anybody good with java?

My assignment is as follows=
write a program that:
-prompts the User for a 2 digit positive integer that does not end in zero
-displays the sum of the numbers divisible by the second digit that are less than the imputed number

Example:
User enters 25, outputs 50 (5+10+15+20)
enter 26, outputs 60 (6+12+18+24)
enter 11, output 55 (1+2+3+4+5+6+7+8+9+10)


Basically, I forgot how to seperate the two digit integer. I know there's something you can do with the string class, or dividing by a multiple of 10, I just cant for the life of me remember what :/
other than that, it should just be a loop. I'm blanking so bad though.
Would something like this work? I'm using the 25 example

//num is the imputed integer
for(I=1; I<=divisors; I++)
// divisors is the quotient of num and the second number, an integer.
{
multiple=secondNum*I;
System.out.println(multiple);
sum=sum+multiple;
}

I know I skipped a lot, but basically secondNum would have the second number. Multiple would be the second number times I, which is incrementing from 1 to the amount of times secondNum goes into the imputed number. Sum is the easy part.
basically, all I need to know is how to get the god damn secondNum

Tip, ask Khult

But you want to know how to separate the two digit integer? What comes to mind is to convert it into a string and use substring. Or wouldn't getCharArray or whatever work?

I'm sure there's something on Google but I'm too lazy and don't want to hand you code that I don't understand. Khult knows a LOT about Java though. I always go to him for help.

I remember something about char array, just not enough to actually use it. The last one we did like this our teach divided by like 100000000, but that was to seperate a phone number. Gaahh, I should of wrote that down, lol

Got it! It was modular division by ten.
Soooooooo,

Sop("please give me a positive number not ending in zero");
//sop is short for system.out.println, cuz I'm lazy ;D
Int inputNum=input.nextInt();
int secondNum=imputNum%10;
int divisors=(inputNum/secondNum)-1;
int multiple=0;
int sum=0;
int I=0;
for(I=1; I<=divisors; I++)
{
multiple=secondNum*I;
Sop(multiple);
sum=sum+multiple;
}
Sop(sum);

Nice! Didn't think of that.