Prime and Composite Numbers | Math | Chapter 3 | Grade 5 | With Exercises Solution
prime and composite numbers chapter 3 grade 5 exercise solution
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Math Sanjal
Definition of Prime Numbers (What do you mean by Prime numbers?)/button
Those numbers which have exactly two factors i.e. 1 and the number itself are called prime numbers. The numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 91, 97, etc are prime numbers.
Question: How many prime numbers are there from 1 to 100?
Answer: There are 26 prime numbers within 1 and 100/from 1 to 100.
For example:
2 = 2 x 1
3 = 3 x 1
5 = 5 x 1
7 = 7 x 1
11 = 11 x 1
13 = 13 x 1
17 = 17 x 1
19 = 19 x 1
23 = 23 x 1
29 = 29 x 1
31 = 31 x 1
37 = 37 x 1
41 = 41 x 1
43= 43 x 1
47 = 47 x 1
...
Also Check:
Easy ways for Testing Divisibility | Divisibility Rules.
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From the above example, you have seen that 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 have only two factors e.i. 1 and the number itself. So, they all are prime numbers.
Factors and Multiples:alert-info
Factors (What are factors?)/button
⥱When two or more numbers are multiplied to give a product, each number is called the factor of the product.
For example:
Let us take a number 10. 10 is divisible by 1, 2 and 5 i.e. 10 = 1 x 2 x 5. So, 1, 2 and 5 are the factors of product 10.
Multiple (What are multiples?)/button
⥱The set of numbers which are divisible by a given number are the multiples of that number.
For example:
Let us take a number 3.
Here,
3 x 1 = 3
3 x 2 = 6
3 x 3 = 9
3 x 4 = 12
3 x 5 = 15
Now, all the deriving numbers 3, 6, 9, 12 and 15 are found by multiplying the given number 3 by other numbers. So, the numbers 3, 6, 9, 12 and 15 are the multiples of 3.
Definition of Composite Numbers (What do you mean by Composite numbers?)/button
Those numbers which have more than two factors are called prime numbers. The numbers 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 92, 93, 94, 95, 96, 98, 99, 100, etc are composite numbers.
Question: How many composite numbers are there from 1 to 100?
Answer: There are 73 composite numbers within 1 and 100/from 1 to 100.
Answer: There are 73 composite numbers within 1 and 100/from 1 to 100.
For example:
4 = 2 x 2 x 1
6 = 2 x3 x 1
8 = 2 x 2 x 2 x 1
9 = 3 x 3 x 1
10 = 2 x 5 x 1
12 = 2 x 2 x 3 x 1
14 = 2 x 7 x 1
15 = 2 x 3 x 5 x 1
16 = 2 x 2 x 2 x 2 x 1
18 = 2 x 3 x 3 x 1
20 = 2 x 2 x 5 x 1
21 = 3 x 7 x 1
22 = 2 x 11 x 1
24= 2 x 2 x 2 x 3 x 1
25 = 5 x 5 x 1
...
From the above example, you have seen that 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25 have more than two factors. So, they all are composite numbers.
4 = 2 x 2 x 1
6 = 2 x3 x 1
8 = 2 x 2 x 2 x 1
9 = 3 x 3 x 1
10 = 2 x 5 x 1
12 = 2 x 2 x 3 x 1
14 = 2 x 7 x 1
15 = 2 x 3 x 5 x 1
16 = 2 x 2 x 2 x 2 x 1
18 = 2 x 3 x 3 x 1
20 = 2 x 2 x 5 x 1
21 = 3 x 7 x 1
22 = 2 x 11 x 1
24= 2 x 2 x 2 x 3 x 1
25 = 5 x 5 x 1
...
From the above example, you have seen that 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25 have more than two factors. So, they all are composite numbers.
Grade 5: Math Subject - Exercise 3.1 (With Solution)/button
1. Determine the factors of the following numbers.
a. 6
Solution:
1 x 6 = 6
2 x 3 = 6
3 x 2 = 6
6 x 1 = 6
Therefore, 1, 2, 3, and 6 are the factors of number 6.
