Question

Let N be the sum of the digits of a natural number A, let B = A+N, Let A’ be the sum of the digits of the number B, and let C = B+A’. Find A if the digits of C are those of A in reverse order.

Answer 1

@amistre64

Answer 2

i got nothing on this one ....

Answer 3

ok sir then Finally I Must close this question :'(

Answer 4

B-A =N
C-B = A'
C-A =N + A'
we know C-A is a multiple of 9
logically speaking, i'd say A is a 2 digit number as i think it is predictable that A can not be of more than 2 digits .
now N + A' should be a factor of 9
assuming A= 10a+b
also assuming a+2b<10 for now

now B = 11a + 2b = 10a + (a+2b)

C = 13a + 4b

we are given
4b + 13a = 10b+a
b=2a

a=1 and b=2 fit in i.e A=12

it was purely accident that i got the ans :|

Answer 5

Thank you sir