If abc-bca = 792, where abc is a 3-digit number, if a+b+c = 18 and a = 9c, what is the number?
We have, abc−cba=792
abc is a 3−digit number.
Hence, abc=100a+10b+c
Similarly, cba=100c+10b+a
So, abc−bca=100a+10b+c−100c−10b−a
=99a−99c
So, 99(a−c)=792
⇒a−c=8
Also, a=9c[given]
∴ 9c−c=8
⇒8c=8
Hence, c=1
So, a=9(1)
Hence, a=9
Also, we are given, a+b+c=18
So, 9+b+1=18
∴ b=8
Hence, the number is 981.