Lets assume the number is xyz

That can also be written as 100*x  +10*y + z

Now the sum of the digits of a three-place number is 16 i.e. x+y+z = 16 ......(i)

and if the digits are reversed and the resulting number is added to

the original, number the sum is 1049.

This can be written as xyz + zyx =1049 or

 100*x  +10*y + z + (100*z+10*y+x) = 1049 or

101*x + 20*y + 101*z = 1049   .............(ii)

Again if the resulting number is subtracted from the original number

the difference is 297.

This can be written as xyz - zyx =297or

100*x  +10*y + z - (100*z+10*y+x) = 297 or

99*x  - 99*z = 297 ......(iii) or

99* (x-z) = 297 i.e. x-z = (297/99)

=> x-z = 3 or x = z+3

Replacing the value of x in (i) and (ii)

x+y+z = 16 becomes z+3+y+z = 16 or y+2z = 13 or

y = 13 - 2*z.......................(iv)

101*x + 20*y + 101*z = 1049 becomes 101*z + 303 + 20*y + 101*z = 1049 or

=>  202*z+20*y = 746 

Using value of y from (iv)

202*z + 20* (13-2*z) = 746 or

162*z + 260 = 746

z= (746-260)/162

z = 486/162 =3

now we know x = z+ 3 and y = 13-2*z

so

x = 6

y = 7

z = 3

XYZ = 673

Check -

673+376 = 1049

673-376 = 297