3 x 2 = 6
6 x 1 = 6
Therefore, 1, 2, 3, and 6 are the factors of number 6.
b. 10
Solution:
1 x 10 = 10
2 x 5 = 10
5 x 2 = 10
10 x 1 = 10
Therefore, 1, 2, 5, and 10 are the factors of number 10.
5 x 2 = 10
10 x 1 = 10
Therefore, 1, 2, 5, and 10 are the factors of number 10.
c. 12
Solution:
1 x 12 = 12
2 x 6 = 12
3 x 4 = 12
4 x 3 = 12
6 x 2 = 12
12 x 1 = 12
Therefore, 1, 2, 3, 4, 6 and 12 are the factors of number 12.
d. 24
Solution:
1 x 24 = 24
2 x 12 = 24
3 x 8 = 24
4 x 6 = 24
6 x 4= 24
8 x 3 = 24
12 x 2 = 24
24 x 1 = 24
Therefore, 1, 2, 3, 4, 6, 8, 12 and 24 are the factors of number 24.
e. 31
Solution:
1 x 31 = 31
31 x 1 = 31
Therefore, 1 and 31 are the factors of number 31.
f. 36
Solution:
1 x 36 = 36
2 x 18 = 36
3 x 12 = 36
4 x 9 = 36
6 x 6= 36
9 x 4 = 36
12 x 3 = 36
18 x 2 = 36
36 x 1 = 36
Therefore, 1, 2, 3, 4, 6, 9, 12, 18 and 36 are the factors of number 36.
g. 42
Solution:
1 x 42 = 42
2 x 21 = 42
3 x 14 = 42
6 x 7 = 42
7 x 6= 42
14 x 3 = 42
21 x 2 = 42
42 x 1 = 42
Therefore, 1, 2, 3, 6, 7, 14, 21 and 42 are the factors of number 42.
h. 49
Solution:
1 x 49 = 49
7 x 7 = 49
49 x 1 = 49
Therefore, 1, 7 and 49 are the factors of number 49.
i. 57
Solution:
1 x 57 = 57
3 x 19 = 57
19 x 3 = 57
57 x 1 = 57
Therefore, 1, 3, 19 and 57 are the factors of number 57.
j. 69
Solution:
1 x 69 = 69
3 x 23 = 69
23 x 3 = 69
69 x 1 = 69
Therefore, 1, 3, 23 and 69 are the factors of number 69.
k. 81
Solution:
1 x 81 = 81
3 x 27 = 81
9 x 9 = 81
27 x 3 = 81
81 x 1 = 81
Therefore, 1, 3, 9, 27 and 81 are the factors of number 81.
l. 87
Solution:
1 x 87 = 87
3 x 29 = 87
29 x 3 = 87
87 x 1 = 87
Therefore, 1, 3, 29, and 87 are the factors of number 87.
m. 100
Solution:
1 x 100 = 100
2 x 50 = 81
4 x 25 = 81
5 x 20 = 81
10 x 10 = 81
20 x 5 = 100
25 x 20 = 100
50 x 2 = 100
100 x 1 = 100
Therefore, 1, 2, 4, 5, 10, 20, 25, 50 and 100 are the factors of number 100.
n. 121
Solution:
1 x 121 = 121
11 x 11 = 121
121 x 1 = 121
Therefore, 1, 11 and 121 are the factors of number 121.
o. 144
Solution:
1 x 144 = 144
2 x 72 = 144
3 x 48 = 144
4 x 36 = 144
6 x 24 = 144
8 x 18 = 144
9 x 16 = 144
12 x 12 = 144
16 x 9 = 144
18 x 8 = 144
24 x 6 = 144
36 x 4 = 144
48 x 3 = 144
72 x 2 = 144
144 x 1 = 144
Therefore, 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72 and 144 are the factors of number 144.
2. Find the first 10 multiples of the given numbers.
a. 3
Solution:
Given number = 3.
