/* For this problem, we'll round an int value up to the next multiple of 10 * if its rightmost digit is 5 or more, so 15 rounds up to 20. Alternately, * round down to the previous multiple of 10 if its rightmost digit is less * than 5, so 12 rounds down to 10. Given 3 ints, a b c, return the sum of * their rounded values. To avoid code repetition, write a separate helper * "public int round10(int num) {" and call it 3 times. Write the helper * entirely below and at the same indent level as roundSum(). */ public int roundSum(int a, int b, int c) { return round10(a) + round10(b) + round10(c); } public int round10(int num) { int rd = num % 10; if(rd >= 5) return num + 10 - rd; return num - rd; }