There are 2 two digit numbers such that one has been obtained by reversing the digits of the other. On subtracting the smaller number from the larger number, I get a number divisible by
Let the number be ab.
Here we have 'a' in tens place and 'b' in units place. So we can write the number as 10a + b. If we reverse it we get 10b+a.
(10a + b) – (10b + a) = 10a + b – 10b – a
= 9a – 9b = 9(a – b).
Hence the number is divisible by 9.
Additionally the resulting number will also be divisible by factors of 9 which is 1 and 3 in this case.
Hence the number obtained will be divisble by 3 and 9.