Phew! It’s only been like three weeks since I last posted but in Internet time that’s forever. Sorry about that. I’ve got about two weeks left in my Master’s program, and I’ve been completely inundated by papers and presentations. My thesis is basically eating me alive. I seriously don’t know how I’m going to get it all done. But that’s school, right? Yay, school!
But I’m not posting to complain! I’m posting because an awesome challenge question popped into my head last night, and because I wanted to tell you what I’m planning to do as soon as school is over and I have more than a few minutes a day to devote to SAT pwning.
- A print version of the Essay Guide
- 7 Deadly Math Problems, Volume 2 (click here for Volume 1)
- Some other cool things I’ve been dreaming up that I don’t even know what to call yet
- Some videos maybe? I dunno.
The point is that things are going to pick back up around here. Soon.
And now that challenge question
As always, first correct response in the comments wins a Math Guide. Full contest rules here, but the important ones are:
- You can’t be anonymous because I need to be able to contact you
- You will have to pay for shipping if you live outside the US
- You can’t win more than once, so please refrain from answering if you’ve won a contest in the past.
- If your comment doesn’t appear on the site right away, don’t panic—you’re just getting stuck in my spam filter because you’re not registered. I receive comments in my email in the order they’re posted.
The sum of n consecutive integers is 1111. What is the greatest possible value of n?
UPDATE: Tutor Delphia (a fellow tutor?!) got it first. Solution below the cut.
This question is, I admit, a bit sadistic. That’s because until the correct solution becomes apparent, it can really feel like you’re being asked to find a needle in trial-and-error haystack. Can you find 2 consecutive integers that add up to 1111? Sure, that’s not so hard. How about 3? How about 4? How about 5? etc.
There are a bunch of consecutive integer sets that will add up to 1111, and there’s no easy algorithm to find them (although I’m sure you could come up with an algorithm if you really put your mind to it). If you’re well-versed in SAT math—and even with my challenge questions I try to adhere to SAT principles—then your spider sense should be tingling. There must be an easier way!
Well, here it is: when you’re dealing with sums of consecutive integers, and that set of consecutive integers includes both negative and positive numbers, then corresponding negatives and positives cancel out. This is obvious, once you think about it a bit:
See how the –2 and 2 cancel out, and the –1 and 1 cancel out, leaving only the 0 and the 3? The greatest number of consecutive integers that sum to 3, it turns out, is 6. Cool, right?
But what happens if we add another consecutive integer?
Now we’ve got a set of 7 consecutive integers with a sum of 7. Is that the greatest number of consecutive integers with a sum of 7? No. Not at all:
That’s right—we can construct a set of 14 consecutive integers with a sum of 7. In general, the largest number of consecutive integers that sum to a given positive integer will be twice that integer.
Therefore, the greatest number of positive integers that sum to 1111 is 2222:
There are 1110 negative integers, 1111 positive integers, and 0.
556 is the greatest possible value of n.
n + (n+1) = 1111
2n + 1 = 1111
2n = 1110
n = 555
n+1 = 556
555 + 556 = 1111
Starting at -1111 and continuing up to 1111, the consecutive integers will add up to 0. (-1111 + 1111 = 0, -1110 + 1110 = 0, etc.) That gives 2220 consecutive integers that add up to 0, the 2221st consecutive integer (1111) is then added to give a total of 1111.
Not sure if you forgot to add 1 to make your list inclusive or if you forgot to include 0 but I think your answer is one too small
Peter is so close!
I say that it is 2222
We need to start at -1110 and go until 1111 (rather than starting at -1111 which will result in a sum of zero).
Also, when counting how many numbers we have, we have to add one from the difference (1111- (-1110))+1 since we are not counting the jumps between the numbers but how many numbers there are inclusive.
Try my fav technique similar but simpler to see that easier:
-2 to 3 is -2,-1,0,1,2,3 which is 6 numbers but 3-(-2) would have only given me 5 so I have to add 1 to make it inclusive.
