An algebra problem by Aashish Chaudhary

Now, we let the 3-digit number be $\overline{abc} = 100a+10b+c$

The digits are in an AP, which would be $a,b,c,\ldots$

This gives us $d=c-b=b-a \implies 2b=a+c\implies$ Eq.(1)

The sum of the digits: $a+b+c = 15\implies$ Eq.(2)

Substitute Eq.(1) into Eq.(2):

$2b+b=15 \implies b = 5$

Substitute the value of $b$ into Eq.(1):

$a+c=10\implies$ Eq.(3)

The reverse of the number = $\overline{cba} = 100c+10b+a$

Now, we know that the reversed number is $594$ less than the original number

$100a+10b+c - (100c+10b+a) = 594\\ 99a-99c = 594\\ 99(a-c) = 594$

$a-c=6\implies$ Eq.(4)

Eq.(3) + Eq.(4):

$a+c+(a-c) = 10+6\\ 2a=16 \implies a=8\\ 8+c=10 \implies c=2$

The original number $\overline{abc} = \boxed{852}$

I knew that the last digit of the difference, 4, would cause carrying over in subtraction since the difference is greater than 499 (if that makes sense). Thus, the last digits could only be 3 & 9, 2 & 8, 1 & 7, and 0 & 6. Making those numbers' digits equal 15 would make the necessary original numbers 933, 852, 771, and 690. Arithmetic progression only allows for 852 to work, thus making it the answer.

Let the digit in the form of xyz Then x+y+z= 15 And y=(x+z)÷2 And 100x+10y+z=100z+10y+x+594 By eq- x+z=10, Y=5, Then 99x-99z=594 X-z=6 Then as x+z=10 and x-z=6 X=8 Y=5 Z=2 so the no. Is 852