Explanation:

24

24 = 1 × 24

24 = 2 × 12

24 = 3 × eight

24 = 4 × 6

24 = 6 × four

prevent right here due to the fact the 4th and 6th took place in advance

So the factors of 24 are 1, 2, three, four, 6, 8, 12 and 24

Explanation:

15

15 = 1 × 15

15 = three × five

15 = five × three

prevent right here due to the fact 3 and five passed off earlier

So the factors of 15 are 1, three, five, and 15

three.

Explanation:

21

21 = 1 × 21

21 = 3 × 7

21 = 7 × 3

stop here because 3 and 7 passed off in advance

So the elements of 21 are 1, 3, 7, and 21

Explanation:

27

27 = 1 × 27

27 = three × 9

27 = nine × three

stop right here due to the fact 3 and nine took place earlier

So the factors of 27 are 1, three, 9, and 27

Explanation:

12

12 = 1 × 12

12 = 2 × 6

12 = 3 × 4

12 = 4 × three

forestall right here due to the fact 3 and four took place in advance

So the elements of 12 are 1, 2, 3, 4, 6, and 12

Explanation:

20

20 = 1 × 20

20 = 2 × 10

20 = four × 5

20 = five × 4

stop here because the 4th and 5th passed off in advance

So the elements of 20 are 1, 2, four, five, 10, and 20

Explanation:

18

18 = 1 × 18

18 = 2 × nine

18 = 3 × 6

18 = 6 × 3

stop right here due to the fact three and six happened in advance

So the elements of 18 are 1, 2, 3, 6, nine, and 18

eight.

Explanation:

23

23 = 1 × 23

23 = 23 × 1

From the 1st and 23rd it befell earlier

So the factors of 23 are 1 and 23

Explanation:

36

36 = 1 × 36

36 = 2 × 18

36 = three × 12

36 = four × 9

36 = 6 × 6

prevent here due to the fact both factors of (6) are the same. So the elements of 36 are 1, 2, 3, four, 6, nine, 12, 18, and 36

Explanation:

the desired multiples are:

5 x 1 = 5

5 × 2 = 10

five × three = 15

five x 4 = 20

five × 5 = 25

So the primary 5 multiples of five are five, 10, 15, 20, and 25

Explanation:

the specified multiples are:

8 x 1 = eight

8 × 2 = sixteen

eight x 3 = 24

8 x 4 = 32

8 x 5 = forty

So the first 5 multiples of eight are 8, sixteen, 24, 32, and 40

Explanation:

the required multiples are:

nine x 1 = 9

nine × 2 = 18

9 × 3 = 27

nine x 4 = 36

nine x five = 45

So the primary 5 multiples of 9 are nine, 18, 27, 36, and forty five

Explanation:

(i) 35 is a a couple of of seven

for this reason option (b)

(ii) 15 is a element of 30

as a result alternative (d)

(iii) sixteen is a a couple of of eight

for this reason option (a)

(iv) 20 is a factor of 20

as a result choice (f)

(v) 25 is a component of 50

therefore choice (e)

Explanation:

nine x 1 = 9

9 × 2 = 18

nine × 3 = 27

9 x four = 36

9 x 5 = forty five

nine x 6 = 54

nine × 7 = 63

nine × 8 = 72

9 × nine = eighty one

nine x 10 = ninety

9 × 11 = ninety nine

∴ All multiples of 9 to 100 are nine, 18, 27, 36, 45, fifty four, 63, 72, eighty one, ninety and 99

Explanation:

The sum of any two bizarre numbers is an excellent range.

Examples: five + 3 = 8

15 + thirteen = 28

what is the sum of any two even numbers?

answer:

The sum of any two even numbers is a fair wide variety

Examples: 2 + 8 = 10

12 + 28 = forty

Explanation:

(a) false. The sum of 3 bizarre numbers is extraordinary.

instance: 7 + 9 + five = 21, i.e. an strange quantity

(b) real. The sum of two bizarre numbers and one even number is even.

instance: 3 + five + 8 = 16, i.e. is a fair quantity.

(c) proper. The product of 3 abnormal numbers is bizarre.

instance: 3 × 7 × 9 = 189, i.e. is an unusual variety.

(d) fake. If a good variety is divisible with the aid of 2, the quotient is even.

instance: eight ÷ 2 = four

(e) false, all top numbers are not odd.

example: 2 is a high wide variety, but also a fair wide variety.

