 8
(4) 63 = 6 × 9 + 9
(5) 87 = 8 × 9 + 15
Let’s now check for three-digit numbers.
100 = 9 × 11 + 1
200 = 9 × 22 + 2
800 = 9 × 88 + 8
Likewise, we can write this for any multiple of 100
250 = 9 × 27 + 7
⇒ (9 × 22 + 2) + (9 × 5) + 5 = 9 × 27 + 7
625 = 9 × 68 + 13
⇒ (9 × 66 + 6) + (9 × 2 + 2) + 5 = 9 × 68 + 13
So any three-digit number is the sum of any multiple of nine, with the sum of the digits.
How can we write 697 in this manner?
Here, the sum of the digits is 22 (6 + 9 + 7), so we can subtract 22 from 697, We get 675.
When we write 675 as a multiple of 9
9 × 75 = 675
So, 697 = 9 × 75 + 22
Now try to write 726 as the sum of any multiple of nine with the sum of its digits.
726 = 700 + 20 + 6
= (9 × 77 + 7) + (9 × 2 + 2) + 6
= (9 × 79) + 15
Can we write 549 like this?
500 + 40 + 9 = (9 × 55 + 5) + (9 × 4 + 4) + 9
= (9 × 59) + 5 + 4 + 9
= 9 × 59 + 18
549 is written as the sum of the multiples of 9 and the sum of the digits, which is 18.
Now write the following numbers as the sum of the multiples of 9 and the sum of the digits.
(1) 464
(2) 387
(3) 825
(4) 949
Answer:
(1) 464 = 400 + 60 + 4
= (9 × 44 + 4) + (9 × 6 + 6) + 4
= (9 × 50) + 4 + 6 + 4
= (9 × 50) + 14
(2) 387 = 300 + 80 + 7
= (9 × 33 + 3) + (9 × 8 + 8) + 7
= (9 × 41) + 3 + 8 + 7
= (9 × 41) + 18
(3) 825 = 800 + 20 + 5
= (9 × 88 + 41) + (9 × 2 + 2) + 5
= (9 × 90) + 8 + 2 + 5
= (9 × 90) + 15
(4) 949 = 900 + 40 + 9
= (9 × 99 + 9) + (9 × 4 + 4) + 9
= (9 × 103) + 9 + 4 + 9
= 9 × 103 + 22
Even if it is two digits, or for any number, this case is true.
4379 = 4000 + 300 + 70 + 9
= (9 × 444 + 4) + (9 × 33 + 3) + (9 × 7 + 7) + 9
= 9 × (444 + 33 + 7) + 4 + 3 + 7 + 9
= (9 × 484) + 23
Now let’s write 8246 like this.
8246 = 8000 + 200 + 40 + 6
= (9 × 888 + 8) + (9 × 22 + 2) + (9 × 4 + 4) + 6
= 9 × (888 + 22 + 4) + 8 + 2 + 4 + 6
= (9 × 914) + 20
In general, we can say that any natural number is equal to the sum of a multiple of nine and the sum of the digits of the number.
Let’s now check the peculiarity of the remainder we got after dividing any of these numbers by 9.
76 ÷ 9
76 = 9 × 7 + 13
That is 76 is a number got by adding 13 to a multiple of 9.
But we didn’t get 13 as a remainder when dividing by 9.
When we again divide 13 by 9, we get 1 and a reminder of 4.
Which means this 4 is the remainder we got by dividing the sum of the digits by 9.
Then what about dividing 365 by 9
Sum of the digits: 3 + 6 + 5 = 14
14 ÷ 9 = 1
Remainder = 5
So we get 5 as the remainder after dividing 365 by 9.
What about the remainder we got after dividing 4257 by 9.
4 + 2 + 5 + 7 = 18
18 ÷ 9
Remainder = 0
We get 0 as the remainder after dividing 4257 by 9.
From this, we understood that:
The remainder on dividing any number by 9 is the remainder on dividing the sum of the digits of the number by 9.
If the sum of the digits of a number is a multiple of 9, then the number itself is a multiple of 9.
From the numbers given below, find which of them are multiples of 9? Also, find the remainder of the numbers that are not multiples of 9.
