Today’s ACT math question of the day is about dividing with remainders.  The remainder, you may or may not recall, is the part “left over” when we divide one integer by another.  For instance, if I divide 4 by 3, I get 1 with 1 left over, so my answer is 1 remainder 1 or 1R1.

One more practice round before we go on to the problem: 14/6 = ?

2R2.  Good job.

Moving on, we are asked to state which of the given numbers is divisible by 3 with no remainder.

Method 1: plug each number into your calculator – carefully! – and divide by 3.  The one with no decimal or other mess in the answer is the correct choice!

Method 2 (no calculator): First, eliminate A, B, and E because the first two digits are evenly divisible by 3 but the second two digits are not, meaning that we will have a remainder. (27/3 = 9, 42/3 = 14, 81/3 = 27).  Now, try C: 44/3 = 14R2, 28/3 = 9R1, 12/3 = 4.  C divides evenly, so we can stop, bubble in C, and move on to the next question.

Method 3: Start with method 2 (to eliminate A, B, and E), then check C and D on your calculator.  By eliminating choices that are definitely wrong before you being your calculations, you can save valuable time.

Method 4: Recall that a number’s digits must sum to a number that is evenly divisible by 3 in order for the number itself to be evenly divisible by 3.  Add the digits of each number; only C (4 + 4 + 8 + 2 = 18) works.

There is absolutely nothing wrong with using your calculator on the math section, but there is nothing wrong with applying the right shortcuts (or divisibility rules) if you see opportunities to do so!