Nice! Another way to do this would to note the numbers we seek is (-1110+-1109+-1108+…1108+1109+1110+1111). Now, the problem becomes how many numbers are in this list. Instead of doing the inclusive, +1 mess, we can simply note that since all the numbers are consecutive, we can just add 1111 to each number in the list. The list will become: 1,2,3,…,2220,2221,2222. Now, we can clearly see that there are 2222 numbers in the list.
I’ll add one more way to think about this:
There are 1110 integers less than 0, 1111 integers greater than 0, and 1 integer that is 0.
1110 + 1111 + 1 = 2222
Yep, that’s it!
Cool! Do I get a free copy? I keep giving my copy away to my students. I need to buy a couple more in addition to the free one. I’m working towards getting a perfect score (40 points to go on the math and I got it on the reading) and I also tutor students.
Yeah man. Send me an email: firstname.lastname@example.org. We’ll coordinate.
I just have to say: I love Beeker. Big thumbs up.
This comes straight from a question out of first edition of the Blue Book: test 1, section 3, problem 20. Unfortunately, this test is now out of print and no longer available. It’s a really cool test showcasing all sorts of key SAT math strats.
Do I get a free copy of your book for having awesome research skills?
Oh well. Great site, anyway!
Took me a while to track my copy down, but yes—there are similarities. However, I’d argue that my question above is much harder than the one in the 1st edition Blue Book. The question in the Blue Book gives an end point, which makes the important leap (that the sum of the consecutive integers from –25 to 25, inclusive, is 0) a fairly easy one to make. By giving only the sum and asking for the greatest value of n, I’m presenting something worthy, I believe, of being called a challenge question. 🙂
aaah… If you restrict them to positive integers, the answer is 22 between 40 …. 61.
Took forever to get that. ;(
Your challenge questions are great, but your solutions are typically unsatisfying or, in cases like this, even dishonest! Here, your solution isn’t really a solution at all: you haven’t *proven* why 2222 is the greatest possible value of n, and that’s the real heart of the question. The question didn’t say: can you think of a really big list of integers that sums to 1111? You’re leading students to believe that, logically, your solution is complete and representative of a full understanding of the problem, when it’s not. You’ve merely exhibited a list of 2222 integers that sum to 1111. The hard part of the question is — how do you know 2222 is the *greatest* value of n? You might say it’s obvious, but that doesn’t teach students anything, nor is it in any way illuminating. Even if you don’t know how to answer that question, you should have mentioned that your solution, as written, is incomplete, and encouraged students to think about why n can’t be larger. (Here is a sketch of a solution: clearly n must be smaller than 2222 if all the integers have the same sign: if they are all positive or all negative. That’s “clear,” because the sum of 2222 consecutive positive integers is obviously greater than 1111, because the set would have to include 1111, and the sum of 2222 consecutive negative integers would have to be negative, which violates the requirement that the list sum to 1111, a positive integer. So it must be that some of the integers are positive and some are negative. Let m stand for how many are negative, in which case the list starts at -m. Then the sum of the first m+1+m=2m+1 integers in the list will give zero, leaving n-(2m+1) positive integers which we want to sum to 1111. Using the sum formula for an arithmetic series, you get that (1/2)(n-(2m+1))(m+1+m+n-(2m+1))=(1/2)(n-(2m+1))(n)=1111. Doubling both sides gives n(n-(2m+1))=2222. Now the heart of the matter is clear: this is actually a DIVISIBILITY problem in disguise. To maximize n, which must be positive by definition, we should minimize n-(2m+1), which means we should make it equal to 1. So m=1110, and the greatest possible value of n really is 2222.)
Thanks for your feedback. I think “dishonest” is a bit harsh, but I appreciate your candor, and your sketch proof, which is quite good. I hope anyone else who finds my brief explanation unsatisfactory scrolls down to find your comment.