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:
(–2) + (–1) + 0 + 1 + 2 + 3 = 3
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?
(–2) + (–1) + 0 + 1 + 2 + 3 + 4 = 7
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:
(–6) + (–5) + (–4) + (–3) + (–2) + (–1) + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 7
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:
(–1110) + (–1109) + … + 0 + … + 1109 + 1110 + 1111 = 1111
There are 1110 negative integers, 1111 positive integers, and 0.
1110 + 1111 + 1 = 2222