1st multiple of 3 = 3 x 1 = 3
2nd multiple of 3 = 3 x 2 = 6
3rd multiple of 3 = 3 x 3 = 9
4th multiple of 3 = 3 x 4 = 12
5th multiple of 3 = 3 x 5 = 15
6th multiple of 3 = 3 x 6 = 18
7th multiple of 3 = 3 x 7 = 21
8th multiple of 3 = 3 x 8 = 24
9th multiple of 3 = 3 x 9 = 27
10th multiple of 3 = 3 x 10 = 30.
Thus, 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30 are the first 10 multiples of 3.
b. 5
Solution:
Given number = 5.
1st multiple of 5 = 5 x 1 = 5
2nd multiple of 5 = 5 x 2 = 10
3rd multiple of 5 = 5 x 3 = 15
4th multiple of 5 = 5 x 4 = 20
5th multiple of 5 = 5 x 5 = 25
6th multiple of 5 = 5 x 6 = 30
7th multiple of 5 = 5 x 7 = 35
8th multiple of 5 = 5 x 8 = 40
9th multiple of 5 = 5 x 9 = 45
10th multiple of 5 = 5 x 10 = 50.
Thus, 5, 10, 15, 20, 25, 30, 35, 40, 45 and 50 are the first 10 multiples of 5.
c. 7
Solution:
Given number = 7.
1st multiple of 7 = 7 x 1 = 7
2nd multiple of 7 = 7 x 2 = 14
3rd multiple of 7 = 7 x 3 = 21
4th multiple of 7 = 7 x 4 = 28
5th multiple of 7 = 7 x 5 = 35
6th multiple of 7 = 7 x 6 = 42
7th multiple of 7 = 7 x 7 = 49
8th multiple of 7 = 7 x 8 = 56
9th multiple of 7 = 7 x 9 = 63
10th multiple of 7 = 7 x 10 = 70.
Thus, 7, 14, 21, 28, 35, 42, 49, 56, 63 and 70 are the first 10 multiples of 7.
d. 8
Solution:
Given number = 8.
1st multiple of 8 = 8 x 1 = 8
2nd multiple of 8 = 8 x 2 = 16
3rd multiple of 8 = 8 x 3 = 24
4th multiple of 8 = 8 x 4 = 32
5th multiple of 8 = 8 x 5 = 40
6th multiple of 8 = 8 x 6 = 48
7th multiple of 8 = 8 x 7 = 56
8th multiple of 8 = 8 x 8 = 64
9th multiple of 8 = 8 x 9 = 72
10th multiple of 8 = 8 x 10 = 80.
Thus, 8, 16, 24, 32, 40, 48, 56, 64, 72 and 80 are the first 10 multiples of 8.
e. 10
Solution:
Given number = 10
1st multiple of 10 = 10 x 1 = 10
2nd multiple of 10 = 10 x 2 = 20
3rd multiple of 10 = 10 x 3 = 30
4th multiple of 10 = 10 x 4 = 40
5th multiple of 10 = 10 x 5 = 50
6th multiple of 10 = 10 x 6 = 60
7th multiple of 10 = 10 x 7 = 70
8th multiple of 10 = 10 x 8 = 80
9th multiple of 10 = 10 x 9 = 90
10th multiple of 10 = 10 x 10 = 100.
Thus, 10, 20, 30, 40, 50, 60, 70, 80, 90 and 100 are the first 10 multiples of 10.
f. 11
Solution:
Given number = 11
1st multiple of 11 = 11 x 1 = 11
2nd multiple of 11 = 11 x 2 = 22
3rd multiple of 11 = 11 x 3 = 33
4th multiple of 11 = 11 x 4 = 44
5th multiple of 11 = 11 x 5 = 55
6th multiple of 11 = 11 x 6 = 66
7th multiple of 11 = 11 x 7 = 77
8th multiple of 11 = 11 x 8 = 88
9th multiple of 11 = 11 x 9 = 99
10th multiple of 11 = 11 x 10 = 110.
