Another Puzzle
Trey Klein sent me another puzzle…
“Find the missing 13th number in this sequence:
77, 341, 923, 1547, 608, 2116, 377, 2263, 518, 1394, 3182, 1645, _____ , 944, 4636, …”
Anyone who solves this, I’d like you to post the sequence of tactics and ideas by which you solved it, in the comments. (Yes, I have solved it, and I’ll put my method in a comment, too.)
COMMENTS WILL CONTAIN SPOILERS, so don’t look unless you want the secret. Trey offered some hints, but I’m against hints, so I won’t publish them.
Well, I’m sure I have it, but there’s one detail I can’t explain. I think the sequence is, well, wrong. I’ll ask that question at the end.
1) first thing I often look at for puzzles like this is to see if there is any odd/even number pattern, which could provide a clue. There wasn’t anything obvious; there’s some clustering, and at a quick glance I would guess that the missing number is odd, but I wouldn’t bet on it. Those first four being odd numbers makes it uncertain.
2) next thing I usually look at is the sequence (D) formed by the difference between successive numbers in the original sequence (S), which could provide a clue:
264, 582, 624, -939 … hey – the first four are all multiples of 3 …, 1508, … rats. But look: even though (D) isn’t always a multiple of 3, (S) is NEVER a multiple of 3. Interesting!
3) okay, but I do notice something else about (D) so far: the absolute value of the first five elements is increasing: … -1739, 1886, -1705 … rats.
4) okay, back to the multiples of 3 issue. The first five (S) are all of the form 3k+2. Then we have a 3k+1 (or should I say 3k-2 ?), a 3k+2, a 3k+1, three more 3k+2 … this isn’t helping.
5) well, although there are obvious declines in (S), the overall trend is towards larger numbers. The local minima and maxima keep increasing. What about factoring in the position in the sequence?
77 = 1*77, 341=2*170+1, 923=3*307+2 … okay, so is the next one 4k+3? …, 1547=4*386+3 … yes it is! …, 608=5*121+3 … rats.
6) could the elements of (S) actually be some other, monotonically increasing sequence, modulo some number larger than 4636? That’s consistent: the declines always occur but I can’t think of any clear way to extract that; I don’t even see a clear pattern yet for the first four numbers. Also in the sequence there is 923, 1547 and then later 944, 4636; I think figuring it out would be very difficult.
7) well, the first two numbers are also divisible by 11, maybe there is some pattern there: 11k+? … no, not that I can see.
is there some pattern to the local minima and maxima? … again, not obvious. 2116 and 2263 are very close. If there is a pattern, I don’t think I have enough info to confirm it: I don’t know if the missing number is small or large.
9 FTW) just what *are* the factors of these numbers?
They look like composite numbers:
7*11, 11*31, 13*71, 7*13*17, (2^5)*19, (2^2)*(23^2) … weird …, 13*29, 31*73 … composite numbers with relatively low-valued factors …, 2*7*37, 2*17*41, 2*37*43 … whoa, those last 3 were very similar, 5*7*47, the missing number, (2^4)*59, (2^2)*19*61. Okay, those last bunch are very suggestive: their largest factors are all consecutive primes, with the missing one fitting nicely. Does that sequence of primes work backwards through the rest? … YES it does, down to 7 as the lowest prime used. OKAY.
So then, the missing number is a multiple of 53.
If we remove that sequence of consecutive primes out of (S), then what does the modified sequence (just the coefficients) look like?
11,31,71,91,32,92,13,73,14,34,74,35,__,16,76.
WELL. The last digit is 111122334445_66. So that’s clear. There’s our odd/even clustering. And the first digit is only from {1379}, the digits relatively prime to 10. The missing coefficient is probably 75 or 95. But some numbers are skipped: 12, 72, 33, … oh of course, the multiples of 3! Fine. So the missing coefficient is bound to be 95. Which means the missing number is 53*95, or …
5035.
The largest number in the sequence given; amusing. I wonder if it would have been easier to solve the puzzle if a smaller number were left out instead.
… but why was 94 skipped? It’s not a multiple of 3. Where we currently have 1645=47*35, I would expect 47*94. Of course, 94 is 2*47, but I can’t see why that is a problem, since we had 2116=23*92 earlier, where 92 is 4*23. So why skip 94?