(f) fake. due to the fact 1 and the wide variety itself are elements of a range of

(g) false. The sum of two prime numbers can also be an bizarre range

example: 2 + 5 = 7, i.e. an unusual variety.

(h) proper. 2 is the best even top range.

(i) fake. because 2 is a high variety.

(j) real. The fabricated from  even numbers is continually even.

instance: 2 × four = 8, i.e. a good number.

Explanation:

prime numbers with the same digits as much as 100 are as follows:

17 and seventy one

37 and 73

seventy nine and ninety seven

Explanation:

2, 3, 5, 7, 11, 13, 17 and 19 are high numbers much less than 20

4, 6, eight, nine, 10, 12, 14, 15, 16 and 18 are composite numbers much less than 20

Explanation:

2, 3, 5 and 7 are top numbers among 1 and 10. 7 is the largest high wide variety amongst them.

Explanation:

(a) three + forty one = 44

(b) 5 + 31 = 36

(c) five + 19 = 24

(d) five + 13 = 18

Explanation:

3 pairs of top numbers whose distinction is 2 are

three, 5

5, 7

11, thirteen

Explanation:

a) 23

1 × 23 = 23

23 × 1 = 23

therefore 23 has handiest two factors 1 and 23. therefore it’s a high quantity

(b) fifty one

1 × 51 = fifty one

three × 17 = fifty one

consequently, fifty one has four elements 1, 3, 17 and 51. So it is not a high quantity, it is a composite quantity.

(c) 37

1 × 37 = 37

37 × 1 = 37

therefore, 37 has  elements 1 and 37. So it is a prime wide variety.

(d) 26

1 × 26 = 26

2 × 13 = 26

consequently 26 has four elements 1, 2, 13 and 26. So it isn't a top quantity, it's miles a composite quantity.

Explanation:

The seven composite numbers among 89 and ninety seven, each of that are top, are 90, ninety one, 92, 93, 94, 95, and 96

Numbers factors

90 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, forty five, ninety

91 1, 7, 13, 91

92 1, 2, four, 23, forty six, 92

ninety three 1, three, 31, 93

94 1, 2, forty seven, 94

ninety five 1, five, 19, 95

ninety six 1, 2, 3, 4, 6, eight, 12, 16, 24, 32, 48, 96

Explanation:

(a) 3 + 5 + 13 = 21

(b) 3 + five + 23 = 31

(c) 13 + 17 + 23 = fifty three

(d) 7 + thirteen + forty one = sixty one

Explanation:

5 pairs of prime numbers less than 20 whose sum is divisible by using five are

2 + three = 5

2 + 13 = 15

3 + 17 = 20

7 + thirteen = 20

19 + eleven = 30

Explanation:

(i) five + three = eight, which is divisible by 4

(ii) 7 + nine = 16 which is divisible via 4

(iii) thirteen + 15 = 28 that's divisible by 4

Explanation:

(a) 24 = 2 × 3 × 4

due to the fact four is composite. thus, top factorization changed into now not completed

(b) 56 = 7 × 2 × 2 × 2

because all elements are paramount. therefore, prime factorization turned into executed

(c) 70 = 2 × five × 7

due to the fact all elements are paramount. consequently, high factorization turned into executed

(d) 54 = 2 × three × nine

due to the fact nine is composite. therefore, top factorization was no longer completed

Explanation:

(a) 572

seventy two are the closing  digits. because seventy two is divisible through 4. consequently 572 is likewise divisible by using four

572 are the final three digits. due to the fact 572 isn't divisible through 8. consequently 572 is not divisible

by eight

(b) 726352

fifty two are the closing two digits. because 52 is divisible by using four. therefore 726352 is divisible with the aid of four

352 are the ultimate 3 digits. because 352 is divisible by means of 8. therefore 726352 is divisible by using 8

(c) 5500

because the last two digits are 00. So 5500 is divisible by way of four

500 are the remaining three digits. due to the fact 500 isn't always divisible through 8. consequently 5500 is not divisible

via 8

(d) 6000

for the reason that last  digits are 00. So 6000 is divisible by using four

because the last 3 digits are 000. So 6000 is divisible through 8

(e) 12159

59 are the closing  digits. because fifty nine isn't always divisible by way of 4. therefore 12159 isn't divisible by means of 4