Thus, 11, 22, 33, 44, 55, 66, 77, 88, 99 and 110 are the first 10 multiples of 11.
g. 12
Solution:
Given number = 12
1st multiple of 12 = 12 x 1 = 12
2nd multiple of 12 = 12 x 2 = 24
3rd multiple of 12 = 12 x 3 = 36
4th multiple of 12 = 12 x 4 = 48
5th multiple of 12 = 12 x 5 = 60
6th multiple of 12 = 12 x 6 = 72
7th multiple of 12 = 12 x 7 = 84
8th multiple of 12 = 12 x 8 = 96
9th multiple of 12 = 12 x 9 = 108
10th multiple of 12 = 12 x 10 = 120.
Thus, 12, 24, 36, 48, 60, 72, 84, 96, 108 and 120 are the first 10 multiples of 12.
h. 15
Solution:
Given number = 15
1st multiple of 15 = 15 x 1 = 15
2nd multiple of 15 = 15 x 2 = 30
3rd multiple of 15 = 15 x 3 = 45
4th multiple of 15 = 15 x 4 = 60
5th multiple of 15 = 15 x 5 = 75
6th multiple of 15 = 15 x 6 = 90
7th multiple of 15 = 15 x 7 = 105
8th multiple of 15 = 15 x 8 = 120
9th multiple of 15 = 15 x 9 = 135
10th multiple of 15 = 15 x 10 = 150.
Thus, 15, 30, 45, 60, 75, 90, 105, 120, 135 and 150 are the first 10 multiples of 15.
3. Write the greatest and least factors of the following numbers except 1 and the number itself.
a. 20
Solution:
1 x 20 = 20
2 x 10 = 20
4 x 5 = 20
5 x 4 = 20
10 x 2 = 20
20 x 1 = 20
Thus, the greatest factor of 20 is 10 and the least factor of 20 is 2.
b. 16
Solution:
1 x 16 = 16
2 x 8 = 16
4 x 4 = 16
8 x 2 = 16
16 x 1 = 16
Thus, the greatest factor of 16 is 8 and the least factor of 16 is 2.
c. 27
Solution:
1 x 27 = 27
3 x 9 = 27
9 x 3 = 27
27 x 1 = 27
Thus, the greatest factor of 27 is 9 and the least factor of 27 is 3.
d. 35
Solution:
1 x 35 = 35
5 x 7 = 35
7 x 5 = 35
35 x 1 = 35
Thus, the greatest factor of 35 is 7 and the least factor of 35 is 5.
e. 33
Solution:
1 x 33 = 33
3 x 11 = 33
11 x 3 = 33
33 x 1 = 33
Thus, the greatest factor of 33 is 11 and the least factor of 33 is 3.
f. 39
Solution:
1 x 39 = 39
3 x 13 = 39
13 x 3 = 39
39 x 1 = 39
Thus, the greatest factor of 39 is 13 and the least factor of 39 is 3.
g. 51
Solution:
1 x 51 = 51
3 x 17 = 51
17 x 3 = 51
51 x 1 = 51
Thus, the greatest factor of 51 is 17 and the least factor of 51 is 3.
h. 34
Solution:
1 x 34 = 34
2 x 17 = 34
17 x 2 = 34
34 x 1 = 34
Thus, the greatest factor of 34 is 17 and the least factor of 34 is 2.
1. Explain with reasons which of the following numbers are divisible by 2?
a. 7
b. 12
c. 16
d. 19
e. 22
f. 25
g. 30
h. 100
Answer: b. 12, c. 16, e. 22, g. 30 and h. 100 are divisible by 2. The reason is that the numbers 12, 16, 22, 30 and 100 are even.
2. Explain with reasons which of the following numbers are divisible by 3?
a. 63
b. 123
c. 147
d. 237
e. 569
f. 456
g. 339
h. 993
Answer: a. 63, b. 123, c. 147, d. 237, f. 456, g. 339 and h. 993 are divisible by 3. The reason is that the sum of each numbers digits is divisible by 3. Like 63=6+3=9, 123=1+2+3=6, 147=1+4+7=12, etc.
