Write the digits from 1 to 9 in a line. If you put times signs after the 2 and 4, a plus sign after the 5, and a minus sign after the 7, you have 12 x 34 x 5 + 67 - 89, which equals 2018. That's six years off from our current year 2012. This example uses four arithmetic symbols. The object is to use just three of the following arithmetic operations: addition, subtraction, multiplication and division, in a line from 1 to 9 to get 2012 exactly. The operations should be performed in order from left to right. There are no tricks to this puzzle. Can you do it?Well, first, let's note that the example originally provided with this puzzle (you can read that in our Sunday post) is different from this. Click on the link above to see the correction in the puzzle.

Enh, I don't much care. Here's our solution:

1234 - 5 - 6 + 789 = 2012

I'll find some photos:

MINUS

MINUS

PLUS

EQUALS

Easy peasy! (Okay, we admit it: we traveled on Sunday, and it wasn't until we were poking at it while waiting for our flight that we solved the sucker.)

Time for ...

Here are this week's picks:

Fewer than 50 51 - 100 -- Skydiveboy 101 - 150 -- DaveJ 151 - 200 -- Dave 201 - 250 -- EKW 251 - 300 -- Ross 301 - 350 -- Magdalen 351 - 400 -- Curtis 401 - 450 451 - 500 501 - 550 -- David 551 - 600 601 - 650 651 - 700 701 - 750 751 - 800 801 - 850 851 - 900 901 - 950 951 - 1,000 | 1,001 - 1,050 1,051 - 1,100 1,101 - 1,150 1,151 - 1,200 1,201 - 1,250 1,251 - 1,300 1,301 - 1,350 1,351 - 1,400 1,401 - 1,450 1,451 - 1,500 1,501 - 1,550 1,551 - 1,600 1,601 - 1,650 1,651 - 1,700 1,701 - 1,750 1,751 - 1,800 1,801 - 1,850 1,851 - 1,900 1,901 - 1,950 1,951 - 2,000 | 2,001 - 2,050 2,051 - 2,100 2,101 - 2,150 2,151 - 2,200 2,201 - 2,250 2,251 - 2,300 2,301 - 2,350 2,351 - 2,400 2,401 - 2,450 2,451 - 2,500 2,501 - 2,750 2,751 - 3,000 3,001 - 3,250 3,251 - 3,500 3,501 - 4,000 4,001 - 4,500 4,501 - 5,000 More than 5,000 More than 5,000 and it sets a new record. |

Our tie-break rule:In the event that a single round number is announced, AND two separate people picked the ranges leading uptoand leading upfromthat round number, the prize will be awarded to whichever entrant had not already won a prize, or in the event that both entrants had won a prize already or neither had, then to the earlier of the two entries on the famous judicial principle of "First Come First Serve," (or in technical legal jargon, "You Snooze, You Lose").

## 7 comments:

Is there a way to do this in some sort of logical or mathematical way, instead of just using random educated guesses or writing a brute-force program? How did you come up with your solution?

WHAT!!!!

Is this the Twilight Zone?

"And, loyal listeners, don't forget to check back with our website at least twice a day until the deadline to make sure you have the latest version of the challenge."

Numbing.

If Will keeps making all these puzzle errors he will end up taking it in the Shortz.

A coworker of mine determined, after running the problem through a computer, that it is impossible to do the problem using three different operations once each. I'm not saying he's definitely right, but he's usually pretty on top of things.

If he's right, I'm more than a little upset because I happened to have heard this problem on the radio and worked on it for a bit. But I operated under the assumption that three

differentoperations were to be used. (I did allow for the possibility that they could be repeated, though.)Would it have been too hard to make the problem clear?

Sigh. I give up. Again.

Phil

Phil,

I am frequently unhappy with the way some of these puzzles are stated, but I really do not understand your complaint as it was very clear that a sign could be used more than one time from the example given. Other than that there are three distinct operations in the solution.

On the other hand there was a major error in the example given that, however, does not affect this puzzle.

I think the original example was correct, given the standard order of operations, and since it came from Ed Pegg, Jr. See, for example, here: http://math.about.com/library/weekly/aa040502a.htm

Of course, we are not all mathematicians, so the original puzzle could have been clearer (and as I pointed after Magdalen's original post, you could get 2012 using 4 operations, assuming the standard order). If you tried the original example, you could have figured it out (probably).

Answering Seth, the first thing I did was to see how close to 2012 I could get using consecutive digits. 1234 gets you 778 away, which is close to 789. Adding 1234 to 789 gave me 2023, which was 11 high with a 5 and 6 to play with, , which is two subtractions. So for me for this puzzle, the solution was straightforward.

Other things you can do is look for prime factors, in this case 2, 2 and 503. (If 2012 was divisible by 9, or 89, or 789, it would give you a possible clue as to where a final multiplication could be.) In fact, I think that the fewer the number of operations, the simpler the answer has to be.

I'm guessing Phil's coworker was correct, as there are only 21,504 possible solutions, a bit much for trial and error, but easy for a computer programmer with the necessary skills (not me).

The gnu puzzle is up and I will go with 1251 this week.

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