The logic is much easier if the number of small bricks is LESS than 5.

Suppose you have 21 small and 2 large. You actually have 1 small and 6 large,

that is, 31 bricks. 31 = 21 + 2*5 = 1 + 6*5

So you divide the number of small bricks by 5 and transfer that amount to # of large bricks.

# of small bricks = # of small bricks MOD 5, so that you get a TRUE INVENTORY of number

of small bricks LESS than 5.

Then to determine if the inventory is sufficient, you divide the size by 5 and compare that against

# of large bricks and then compare size MOD 5 to # of small bricks.... Here's the code

class Bricks

{

public int makeBricks( int numSmallBricks, int numLargeBricks, int size)

{

int iReturn=0;

//fixes the inventory so that numSmallBricks < 5

numLargeBricks += (numSmallBricks/5);

numSmallBricks %= 5;

if (

((size / 5) <= numLargeBricks) &&

(( size % 5) <= numSmallBricks)

)

{

iReturn = 1;

}

return (iReturn);

}

public static void main(String args[])

{

Bricks myBricks = new Bricks();

System.out.println(myBricks.makeBricks(3,2,10)); //yes

System.out.println(myBricks.makeBricks(10,0,10)); //yes

System.out.println(myBricks.makeBricks(5,2,25)); //no

}

} //class

//end of code

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