3. Explain with reasons which of the following numbers are divisible by 4?
a. 500
b. 497
c. 92
d. 940
e. 2510
f. 1824
g. 2356
h. 2808
Answer: a. 500, c. 92, d. 940, f. 1824, g. 2356 and h. 2808 are divisible by 4. The reason is that the last two digits of each number is divisible by 3. Like 92 is divided by 4 (4x23), 40 is divided by 4 (4x10), etc.
4. Explain with reasons which of the following numbers are divisible by 5?
a. 255
b. 299
c. 505
d. 1500
e. 2510
f. 3105
g. 2955
h. 3000
Answer: a. 255, c. 505, d. 1500, e. 2510, f. 3105, g. 2955 and h. 3000 are divisible by 5. The reason is that the end digit of each number is 5 or 0. Like 255= last digit is 5, 1500=last digit is 0, etc. I.e. to be divisible by 5, the end/last digit of a number must be 5 or 0.
5. Explain with reasons which of the following numbers are divisible by 6?
a. 72
b. 900
c. 910
d. 196
e. 2190
f. 1530
g. 2748
h. 5808
Answer: a. 72, b. 900, e. 2190, e. 2510, f. 1530, g. 2748 and h. 5808 are divisible by 6. The reason is that the numbers are even and divisible by 3. Like 72= 72 is divisible by 3 (3x24), 900= 900 is divisible by 3 (3x300), etc.
6. Explain with reasons which of the following numbers are divisible by 7?
a. 174
b. 315
c. 2,107
d. 1,645
e. 890
f. 182
g. 17,948
h. 1,610
Answer: b. 315, c. 2,107, d. 1,645, f. 182, g. 17,948 and h. 1,610 are divisible by 7. The reason is that if we double last digit of a number and subtract it from rest digits, the product number is divisible by 7. Like 315= 31-5x2=31-10=21 where 21 is divisible by 7, 2107=210-7x2=210-14=196 where 196 is divisible by 7, etc.
7. Explain with reasons which of the following numbers are divisible by 9?
a. 729
b. 505
c. 555
d. 4104
e. 5120
f. 5130
g. 53010
h. 4301
Answer: a. 729, d. 4104, f. 5130 and g. 53010 are divisible by 9. The reason is that sum of all digits of each number is divisible by 9. Like 729= 7+2+9=18 where 18 is divisible by 9 (9x2), 4104=4+1+0+4=9 where 9 is divisible by 9 (9x1), etc.
8. Explain with reasons which of the following numbers are divisible by 10?
a. 450
b. 555
c. 975
d. 1000
e. 1550
f. 1690
g. 1710
h. 9999
Answer: a. 450, d. 1000, e. 1550, f. 1690 and g. 1710 are divisible by 10. The reason is that the ending digit of each number is 0. Like 450= end digit is 0, etc. It means the numbers having end digits 0 is divisible by 10.
9. Explain with reasons which of the following numbers are divisible by 11?
a. 1353
b. 1595
c. 2815
d. 5280
e. 3916
f. 1221
g. 2849
h. 4059
Answer: a. 1353, b. 1595, d. 5280, e. 3916, f. 1221, g. 2849 and h. 4059 are divisible by 11. The reason is that the difference of sum of odd places digits and sum of even places digits must be 0 or multiple of 11. Or subtract end digit from the rest digit and the number must be divisible by 11. Like 1353=1+5=6 and 3+3=6 - difference is 0, so it is divisible by 11. Another 1353=135-3=132 where 132 is divisible by 11 (11x12).
10. Check which of these numbers are divisible by the given numbers.
a. 7192; 465; 1,357, 56,340 by 2, 5 or 10.
Answer:
By 2, 7192 and 56340
By 5, 465 and 56,340
By 10, 56340.
b. 6,342; 9,990; 1,721 by 3 or 9.
Answer:
By 3, 6,342 and 9,990
By 9, 9,990
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Math Sanjal
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