At the end of every SAT Math section, the test makers try to come up with an extremely difficult problem that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

What is the remainder when 3^200 is divided by 5?

A) 0

B) 1

C) 2

D) 3

E) 4

At first you might think, “Hey, this is easy! I’ll just plug it into my calculator–”

3^200 = 2.656139889… x 10^95

“?”

Question mark indeed. You see, these SAT test writers are truly diabolical. They know how much you love your calculator, so they’re always scheming up ways to separate you from it, in much the same way that Loki is always trying to separate Thor from his hammer. Of course, even without his hammer, Thor is pretty strong, and don’t worry – so are you! You’re just going to have to use your other secret weapon…your brain!

…

Let me show you how. In this case, they’ve made it so that you need to know the ones digit of an insanely large number that is too big for your calculator to display on your screen. However, we can still figure out what the ones digit of 3^200 is. Are you just going multiply it out by hand? No! Of course not! That would take too long. All you need to do is recognize a pattern. Consider the first few powers of 3:

3^1 = **3**

3^2 = **9**

3^3 = 2**7**

3^4 = 8**1**

3^5 = 24**3**

3^6 = 72**9**

3^7 = 218**7**

3^8 = 656**1**

3^9 = 1968**3**

…

You may have noticed that the ones digits follow a pattern: 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1… over and over again. Every 4 powers the pattern repeats. Now we just need to figure out if 3^200 ends in 3, 9, 7, or 1. To do this, we just need to divide 200 by 4 and see what the remainder is. To see how this works, consider the following chart:

Consider: | The Power is: | Power/4 = | Remainder = | So ones digit = |

3^1 = 3 | 1 | 1/4 = .25 | 1 | 3 |

3^2 = 9 | 2 | 2/4 = .5 | 2 | 9 |

3^3 = 27 | 3 | 3/4 = .75 | 3 | 7 |

3^4 = 81 | 4 | 4/4 = 1 | 1 | |

3^5 = 243 | 5 | 5/4 = 1.25 | 1 | 3 |

3^6 = 729 | 6 | 6/4 = 1.5 | 2 | 9 |

3^7 = 2187 | 7 | 7/4 = 1.75 | 3 | 7 |

3^8 = 6561 | 8 | 8/4 = 2 | 1 |

200/4 = 50

200 divides evenly by 4, so there is a remainder of 0, which means the ones digit of 3^200 must be a 1. Any number that ends in a 1 has a remainder of 1 when divided by 5:

1/5 = 0 R1

11/5 = 2 R1

21/5 = 4 R1

Thus, the answer is choice B. If you know what to do, it takes only about 30 seconds to solve this problem. So you see, with practice, even the hardest problems on the SAT become easy. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until then, keep up the good work and happy studying!