159 are the last three digits. because 159 isn't divisible through eight. consequently 12159 isn't divisible by way of 8

(f) 14560

60 is the ultimate two digits. because 60 is divisible by means of 4. consequently 14560 is divisible by using 4

560 are the last 3 digits. because 560 is divisible by way of eight. consequently 14560 is divisible with the aid of 8

(g) 21084

84 are the remaining  digits. due to the fact eighty four is divisible by 4. consequently 21084 is divisible by means of four

084 is the closing three digits. due to the fact 084 isn't always divisible through 8. therefore 21084 is not divisible

with the aid of 8

(h) 31795072

seventy two are the last  digits. because seventy two is divisible through 4. therefore 31795072 is divisible by means of 4

072 are the last three digits. due to the fact 072 is divisible via 8. consequently 31795072 is divisible by 8

(i) 1700

due to the fact the closing two digits are 00. So 1700 is divisible by using 4

700 are the last 3 digits. due to the fact 700 isn't divisible by way of eight. therefore 1700 is not divisible

by eight

(j) 2150

50 is the remaining  digits. due to the fact 50 isn't divisible with the aid of 4. consequently 2150 is not divisible by using four

one hundred fifty are the remaining three digits. due to the fact 150 isn't always divisible by means of 8. consequently 2150 isn't always divisible

Explanation:

(a) 297144

due to the fact the ultimate digit of the number is four. So the quantity is divisible via 2

including all the digits of the number offers 27 that's divisible by three. So the wide variety is divisible by using 3

∴ The range is divisible via both 2 and 3. So the number is divisible by way of 6

(b) 1258

because the closing digit of the quantity is eight. So the wide variety is divisible by way of 2

adding all the digits of the quantity gives sixteen which isn't divisible via three. So the quantity isn't divisible by using three

∴ The quantity is not divisible via 2 or three. So the quantity isn't always divisible by way of 6

(c) 4335

because the last digit of the number is five which isn't always divisible with the aid of 2. So the quantity isn't always divisible by 2

adding all the digits of the quantity offers 15 that's divisible via 3. So the variety is divisible with the aid of three

∴ The wide variety isn't always divisible with the aid of 2 or 3. So the range isn't divisible by way of 6

(d) 61233

due to the fact the ultimate digit of the wide variety is three, which isn't divisible by 2. So the quantity isn't always divisible by 2

adding all of the digits of the wide variety offers 15 that is divisible by way of three. So the wide variety is divisible with the aid of three

∴ The variety isn't always divisible with the aid of 2 or three. So the variety is not divisible through 6

(e) 901352

because the final digit of the wide variety is two. So the range is divisible with the aid of 2

including all of the digits of the quantity offers 20 which is not divisible with the aid of three. So the range isn't divisible by way of three

∴ The number isn't divisible by 2 or three. So the variety is not divisible by way of 6

(f) 438750

due to the fact the final digit of the variety is zero. So the quantity is divisible by means of 2

adding all of the digits of the quantity gives 27 that is divisible by 3. So the quantity is divisible by using three

∴ The quantity is divisible via each 2 and three. So the number is divisible by using 6

(g) 1790184

because the remaining digit of the variety is four. So the range is divisible by means of 2

including all of the digits of the quantity offers 30 which is divisible by using 3. So the quantity is divisible by way of 3

∴ The range is divisible with the aid of each 2 and 3. So the wide variety is divisible by using 6

(h) 12583

because the closing digit of the wide variety is 3. So the range isn't divisible by way of 2

adding all of the digits of the range offers 19 which isn't always divisible via three. So the wide variety is not divisible with the aid of three

∴ The number isn't divisible by using 2 or three. So the quantity isn't divisible by using 6

(i) 639210

due to the fact the closing digit of the variety is 0. So the range is divisible by means of 2

including all of the digits of the variety gives 21 which is divisible via 3. So the number is divisible by 3

∴ The quantity is divisible with the aid of each 2 and 3. So the quantity is divisible through 6

(j) 17852

due to the fact the ultimate digit of the number is two. So the wide variety is divisible through 2

including all of the digits of the wide variety gives 23 which is not divisible by three. So the variety isn't divisible via three

∴ The wide variety isn't divisible via 2 or 3. So the number isn't always divisible through 6

Explanation:

