প্রশ্নে বলা হচ্ছে, তিন সংখ্যা বিশিষ্ট একটি সংখ্যার তিনটির যোগফল 10. মাঝের অঙ্কটি অন্য দুটি সংখ্যার সমষ্টির সমান এবং ঐ দুটি সংখ্যা স্থান বিনিময় করলে যে সংখ্যাটি পাওয়া যায় তা মূল সংখ্যার চেয়ে 99 বেশি হলে সংখ্যাটি কত?

$$\begin{gathered} Let, the Hundredth digit is x, \\ Tenth digits is y \\ And Unit digits is z \\ \therefore The number is (100x+10y+z) \\ According to question, \\ x+y+z=10................ \left(i\right) \\ y=x+z............... \left(ii\right) \\ 100x+10y+z+99 =100z+10y+x................ \left(iii\right) \end{gathered}$$

From equation (iii) 

$$\begin{gathered} 100x+10y+z+99=100z+10y+x \\ \Rightarrow 100x+z+99+10y-10y=100z+x \\ \Rightarrow 100x+z+99=100z+x \\ \Rightarrow 100z-z+x-100z=99 \\ \Rightarrow 99z-99x=99 \\ \Rightarrow 99\left(z-x\right)=99 \\ \Rightarrow \left(z-x\right)=\frac{99}{99} \\ \Rightarrow z-x=1 \\ \therefore z=x+1........................\left(iv\right) \end{gathered}$$

putting the value of z in equation (ii) 

$$\begin{gathered} y=z+x \\ \Rightarrow y=x+x+1 \\ \Rightarrow y=2x+1................\left(v\right) \end{gathered}$$

putting the value of y & z in equation (i) 

$$\begin{gathered} x+y+z=20 \\ \Rightarrow x+2x+1+x+1=10 \\ \Rightarrow 4x+2=10 \\ \Rightarrow 4x=10-2 \\ \Rightarrow 4x=8 \\ \Rightarrow x=\frac{8}{4} \\ \therefore x=2 \end{gathered}$$

putting the value of x in equation (v) 

$$\begin{gathered} y=2x+1 \\ \Rightarrow y=2\times 2+1 \\ \Rightarrow y=4+1 \\ \therefore y=5 \end{gathered}$$

putting the value of x in equation (iv) 

$$\begin{gathered} z=x+1 \\ \Rightarrow z=2+1 \\ \therefore z=3 \end{gathered}$$

The number will be $$\begin{gathered} = 100x+10y+z \\ =\left(100\times 2\right)+\left(10\times 5\right)+\left(3\times 1\right) \\ =200+50+3 \\ =253 \end{gathered}$$

প্রশ্নে বলা হচ্ছে, দুই অঙ্ক বিশিষ্ট কোন সংখ্যা তার অঙ্ক দুটির যোগফলের 4 গুণ। অঙ্ক দুটিকে স্থান পরিবর্তন করলে যে সংখ্যা পাওয়া যায় তা মূল সংখ্যার চেয়ে 18 বেশি। মূল সংখ্যাটি কত?

Let, Tenth digit is x and unit digit is y

∴ The number = (10x + y)

According to first condition, $4(x + y) = 10x + y . . . . . . . . . . . . \left(i\right)$

And second condition, $(10x + y) + 18 = 10y + x . . . . . . . . . . . . . \left(ii\right)$

From equation (ii), $10x + y + 18 = 10y + x$

$$\begin{gathered} \Rightarrow 9x-9y=-18 \\ \Rightarrow 9y-9x=18 [Multiplying by - 1] \\ \Rightarrow 9(y-x) = 18 \\ \Rightarrow (y-x) = \frac{18}{9} = 2 \\ \Rightarrow y-x = 2 \\ \therefore y = x+2 . . . . . . . . . \left(iii\right) \end{gathered}$$

From equation (i), we get $4(x + y) = 10x + y$

$$\begin{gathered} \Rightarrow 4x+4y = 10x + y \\ \Rightarrow 3y = 6x \\ \Rightarrow y = 2x \\ \Rightarrow x+2 = 2x [Putting the value of y] \\ \therefore x = 2 \end{gathered}$$

Now, putting the value of x in equation (iii) we get

$$\begin{gathered} y= x+2 \\ \Rightarrow y = 2+2 = 4 \\ \therefore y = 4 \\ The number is (10x + y) = (10 \times 2)+4=20+4=24. ans. \end{gathered}$$

Let the unit digit is b and tenth dight is a So the number = 10a + b

According to first condition, a + b = 16 ........(i)

According to second condition, 

10a + b + 18 = 10b + a

= 10b + a - 10a - b = 18

= 9b - 9a = 18

= b - a = 2 ........ (ii)

By adding equation (i) and (ii),

a + b + b - a = 16 + 2

= 2b = 18

= b = 9

From equation (i),

a + b = 16

= a + 9 = 16

= a = 7

The original number 

= 10a + b

= $10 \times 7$ + 9

= 79 (Answer)