Solution:
91 is divisible by 7. So, it is not a prime number.

Solution:

Clearly, 7261 must be replaced by 2961,
which is possible if 6648 is replaced by 6218,
which in turn is possible if 5599 is replaced by 5556.
Thus, the correct number is 555681

Solution:
At last thief is left with one diamond.
Hence, the number of diamonds before he gave some diamonds to the third watchman,
$$\eqalign{ \Rightarrow x - \left( { {\frac{x}{2}} + 2} \right) = 1 \cr {\text{or, }}{\kern 1pt} \frac{{ {x - 4} }}{2} = 1 \cr {\text{or, }}{\kern 1pt} x = 6 \cr} $$
Hence, he had 6 diamonds before he gave 5 to the third watchman.
Similarly number of diamonds before giving to second watchman,
$$\eqalign{ \frac{{ {x - 4} }}{2} = 6 \cr {\text{or,}}\,{\kern 1pt} x = 16 \cr} $$
And number of diamonds before giving to the first watchman,
$$\eqalign{ \frac{{ {x - 4} }}{2} = 16 \cr {\text{or, }}{\kern 1pt} x = 36 \cr} $$
∴ The thief has stolen 36 diamonds originally

Solution:
$$\eqalign{ x = 13p + 11\,{\text{and}}\,x = 17q + 9 \cr \therefore 13p + 11 = 17q + 9 \cr \Rightarrow 17q - 13p = 2 \cr \Rightarrow q = \frac{{2 + 13p}}{{17}} \cr {\text{Thr}}\,{\text{least}}\,{\text{value}}\,{\text{of}}\,p\,{\text{for}}\,{\text{which}} \cr q = \frac{{2 + 13p}}{{17}}\,{\text{is}}\,{\text{a}}\,{\text{whole}}\,{\text{number}}\,{\text{is}}\,p = 26 \cr \therefore x = \left( {13 \times 26 + 11} \right) \cr \,\,\,\,\,\,\,\,\,\,\, = \left( {338 + 11} \right) \cr \,\,\,\,\,\,\,\,\,\,\, = 349 \cr} $$

Solution:
Length of the room
= 15 m 17 cm
= 15 × 100 + 17
= 1517 cm.
Breadth of the room
= 9 m 2 cm
= 902 cm.
The HCF of the 1517 and 902 will be size of square tiles. HCF 1517 and 902 = 41 cm.
Area of the room
= length × breadth
= 1517 × 902 cm²
Area of tiles = 41 × 41 cm² So, number of tiles required
= $$\frac{{1517 \times 902}}{{41 \times 41}}$$
= 814 tiles

Solution:
$$\eqalign{ {\text{Let the unit digit }} \cr {\text{Ten digit }} = x - 2 \cr \therefore {\text{ Number :}} \cr = {\text{ }}10(x - 2){\text{ }} + {\text{ }}x \cr = {\text{ }}10x{\text{ }} - {\text{ }}20{\text{ }} + {\text{ }}x \cr = {\text{ }}11x{\text{ }} - {\text{ }}20 \cr} $$
New number obtained after reversing the digits
$$\eqalign{ = 10x + x - 2{\text{ }} \cr = {\text{ }}11x - 2 \cr {\text{According to the question,}} \cr {\text{3}}\left( {11x - 20} \right) + \frac{6}{7}\left( {11x - 2} \right) = 108 \cr 7\left( {11x - 20} \right) + 2\left( {11x - 2} \right) = 36 \times 7 \cr 77x - 140 + 22x - 4 = 252 \cr 99x = 252 + 144 \cr x = \frac{{396}}{{99}} = 4 \cr {\text{Number :}} \cr = {\text{ }}11x - 20{\text{ }} \cr = {\text{ }}11 \times 4 - 20 \cr = {\text{ }}24 \cr {\text{Sum of digit }} = 2 + 4 = 6 \cr} $$

Solution:
A + B + C + D = 60 -----------> (1)

Now According to Question,
A = $$\frac{1}{2}$$ (B + C + D)
2A = B + C + D
Put the value of B + C + D in Equation 1
A + 2A = 60
3A = 60
A = 20

Now again according to Question
B = $$\frac{1}{3}$$ (A + C + D)
A + C + D = 3B
By Putting the Value of A + C + D in Equation 1
we get,
B + 3B = 60
4B = 60
B = 15

Now again according to Question
C = $$\frac{1}{4}$$ (A + B + D)
A + B + D = 4C
By Putting the Value of A + B + D in equation 1
We get C + 4C = 60
5C = 60
C = 12

Since A + B + C + D = 60
20 + 15 + 12 + D = 60
D = 60 - 47 = 13
Amount of D = Rs. 13