(a) 5445

Sum of digits in bizarre places = 5 + four

= 9

Sum of digits in even places = 4 + five

= nine

difference = nine – 9 = 0

because the difference among the sum of the digits within the ordinary locations and the sum of the digits within the even locations is 0. consequently, 5445 is divisible through 11

(b) 10824

Sum of digits in abnormal locations = four + 8 + 1

= thirteen

Sum of digits in even places = 2 + 0

= 2

distinction = 13 – 2 = 11

because the distinction between the sum of the digits within the odd locations and the sum of the digits in the even locations is eleven, which is divisible by using 11. consequently, 10824 is divisible through 11

(c) 7138965

Sum of digits in atypical locations = five + nine + three + 7 = 24

Sum of digits in even locations = 6 + 8 + 1 = 15

difference = 24 – 15 = nine

because the difference among the sum of the digits within the peculiar locations and the sum of the digits in the even locations is nine, which isn't always divisible by means of eleven. therefore, 7138965 isn't divisible by means of 11

(d) 70169308

Sum of digits in abnormal locations = 8 + three + 6 + zero

= 17

Sum of digits in even places = zero + nine + 1 + 7

= 17

distinction = 17 – 17 = zero

due to the fact the difference among the sum of the digits inside the abnormal locations and the sum of the digits inside the even locations is 0. consequently, the variety 70169308 is divisible through 11

(e) 10000001

Sum of digits in extraordinary places = 1

Sum of digits in even locations = 1

distinction = 1 – 1 = 0

due to the fact the distinction between the sum of the digits within the ordinary places and the sum of the digits inside the even locations is 0. consequently 10000001 is divisible via 11

(f) 901153

Sum of digits in bizarre places = three + 1 + 0

= four

Sum of digits in even places = 5 + 1 + nine

= 15

difference = 15 – 4 = 11

due to the fact the difference among the sum of the abnormal digits and the sum of the even digits is eleven, that is divisible through eleven. therefore, 901153 is divisible by using 11

Explanation:

(a) __ 6724

Sum of the given digits = 19

The sum of its digits should be divisible via 3 for the quantity to be divisible through 3

because 21 is the smallest a couple of of 3 that comes after 19

So the smallest wide variety = 21 – 19

= 2

Now 2 + three + three = 8

but 2 + three + three + 3 = eleven

Now if we positioned 8 the sum of the digits may be 27 which is divisible through three

therefore, the range could be divisible by way of 3

So the largest wide variety is eight

(b) 4765 __ 2

Sum of digits entered = 24

The sum of its digits have to be divisible by using 3 for the quantity to be divisible through 3

because 24 is already divisible by means of three. So the smallest variety that can be changed is zero

Now 0 + three = three

3 + three = 6

three + 3 + three = nine

three + three + three + three = 12

If we put nine, the sum of its digits will be 33. due to the fact 33 is divisible by using three.

consequently, the variety may be divisible by 3

So the largest quantity is nine

Explanation:

(a) 92__ 389

let "a" be placed right here.

The sum of its digits in atypical places = nine + three + 2

= 14

Sum of its digits in even places = 8 + a + nine

= 17 + a

distinction = 17 + and – 14

= 3 + a

The difference have to be 0 or a a couple of of 11, then the variety is divisible by way of 11

If 3 + a = 0

a = -three

however it cannot be a negative

We take the nearest a couple of of eleven this is close to three

it is eleven, that's near to three

Now three + a = eleven

a = 11 – 3

a = eight

So the specified digit is eight

(b) 8__ 9484

allow "a" be placed right here.

The sum of its digits in bizarre places = four + four + a

= eight + a

Sum of its digits in even places = eight + nine + 8

= 25

distinction = 25 – (8 + a)

= 17 – a

The difference ought to be 0 or a more than one of 11, then the wide variety is divisible by using 11

If 17 – a = 0

a = 17 (which isn't viable)

Now take a more than one of 11.

