\(\because\) 6 = 2 x 3 and 10 = 5 x 2

Hence, missing numbers are 3 and 2.

1, 2, 4

Factors of 20 and 28 are as follows:
20 = 1 x 20; 20 = 2 x 10; 20 = 4 x 5 and 28 = 1 x 28; 28 = 2 x 14; 28 = 4 x 7
Now, all factors of 20 = 1, 2, 4, 5, 10, 20 and all factors of 28 = 1, 2, 4, 7, 14, 28
\(\therefore\)Common factors = 1, 2, 4
Hence, common factors of 20 and 28 are 1, 2 and 4.

Yes

We have, 10824,
\(\begin{matrix} 1 \\ \downarrow \\ O \end{matrix}\begin{matrix} 0 \\ \downarrow \\ E \end{matrix}\begin{matrix} 8 \\ \downarrow \\ O \end{matrix}\begin{matrix} 2 \\ \downarrow \\ E \end{matrix}\begin{matrix} 4 \\ \downarrow \\ O \end{matrix}\)
Sum of digits at odd places from right = 4 +8 + 1= 13
Sum of digits at even places from right = 2 + 0 = 2
Now, difference = 13 - 2 = 11
So, 10824 is divisible by 11.

In the number 9524624, the last two digit number is 24, which is divisible by 4. Hence, 9524624 is divisible by 4.

0

Prime numbers between 1 and 10 are 2,3,5 and 7.
Therefore, the greatest prime number between 1 and 10 is 7.

Here, we have 34
34 = 31 + 3 or 29 + 5

Yes

11, 22, 33, 44, 55, 66, 77, 88, 99, 110.

1, 2, 3, 6, 7,14,21,42

Multiples of 8
8 x 1= 8; 8 x 2 = 16; 8 x 3 = 24; 8 x 4 = 32; 8 x 5 = 40
\(\therefore\)First five multiples of 8 are 8, 16, 24, 32 and 40.

We have, 21
21 = 1 x 21; 21 = 3 x 7
\(\therefore\)Factorsof 21 are 1, 3, 7 and 21.

Factor is the exact divisor of that number. So, for finding factors, we use factorisation method.
Factors of 12 = 1, 2, 3, 4, 6 and 12.

Now, for finding multiples, we multiply 12 with natural numbers.
So, first four multiples of 12 are 12 x 1, 12 x 2, 12 x 3,12 x 4
i.e. 12, 24, 36 and 48.

10080

We have to find the LCM of 9,6,4, 10 and 8.

LCM = 2 x 2 x 3x3x5x2=360
In 1 h, the ring will toll together
\(3600\over 360\)times=10times

We have, 136 

\(\therefore\)136=1x2x2x2x17

We have, 30 and 415
30 = 1 x 30; 30 = 2 x 15; 30 = 3 x 10; 30 = 5 x 6
Factors of 30 are 1,2,3,5,6,10,15 and 30 and 415 = 1 x 415; 415 = 5 x 83
Factors of 415 are 1, 5, 83 and 415.
\(\because\)Common factors of 30 and 415 are 1 and 5.
So, 30 and 415 are not co-prime numbers.

We have, 1258
(i) Divisibility by 2
\(\because\)Unit digit of number = 8, so 1258 is divisible by 2.
(ii) Divisibility by 3
Sum of digits of given number = 1+ 2 + 5 + 8 = 16 is not divisible by 3, so 1258 is not divisible by 3. Now, we see that 1258 is divisible by 2 but not divisible by 3.
Hence, it is not divisible by 6.

99360

Required second of time = LCM of 42, 56 and 63

LCM = 2 x 3 x 7 x 2 x 2 x 3 = 504 s
Hence, after 504 s, they will be together at starting point.

(a) The steps measure of each person is 40 cm 42 cm and 46 cm. So, the minimum distance each should walk is the LCM of 40, 42 and 46.

LCM of 40, 42 and 46 = 2 x 20 x 21 x 23
=19320 cm
= 193.2 m
(b) Morning walk benefits us in many ways. It maintain our good health. Some benefits of morning walk are listed below:
(i) Strengthen our heart
(ii) Delays or prevents major diseases or illness
(iii) Reduces blood pressure and the risk of stroke
(iv) Reduces cholestrol
(v) Boost immune system

The given postal charges for three parcels are Rs20, Rs28, Rs 36.
The greatest denomination of stamps is the HCF of 20, 28 and 36.


HCF of 20 and 28 is 4.
Now, we.find the HCF of 4 and 36.
HCF of 4 and 36 is 4.
Hence, HCF of 20, 28 and 36 is 4.
\(\therefore\) Fatima would by stamp of denomination of Rs4.

Let the required unknown digit be x.
Then, the number becomes
\(\begin{matrix} 9 \\ \downarrow \\ E \end{matrix}\begin{matrix} 2 \\ \downarrow \\ O \end{matrix}\begin{matrix} x \\ \downarrow \\ E \end{matrix}\begin{matrix} 3 \\ \downarrow \\ O \end{matrix}\begin{matrix} 8 \\ \downarrow \\ E \end{matrix}\begin{matrix} 9 \\ \downarrow \\ O \end{matrix}\)
where, O = Odd and E = Even
Sum of digits at odd places from right = 9 + 3 + 2 = 14
Sum of digits at even places from right = 8 + x + 9 = 17+x
\(\because\)Number is divisible by 11.
\(\therefore\)Difference of digits will be 0 or 11.
\(\Rightarrow\) (17 + x) -14 = 0 or 11
\(\Rightarrow\)17 + x - 14 = 0 or 11 \(\Rightarrow\) x +3 = 0 or 11
Taking difference 0, x +3 = 0
\(\Rightarrow\)x = 0 - 3 = - 3 (not possible)
Taking difference 11, x + 3 = 11 \(\Rightarrow\) x = 11- 3 = 8
So, required digit to write in the blank space is 8.