(1) 49
(2) 378
(3) 538
(4) 1489
(5) 12489
(6) 3690
(7) 42345
(8) 52374
Answer:
(1) 49
Sum of the digits = 4 + 9 = 13
⇒ not a multiple of 9
13 ÷ 9
⇒ Remainder = 4
(2) 378
Sum of the digits = 3 + 7 + 8 = 18
⇒ multiple of 9
(3) 538
Sum of the digits = 5 + 3 + 8 = 16
⇒ not a multiple of 9
16 ÷ 9
⇒ Remainder = 7
(4) 1489
Sum of the digits = 1 + 4 + 8 + 9 = 22
⇒ not a multiple of 9
22 ÷ 9
⇒ Remainder = 6
(5) 12489
Sum of the digits = 1 + 2 + 4 + 8 + 9 = 24
⇒ not a multiple of 9
24 ÷ 9
⇒ Remainder = 6
(6) 3690
Sum of the digits = 3 + 6 + 9 + 0 = 18
⇒ multiple of 9
(7) 42345
Sum of the digits = 4 + 2 + 3 + 4 + 5 = 18
⇒ multiple of 9
(8) 52374
Sum of the digits = 5 + 2 + 3 + 7 + 4 = 21
⇒ not a multiple of 9
21 ÷ 9
⇒ Remainder = 3
Remainder on Diving by 3
Same as in the case of 9, we can find the remainder on dividing by 3 simply by adding the digits in that number.
The remainder on dividing any number by 3 is the remainder on dividing the sum of the digits of the number by 3.
In other words, we can say that if the sum of the digits of a number is a multiple of 3, then the number itself is a multiple of 3.
Check whether 1427 is a multiple of 3?
The sum of digits in 1427 is 1 + 4 + 2 + 7 = 14
Here, 14 is not a multiple of 3.
So, as the remainder we got by dividing 14 by 3 is 2, the remainder we get after dividing 1427 will also be 2.
Check whether the numbers given below is the multiple of 3, else find the remainder when dividing the number by 3?
(a) 527
(b) 648
(c) 1397
(d) 12486
Answer:
(a) 527
5 + 2 + 7 = 14
⇒ not a multiple of 3
⇒ remainder = 2
(b) 648
6 + 4 + 8 = 18
⇒ multiple of 3
(c) 1397
1 + 3 + 9 + 7 = 20
⇒ not a multiple of 3
⇒ remainder = 2
(d) 12486
1 + 2 + 4 + 8 + 6 = 21
⇒ multiple of 3
From the numbers given below, which of them are multiples of 3 and 9?
(a) 999
(b) 1647
(c) 3984
(d) 4788
(e) 5642
(f) 3864
(g) 24329
(h) 49995
Answer:
(a) 9 + 9 + 9 = 27
⇒ multiple of 3 and 9
(b) 1 + 6 + 4 + 7 = 18
⇒ multiple of 3 and 9
(c) 3 + 9 + 8 + 4 = 24
⇒ multiple of 3 but not a multiple of 9
Remainder on dividing by 9 = 6
(d) 4 + 7 + 8 + 8 = 27
⇒ multiple of 3 and 9
(e) 5 + 6 + 4 + 2 = 17
⇒ not a multiple of 3 and 9
Remainder on dividing by 3 = 2
Remainder on dividing by 9 = 8
(f) 3 + 8 + 6 + 4 = 21
⇒ multiple of 3 and not a multiple of 9
Remainder on dividing by 9 = 3
(g) 2 + 4 + 3 + 2 + 9 = 20
⇒ not a multiple of 3 and 9
Remainder on dividing by 3 = 2
Remainder on dividing by 9 = 2
(h) 4 + 9 + 9 + 9 + 5 = 36
⇒ multiple of 3 and 9
Playing with Nine
Any number is a multiple of 9 added to the sum of the digits of the number.
If we subtract from a number the sum of its digits, we get a multiple of 9.
If we subtract 19 from 487
487 – 19 = 468, which is a multiple of 9.
Also, if the sum of the digits of two numbers are same, then their difference will be a multiple of 9.
Example: The sum of the digits of the numbers 4237 and 3544 is 16.
The difference between these numbers
4237 – 3544 = 693
Since the sum of the digits of the number 693 is 18, 693 is a multiple of 9.
Which of the following differences of number pairs are multiples of 9?
(1) 3487, 4583
(2) 5684, 9284
(3) 6247, 3584
(4) 7254, 5814
Answer:
The difference in the sum of the digits of the 1st and 3rd pair is different.
So their difference is also not a multiple of 9.
The difference of the sum of the digits of the 2nd and 4th pair is the same.
So their difference is also a multiple of 9.