Let the two odd numbers be 5 and 7.
Their sum = 5 + 7 = 12 (even number)
Taking one more example.
Let the two odd numbers be 3 and 7.
Their sum = 3 + 7 = 10 (even number)
Hence, we can say that the sum of any two odd numbers is an even number

Yes

11, 22, 33, 44, 55;
13, 26, 39, 52 and 65

We have, 23
23 = 1x 23
\(\therefore\)Factors of 23 are 1 and 23.

We have, 20
20 = 1 x 20; 20 = 2 x 10; 20 = 4 x 5
\(\therefore\)Factors of 20 are 1, 2, 4, 5, 10 and 20.

4

2

For the HCF of 58, 70.

\(\therefore\) HCF of 58 and 70 is 2.

Factors of 32 = 1,2,4,8, 16 and 32

4335 is not divisible by 6.

9:14 am

Required length = HCF of 7 m, 3 m 85 cm and 12m 95 cm or HCF of 700 cm, 385 cm and 1295 cm.

Hence, required length is 37 cm.

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.