let's take eleven

17 – a = eleven

a = 17–11

a = 6

So the required digit is 6

Explanation:

a) 20 and 28

1, 2, 4, five, 10, and 20 are elements of 20

1, 2, four, 7, 14, and 28 are elements of 28

commonplace elements = 1, 2, 4

b) 15 and 25

1, three, five, and 15 are factors of 15

1, 5, and 25 are factors of 25

common elements = 1.5

(c) 35 and 50

1, five, 7, and 35 are elements of 35

1, 2, five, 10, 25, and 50 are elements of fifty

commonplace elements = 1.5

(d) 56 and a hundred and twenty

1, 2, four, 7, eight, 14, 28, and 56 are factors of 56

1, 2, 3, four, 5, 6, 8, 10, 12, 15, 20, 24, 30, forty, 60, and a hundred and twenty are factors of one hundred twenty

common factors = 1, 2, four, 8

Explanation:

a) four, eight and 12

1, 2, 4 are factors of 4

1, 2, four, 8 are factors of 8

1, 2, 3, 4, 6, 12 are elements of 12

commonplace factors = 1, 2, four

b) five, 15 and 25

1, five are factors of 5

1, three, five, 15 are factors of 15

1, five, 25 are elements of 25

not unusual factors = 1,five

Explanation:

a) 6 and 8

6, 12, 18, 24, 30 are multiples of 6

eight, sixteen, 24, 32 are multiples of 8

The three commonplace multiples are 24, 48, 72

b) 12 and 18

12, 24, 36, 48 are multiples of 12

18, 36, 54, seventy two are multiples of 18

The three common elements are 36, seventy two, 108

Explanation:

Multiples of three are three, 6, nine, 12, 15

Multiples of 4 are 4, eight, 12, sixteen, 20

commonplace multiples are 12, 24, 36, forty eight, 60, seventy two, 84 and 96

Explanation:

a) 18 and 35

The elements of 18 are 1, 2, 3, 6, nine, 18

The factors of 35 are 1, 5, 7, 35

not unusual aspect = 1

due to the fact their not unusual issue is 1. The given two numbers are therefore high numbers

b) 15 and 37

The factors of 15 are 1, 3, five, 15

The factors of 37 are 1.37

not unusual elements = 1

due to the fact their commonplace element is 1. The given two numbers are consequently top numbers

(c) 30 and 415

factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30

The elements of 415 are 1, 5, 83, 415

not unusual factors = 1.five

considering the fact that their not unusual factor is other than 1. So the given two numbers are not co-top

(d) 17 and 68

The factors of 17 are 1, 17

The elements of 68 are 1, 2, 4, 17, 34, sixty eight

not unusual factors = 1, 17

on account that their commonplace thing is apart from 1. So the given two numbers are not co-high

(e) 216 and 215

The elements of 216 are 1, 2, 3, 4, 6, eight, nine, 12, 18, 24, 27, 36, 54, 72, 108, 216

The elements of 215 are 1, 5, forty three, 215

common factors = 1

because their common component is 1. The given  numbers are therefore high numbers

(f) eighty one and 16

The factors of 81 are 1, 3, 9, 27, 81

The elements of 16 are 1, 2, four, 8, sixteen

common factors = 1

due to the fact their common aspect is 1. The given  numbers are therefore top numbers

Explanation:

The elements of 5 are 1.five

The factors of 12 are 1, 2, 3, four, 6, 12

Their commonplace aspect = 1

because their not unusual element is 1. The given  numbers are high numbers and also are divisible with the aid of their product 60

The elements of 60 are 1, 2, 3, 4, five, 6, 10, 12, 15, 20, 30, 60

Explanation:

due to the fact the wide variety is divisible through 12. consequently it is also divisible by way of its elements i.e. 1, 2, three, four, 6, 12

consequently 1, 2, 3, 4 and six are numbers apart from 12 with the aid of which this range is likewise divisible

forty.

Explanation:

(a) fake, 6 is divisible through 3 however no longer divisible by using nine

(b) real due to the fact 9 = 3 × 3. So if a number of is divisible by nine, it's going to additionally be divisible through three

(c) fake. due to the fact 30 is divisible by way of each 3 and six, but no longer by using 18

(d) authentic because 9 × 10 = 90. consequently, if a number of is divisible via both nine and 10, it is divisible by using 90

(e) false. due to the fact 15 and 32 are high numbers together and additionally composite numbers

(f) false because 12 is divisible by way of 4 but now not divisible with the aid of 8

(g) real due to the fact 2 × 4 = eight. So if a number of is divisible with the aid of eight, it'll also be divisible with the aid of 2 and four

(h) actual because 2 divides four and eight and also divides 12 (4 + 8 = 12)

(i) fake due to the fact 2 divides 12 however does now not divide 7 and 5