What percent of the first twenty integers in the following series is less than 1,000?

-1, 8, -27, 64…

(A) 30%

(B) 45%

(C) 50%

(D) 70%

(E) 75%

**Approach**

The following question can seem daunting. First off, we need to decode what the question is asking. Then, we need to find a way to quickly answer the question. A mistake many students make is thinking that writing anything out is a complete waste of time. You may be tempted to think the same, especially in a problem like the one above that mentions twenty integers in a series. Remember, however, that there is always a pattern. To really see that pattern, you will have to write out a bit of the problem.

First off, the problem is asking for the number of integers less than 1,000. We already have three (-1, 8,- 27, 64). You will notice that this sequence cubes each off the consecutive integers, starting with -1. Then, it alternates with a negative sign.

Now, let’s continue the sequence: -1, 8, -27, 64, -125, 216…

At this point, you may already be thinking, “How did you know that was -216?” Good question. The key is you don’t really need to know that. We simply need to know if the number is less than 1,000. You may not know that 6^3 = 216, but it surely less than 1,000. You may also notice that it has a negative sign in front of it, so it has to be less than 1,000. In fact, half of the numbers in the series are negative (remember the sign is alternating, with the negative in front of the odd numbers cubed), so at least half of the numbers have to be less than 1000. This insight helps you automatically get rid of (A) and (B).

**Almost There…**

Now, we just have to find out what numbers are less than 1,000. Well, you should know that 10^3 = 1,000 (it is an important arithmetic fact that you should tuck away for safekeeping). Every even number including 10 will not be less than 1,000 (e.g. 12^3 > 1,000, 14^3). 1-9 (but not 10, because 10^3 is not less than 1,000; it is 1,000). Then, we have the odd numbers above 10: 11, 13, 15, 17, 19. That yields 14 out of 20 or 14/20, which works out to be 70%. Answer (D)

What I’ve tried to do with this question is walk you through my thought process. Now, let’s see if you can apply any of the above to the problem below. Good luck!

Set A consists seven consecutive digits. Which of the following could be the average of the largest and smallest terms of the sequence?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

(F) 5

(G) 6

(H) 7

(I) 8

(J) 9

**Approach**

Obviously, this question seems horrible, and you could spend an entire hour racking your brain just to come up with the correct response. The answer, though, isn’t nearly that difficult, and only requires a little logic. First off, what is the question asking? Ostensibly, it states that, “the average…largest and smallest numbers…”. By experimenting with numbers 1, 2, 3, 4, 5, 6, 7, you can see that the average of the sum of 1 and 7 is 4, the middle number of the sequence.

However, many student never simply write out a sequence above, thinking, “Oh no, that will take way too long.” Meanwhile, they spend a minute looking at the computer screen, re-reading the question and hoping that some solution will magically fall from the air.

Usually, this won’t happen. However, writing out the series (it is only seven numbers, after all) a second time, 2, 3, 4, 5, 6, 7, 8, will show you that the middle number always equals the average of the smallest number and largest number.

Now, we can reword the question: which of the following numbers could be the middle digit of a series of seven consecutive integers?

Well, any of them. Note the question doesn’t say the integers have to be either positive or negative. They are simply consecutive. So, even zero could be the middle number, -3, -2, -1, 0, 1, 2, 3.

*Takeaway*

*Always write out enough of the problem to find a pattern.*

*Don’t worry, you usually won’t have to write too much. A pattern should reveal itself.*

*Dispel the notion that writing anything out is a waste of time. As to what writing “just enough” is, you’ll get the hang of it after you do several problems. *

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