On this one point, I am stumped. I’m confident the missing number is 5035. But I think there’s a bug in the sequence.
[James' reply: There is no bug in the sequence. When you solve the puzzle, everything will make sense.]
Comment by Zac Thompson — March 15, 2010 @ 4:07 am
oops, in 6) above I meant to say “the declines always occur before an odd/even switch, which would happen if (S) were modulo an odd number, but I can’t think of any clear way to extract that.”
Comment by Zac Thompson — March 15, 2010 @ 4:10 am
Oh, I see. It’s just adjacent primes with the digits of the larger one reversed. Much simpler. The missing multiples of 3 should have helped me there, along with the first digit being relatively prime to 10. 49 is composite, so 94 is not used as a factor. The missing number is still 5035, of course. Good puzzle.
Lots of cool puzzles in the archives of the WPC here:
http://wpc.puzzles.com/history/index.htm
[James' reply: Excellent! Yeah you were just one tiny insight short. When I solved it, I noticed the reversal once I separated the factors, but only for the first two terms, because after that there were too many factors. But when I thought to combine the factors other than the consecutive primes together that part of the pattern became clear.]
Comment by Zac Thompson — March 15, 2010 @ 10:40 pm
Respected Sir,
On the basis of my calculations the answer is 5035.
My Approach: I started with playing with numbers. I tried different mathematical operator to get some meaningful result but failed. I was gazing at numbers with hope that they would give some hint and they did. After watching the number, I decided to find the factors of each number in the series and here the answer was lying.
After factorization it looks like:
77 = 7*11,
341 =11*31,
923 = 13 * 71,
1547 = 17 * 91,
608 = 19 * 32,
2116 = 23*92,
377 = 29 * 13,
2263 = 31 * 73,
518 = 37 * 14,
1394 = 41 * 34,
3182 = 43 * 74,
1645 = 47 * 35,
x = y * z,
944 = 59 * 16,
4636 = 61 * 76
After getting this, it was easy to find y and z. y is reverse of second factor of previous number and z is the reverse of first factor of next number. Hence in the case y = 53 and z = 95 so x = 5035.
I am sure I got the right answer but still waiting for your approval.
With Regards
Mohit Verma
[James' Reply: Good job, Mohit. You are correct. Ah the power of play!]
Comment by Mohit Verma — March 22, 2010 @ 4:06 am
Another thing I noticed – the prime numbers. They are appearing in order like 7, 11, 13, 17, 19, 23….after 47 it must be 53 and then second multiple was reverse of next prime number i.e. reverse of 59.
With Regards
Mohit Verma
Comment by Mohit Verma — March 22, 2010 @ 4:29 am
first random idea: reduce numbers to prime factors (since 77 = 7 * 11 and 341 = 31 * 11 and that looks too interesting to be accidental)
77 = 7*11
341 = 31 * 11
923 = 13 * 71
1547 = 17 * 91
608 = 19 * 2^5 wow, that’ll screw things up
2116 = 2^2 * 23^2
377 = 13*29
ok, so I think this prime fatorization thing is going somewhere, let’s keep going
2263 = 31*73
518 = 2 * 7 * 39
1394 = 2 * 17 * 41
3182 = 2 * 37 * 43
1645 = 5 * 7 * 47
———-
heyyy .. 13, 17, 19, 23, 29, 31, 29, 41, 43, 47 – those are prime numbers in sequence – embarrassing that it took me this long to see it
944 = 2^4 * 59
4636 = 61 * 2^2 * 19
So the blank is going to be 53 times something. what’s the something?
11
31
71
91
2^5 = 32
2^2*23 = 92
13
73
2*7 = 14
2*17 = 34
2*37 = 74
5*7 = 35
______
2^4 = 16
2^2*19 = 76
wellll, OK, that’s a pattern on the units digit preceded by an odd number. So the blank should be either 55, 75, or 95. 75 = 3*5*5*5 which just looks different from the rest, eh? So I guess it’s between 55 = 5*11, which has the problem that the first digit is ‘5′ which doesn’t happen elsewhere in the pattern, and 95 = 5*19.
