Step-by-step explanation:
First of all, we can easily find that, for example, 63, has that particular property, therefore Jacinto is wrong.
Now we proceed to show that every (two digit) number is divisible by 3 if and only if it's sum of digits is.
Let n be a natural two digits number, and let a and b be the digits of said number, with
We can write n as
Let's call r1 the remainder of 10a/3, with, of course,
(for example, if a=1 then r1=1, because 10=3*3+1)
Now, each power of 10 leaves a remainder of 1 when divided by 3, thus r1 is equal to the remainder of a/3.
As an example, if a=3, then r1=0, because 30=3*10.
And let's call r2 the remainder of b/3.
In order for n to be divisible by 3, r1+r2 must be equal to a multiple of 3 (otherwise the division would leave a remainder).
Now, a can be written as
And similarly, b can be written as
Therefore, a+b is equal to (a multiple of 3)+r1+r2.
But since we already showed that r1+r2 needs to be equal to a multiple of 3, we have that 3 divides n if and only if the sum of its digits is a multiple of 3.
With a similar reasoning we can find that the same holds for 9.
The largest two digits multiple of 9 is 99, but it doesn't work, because 9+9=18, and 99 is odd.
Next up we have 90, which is of course good, since 90/9=10.
It is the largest two digits integer to have this property, since it's easily checked that others do not work