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Have you started studying Number Properties yet? Most people find this topic on the more difficult side in general, particularly the area of divisibility and prime. We did learn all of these basic concepts years ago, when we were about 8 or 10 years old \u201c number properties\u009d refers to all of the building blocks we use later in school to do algebra, geometry, and more advanced math.<\/p>\n
However, most of what we learned in school was at a much more basic level (we were only 10 after all!) and we also didn’t have to understand the number properties theory or answer questions that were anything like some of the bizarre-seeming questions we find on standardized tests.<\/p>\n
Let’s try this problem first (\u00a9 Manhattan Prep) from our GRE Number Properties Strategy Guide. Set your timer for 2 minutes.<\/p>\n
The quantity 33<\/sup>44<\/sup>55<\/sup>66<\/sup> \u201c 36<\/sup>45<\/sup>54<\/sup>63<\/sup> will end in how many zeros?<\/p>\n
(A) 3<\/p>\n
(B) 4<\/p>\n
(C) 5<\/p>\n
(D) 6<\/p>\n
(E) 9<\/p><\/blockquote>\n
Got your answer? Great, let’s get started. You want to know what the correct answer is? Let me ask you a couple of questions first.<\/p>\n
Are you confident about your answer? Did you end up having to guess? Did you give up without guessing? (If the last, make a guess right now. You can’t keep reading till you do. Well, obviously I can’t stop you, but I’m serious \u201c make a guess.)<\/p>\n
<\/p>\n
Why do I care so much that you make a guess? Because you are almost certainly going to have to guess on the real test (unless you’re going to get a 99th<\/sup> percentile score!), so start practicing right now. First, you have to get over what I call the guessing hump.\u009d People are reluctant to guess, so they try to avoid it, and they end up wasting 30 or 60 seconds continuing to try the problem, or they just aren’t sure what to guess and it takes them forever to decide. You’re guessing: it doesn’t matter what<\/em> you pick; it only matters that you pick it quickly<\/em> and move on. : )<\/p>\n
Next, while studying, you can actually learn a very useful skill called educated guessing. That’s a fancy way of saying: cross off some wrong answers, if you can, before you make your guess. You’ve just improved the odds that you might get lucky, even though you don’t know how to do the problem.<\/p>\n
Okay, let’s get back to the problem. What are we going to do with that thing? The first thing I notice is that there are many exponents and the numbers are ridiculously large \u201c this is not<\/em> one we’re going to punch into the calculator. It’d be too easy to miss one of the multiplication steps and get the wrong number (and the number might be so large that the calculator can’t handle it).<\/p>\n
So what can I do? Hmm. They ask how many times the number 0 will show up at the end of whatever this big number is. What’s an example of a number with a 0 at the end? 10. 20. 60. 100. Okay.<\/p>\n
Next, I notice that the problem has a bunch of small numbers that are either primes already or could easily be broken into primes \u201c since I’m not going to multiply this out, maybe that’ll help. How do we make\u009d our first example, 10? That’s two times five. What about 20? That’s just 10 times 2: 2 \u00c3\u2014\u00a02 \u00a0\u00c3\u2014\u00a05. 30 = 2 \u00a0\u00c3\u2014\u00a03 \u00a0\u00c3\u2014\u00a05. 100 =10 \u00a0\u00c3\u2014\u00a010 = 2 \u00a0\u00c3\u2014\u00a05 \u00a0\u00c3\u2014\u00a02 \u00a0\u00c3\u2014\u00a05.<\/p>\n
I’m starting to notice a pattern. All the examples that contained just one 2 \u00c3\u2014\u00a05 ended in one zero. Once I jumped to 100, I had two 2 \u00c3\u2014\u00a05 pairs, and 100 ends in two zeros. Interesting.<\/p>\n
Very Important Side Note: Actually, I didn’t have to figure this out when I first tried this problem about 10 minutes ago. That’s because I’d already seen something like this before, and I figured it out then. In other words, I already discovered and memorized the rule\u009d that something that ends in zero must contain at least one 2 \u00c3\u2014\u00a05 pair. Two zeros, two 2 \u00c3\u2014\u00a05 pairs. And so on. This is what you’re trying to do when you’re studying: not<\/span> figure everything out in the moment, but taking time to figure out how things work after<\/em> you’ve tried the problem and then using that new knowledge or understanding to recognize the same thing on a future problem<\/em>.<\/p>\n
Also note: remember earlier when I said I noticed that the given numbers were primes or could be broken down into primes? I was even more excited about that because I already knew that the 2 \u00c3\u2014\u00a05 combo was going to be important \u201c and that pair, of course, is made of two primes, 2 and 5.<\/p>\n
