Answer :
Let x, y and z be the digits in a three-digit number A = xyz.
Reversing the digits, we get B = zyx.
Sum of the digits in A is (100x + 10y + z).
Sum of the digits in B is (100z + 10y + x).
Difference S = A + B = (100x + 10y + z) - (100z + 10y + x).
S = 100x-x + 10y-10y + z-100z = 99x-99z
99(x-z) = 9x11(x-z)
The number is divisible by 9 and 11.
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
view all courses
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation


RELATED QUESTIONS :
Fill in the blank
NCERT - Mathematics ExemplarFill in the blank
NCERT - Mathematics ExemplarIf abc is a three
NCERT - Mathematics ExemplarLet abc be a thre
NCERT - Mathematics ExemplarThe sum of all nu
NCERT - Mathematics ExemplarFill in the blank
NCERT - Mathematics ExemplarFill in the blank
NCERT - Mathematics ExemplarFill in the blank
NCERT - Mathematics ExemplarFill in the blank
NCERT - Mathematics ExemplarLet abc be a thre
NCERT - Mathematics Exemplar