Let’s consider both simple and complex applications of remainders. The most straightforward scenario is for the question to ask what the remainder is in a given context.Let’s start by looking at an official Data Sufficiency question of moderate difficulty:
What is the remainder when x is divided by 3?
1) The sum of the digits of x is 5
2) When x is divided by 9, the remainder is 2
Pretty straightforward question. In Statement 1, we could approach by simply picking numbers. If the sum of the digits of x is 5, x could be 14. When 14 is divided by 3, the remainder is 2. Similarly, x could be 32. When 32 is divided by 3, the remainder will again be 2. Or x could be 50, and still, the remainder when x is divided by 3 will be 2. So no matter what number we pick, the remainder will always be 2. Statement 1 alone is sufficient.
Note that if we know the rule for divisibility by 3 – if the digits of a number sum to a multiple of 3, the number itself is a multiple of 3 – we can reason this out without picking numbers. If the sum of the digits of x were exactly 3, the remainder would be 0. If the sum of the digits of x were 4, then logically, the remainder would be 1. Consequently, if the sum of the digits of x were 5, the remainder would have to be 2.
Again, in Statement 2, we can pick numbers. We’re told that when x is divided by 9, the remainder is 2. To quickly generate a list of numbers that we might test, we can start with multiples of 9: 9, 18, 27, 36, etc. Then, we can add two to each of those multiples of 9 to get the following list of numbers: 11, 20, 29, 38, etc. All of these numbers will give us a remainder of 2 when divided by 9. Now we can test them. If x is 11, when x is divided by 3, the remainder will be 2. If x is 20, when x is divided by 3, the remainder will be 2. We’ll quickly see that the remainder will always be 2, so Statement 2 is also sufficient on its own. The answer to this question is D, either statement alone is sufficient. That’s not too bad.
Take this more challenging question:
June 25, 1982 fell on a Friday. On which day of the week did June 25, 1987 fall. (Note: 1984 was a leap year.)
Consider a very simple case. Say that June 1 is a Monday, and I want to know what day of the week it will be 14 days later. Clearly, that would also be a Monday. And if I asked you what day of the week it would be 16 days later, you’d know that it would be a Wednesday – two days after Monday. Put another way – because we’re dealing with weeks, or increments of 7 – all we need to do is divide the number of days elapsed by 7 and then find the remainder in order to determine the day of the week. 16 divided by 7 gives a remainder of 2, so if June 1 is a Monday, 16 days later must be 2 days after Monday.
Suddenly the aforementioned question is considerably more approachable. From June 25, 1982 to June 25, 1983 a total of 365 days will pass. 365/7 gives a remainder of 1, so if June 25, 1982 was a Friday, June 25 1983 will be a Saturday. From June 25, 1983 to June 25, 1984, 366 days will pass because 1984 is a leap year. 366/7 gives a remainder of 2, so if June 25, 1983 was a Saturday, June 25, 1984 will be 2 days later, or Monday. We already know that in a typical 365 day year, the remainder will be 1, so June 25, 1985 will be Tuesday, June 25, 1986 will be Wednesday and June 25, 1987 will be Thursday, which is our answer.
Takeaway: the challenge isn’t necessarily that questions are asking you to do difficult math, but that it can be hard to figure out what the questions are asking you to do. When you encounter something that seems unfamiliar or strange, remind yourself that virtually every problem you encounter will involve the application of a concept considerably simpler than the nebulous wording the question might suggest.