So all of this told me my next step: figure out how many 2 \u00c3\u2014\u00a05 pairs I’ve got. Oh, wait the problem did throw one curveball at me. I’ve got two numbers, not just one, and there’s an annoying subtraction sign in the middle there. I’m not sure I know a rule for this whole 2 \u00c3\u2014\u00a05 thing while also dealing with subtraction.<\/p>\n
So I decided to factor out everything possible, hoping that whatever was left in the parentheses (with the subtraction sign) would be on the small side.<\/p>\n
33<\/sup>44<\/sup>55<\/sup>66<\/sup> \u201c 36<\/sup>45<\/sup>54<\/sup>63<\/sup><\/p>\n
33<\/sup>44<\/sup>54<\/sup>63<\/sup>(30<\/sup>40<\/sup>51<\/sup>63<\/sup> \u201c 33<\/sup>41<\/sup>50<\/sup>60<\/sup>)<\/p>\n
Anything raised to the zero power is 1, so I can ignore those bits:<\/p>\n
33<\/sup>44<\/sup>54<\/sup>63<\/sup>(51<\/sup>63<\/sup> \u201c 33<\/sup>41<\/sup>)<\/p>\n
Bingo! For the first part, I can figure out how many 2 \u00c3\u2014\u00a05 pairs I’ve got. For the second part, I can just do that math on my calculator.<\/p>\n
Let’s see. In the first part, 33<\/sup>44<\/sup>54<\/sup>63<\/sup>, there are four 5s. Are there also four 2s? Yes, the 4 can be broken down into 2 \u00c3\u2014\u00a02, and since that is raised to the 4th<\/sup> power, there are many more than four 2s there. Great, so we have four 2 \u00c3\u2014\u00a05 pairs there, leading to four zeros at the end of whatever that part of the number is. What about the other part, over there in the parentheses?<\/p>\n
This is where you could pull out the calculator but we can avoid it, actually! (I hate using the calculator \u201c it’s really clunky and hard to use and half the time I don’t click a button hard enough and have to restart the whole calculation. So I avoid it whenever I can think of an easier way to do the math on paper.)<\/p>\n
All I really care about here is whether there are zeros at the end of this number, too. If the unit’s digit is zero, for example, then multiplying that number by the first part will result in 4 + 1 = 5 zeros at the end of the final number. If the unit’s digit is anything other than zero, then multiplying that number by the first part will result in 4 zeros at the end of the final number.<\/p>\n
Not sure how that works? Try it with some easier numbers. What is (10)(3)? In this example, we have one zero in the first number and the second number does not end in zero. The result is 30 \u201c one zero. What about (10)(20)? In this case, we’ve got one zero at the end of the first number and another one at the end of the second, resulting in a total of two zeros: 200. We already know we have four zeros at the end of the first number; we still need to figure out the second.<\/p>\n
First, let’s tackle 51<\/sup>63<\/sup>: we can use the same trick we used above. We have one 5 and (at least) one 2, so we’ve got a 2 \u00c3\u2014\u00a05 pair; this number will end in zero. Next, 33<\/sup>41<\/sup> = 27 \u00c3\u2014\u00a04. Only multiply the unit’s digits: 7 \u00c3\u2014\u00a04 = 28, so this number will end in 8. If we take some number that ends in 0 and subtract some other number that ends in 8, will the resulting number have zero as the unit’s digit? No. (The unit’s digit will be 2.) So now we’ve got:<\/p>\n
(number that ends in 4 zeros)(number that ends in 2)<\/p>\n
If we multiply those two numbers together, how many zeros will be at the end? Four. (Not sure? Try an example. What’s 10,000 \u00c3\u2014\u00a02? 20,000 which ends in 4 zeros.<\/p>\n
The correct answer is B.<\/p>\n
Takeaways for Number Properties:<\/h3>\n
\u00a0<\/p>\n
(1) The topic of number properties lends itself well to disguised\u009d problems \u201c it’s not going to be immediately obvious what the problem is testing or how to go about solving it. Your best defense is a strong offense: when studying<\/em> these, make sure that you are actually figuring out both the theory being tested and how to recognize<\/em> that this is what’s being tested the next time you see something similar<\/em>.<\/p>\n
(2) Get a bunch of flashcards. On one side, write When I see:\u009d and on the other write I’ll think \/ do:\u009d. When you learn that ending in zero\u009d means the number must have at least one 2 \u00c3\u2014\u00a05 pair, write that on a flashcard. The when I see\u009d side should show what you’d see in the actual problem \u201c this is the clue. The I’ll think \/ do\u009d side should tell you what to do next!<\/p>\n
(3) Drill those flashcards. Next time you see a problem talking about numbers that end in zero, you’ll recognize it and have an idea of how to start, at the very least!<\/p>\n","protected":false},"excerpt":{"rendered":"
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