Question

in a three digit number the digit at ones place is equal to the sum of digits at tens and hundreds place. the sum of all digits is 10 and if the number is subtracted from the number formed after reversing the three digits,the difference is 297. find the number ​

Answer 1

Answer:

let the digit at hundred's place be'x' , ten's place be'y' and one's place be 'z'

the original number is 100x+10y+z.

reverse order is 100z+10y+x

according to the question,

z = x+y -----------eq(1)

x+y+z = 10

from eq (1),  

x+y+x+y = 10         (x+y=z)

2x+2y = 10 -----------eq(2)

divide the eq(2) by 2,    

x+y = 5

we know that x+y = z so,    z=5

according to the question,

100x+10y+z - [100z+10y+x] = 297

100x+10y+z-100z -10y-x = 297

99x-0y-99z= 297

99x - 99×5 = 297              (z=5)

99x - 495 = 297

x= 297+495/99 = 8, x=8

we know that x+y=5

8+y=5, y= -3

x=8, y= -3, z=5

hope this helps