Any integer that divides into another integer evenly is a factor. For example, 3 and 2 are factors of 6, 12, 18, etc. The number 16 has 2, 4 and 8 as factors. The number is always a factor of itself. The number one is a factor for all whole numbers.
Now we’ll review our SAT Test Quick Division Answers. This article series contains the following articles related to Quick Division on the SAT Test:
1. Which of the following numbers can be evenly divided into 13,548,915,486?
Note how tedious it would be to divide 13,548,915,485 by each of the answer choices. That must not be what the SAT is after. First start by eliminating the easy ones, (A) 2 and (E) 10, since the number is not even nor does it end in 0. Next, notice that you have two variants on the “3” rule – 6 and 9. Since the number is odd (ends in 5), we know 6 doesn’t work, so you can eliminate (C). You’re left with 9 and 4. My opinion is that the “4” rule is easier to apply since it only involves the last two digits. But the last two digits add up to 13 which is not divisible by 4, so eliminate (B).
If this were the actual SAT, you would mark (D) and move on, even without testing it. You’ve eliminated all the other choices and one of them has to be right, so why take the time to test it? But for the purpose of study, let’s look at (D) more closely. The “9” rule says add all the digits and see if the sum is divisible by 9.
1 + 3 +5 + 4 + 8 +9 + 1 + 5 +4 + 8 + 6 = 54 which is divisible by 9 (9 x 6).
2. A crate has 3,448 bolts. The bolts could be evenly divided into which of the following number of boxes?
This is just another way of asking a question similar to the first one. You can immediately eliminate (C) 5 and (E) 10. Try the “3” rule next because it applies to two answer choices, (A) and (C). 3 + 4 + 4 + 8 = 19, which is not divisible by 3. Since it is not divisible by three, it will not be divisible by 6 either, so you can eliminate both (A) and (C).
The answer must be (D), but let’s see why. The last three numbers, 448, are divisible by 8. Notice how this is a hard rule to test, so we left if for last. It was easier to quickly eliminate the others.
3. A box has 63 red blocks, 47 blue blocks and 95 yellow blocks. If you added 1 more of each type of block to the box, each of the three colors of blocks could be evenly divided among a group of children. What is the greatest possible number of children in the group:
This one is tricky. First, make a list of the numbers + 1, as the question stem asks you to add one block to each color group:
Red = 64
Blue = 48
Yellow = 96
The “8” rule jumps out since there are three multiples of 8 in the answer choices, (A) 8, (D) 16 and (E) 32. And looking at the number of blocks, they are all divisible by 8. However, the question is asking for the largest group, so we have to see if the answers larger than 8 work.
Since the question is asking for the greatest possible number of children, start with the highest answer choice, (E) 32. If you double 32 you have 64, which is the amount of the red blocks, but too many for the blue blocks. In other words, you could evenly divide 64 (each child would get two blocks) but not 48. So eliminate (E).
How about (D) 16, the next most logical answer? Since we know each set of blocks is divisible by 8, possibly 16 could work. And when we do the calculations 16 is the answer: 64 ÷ 16 = 4; 48 ÷ 16 = 3; 96 ÷ 16 = 6.
Now return to our SAT page to read another lesson.