I’ve spent 10-12 minutes on this and I don’t think I’m going to get any closer than that it’s either 53 * 55 or 53 * 95 and it seems like 53 * 95 = 5035 is slightly better.
Now I’m going to read the other comments
Comment by JMike — March 23, 2010 @ 2:23 pm
After reading the other comments, I am of course shamed and horrified to have missed such an easy insight.
Something I want to point out that’s similar to that other numeric problem you posted a while ago: the thing that I’m really interested in, and wish I understood further, is what lies behind that initial insight that causes the problem to open (most of) its guts to you. I looked at this problem, and by the time I was aware of having thoughts about it, the little subvocalization thread over there in the back of my head was saying “look at the prime factorization of the numbers”. (a) How did it do that, and (b) how did it do that in such a way that “the rest of me” was taken by surprise? That’s the real thing, to me anyway. The rest is just working out the details. (And I’m aware of throwing in the word “just” there in a way that I shouldn’t, cf. either Gerald Weinberg or Robert Pirsig or both.)
[James' Reply: You seemed to have gone right to the factorization. I wandered through several other by-ways, first, before I remembered that Trey had once before hit me with a factors problem, so I visited easycalculation.com and played with prime factors.
Nothing looked strange except 11 showing up as a factor in the first two terms, and 71 and 17 seeming interesting because they were reversed.
A few minutes later the pattern clicked.
I think for me, it's like this: I visit specific patterns I've seen before. If those don't work, break it down with some common forms of analysis (deltas, scatterplots, cycles, and visiting the Online Encyclopedia of Integer Sequences), looking for things that seem like repetitions or strange consistencies.]
Comment by JMike — March 23, 2010 @ 2:30 pm
I first starting by summing the individual digits to see if any patterns emerged. Failing to see any, I wanted to see what kind of factors I was dealing with. I used the following bash command and then looked at the output.
for n in 77 341 923 1547 608 2116 377 2263 518 1394 3182 1645 944 4636; do echo “Testing $n…” >> out; for ((a=2;a> out; fi; done; done;
The beginning of the “out” file looked like this:
Testing 77…
7
11
Testing 341…
11
31
Testing 923…
13
71
Testing 1547…
7
13
17
91
119
221
It was somewhat obvious that primes were involved. Looking at the factors laid out in this fashion actually made it simpler to recognize the reversal pattern. “Hmm, 31 and then 13; 71 and then look, 17; There must be something to this reversal.”
After going through the rest of the list and looking at the common elements from number to number in the sequence, I found this to be ascending primes, with the higher number prime reversed.
So, missing number is: 5035.
Cool puzzle! Also, very interesting to read everyone’s approaches and thoughts.
Keep these coming!
Comment by Adam — March 23, 2010 @ 6:59 pm
That didn’t paste so well -
for n in 77 341 923 1547 608 2116 377 2263 518 1394 3182 1645 944 4636; do echo “Testing $n…” >> out; for ((a=2;a> out; fi; done; done;
was the bash command
Comment by Adam — March 23, 2010 @ 7:00 pm
So, should the first three numbers in the sequence be 6, 15, and 35 ?
[James' Reply: All you need to tell me is the missing number.]
Comment by Lance — April 13, 2010 @ 1:19 pm
You can try another one here:
http://www.testalways.com/2010/07/09/160/
Comment by Eusebiu Blindu — July 11, 2010 @ 10:28 am
First I checked to see if there’s any connection between the even and odd numbers, but I didn’t notice anything.
Then I checked to see if I can find a connection to the ascending / descending order of the numbers, but I didn’t notice anything
Then I summed the digits of each number and noticed that sum(3rd)=sum(9th), sum(4th)=sum(10th), sum(5th)=sum(11), but all the other pairs didn’t match, so I dropped this path too
I noticed that 77=7×11, then I looked at 341. I know that only 1×1 and 3×7 give a 1 and I found out that 341 = 31×11.
Then I though to divide 923 to 13 (which in my mind was 31 inverted) => 923 = 71×13
Then I divided 1547 to 17 (71 inverted)=> 1547 = 91×17. I was surprise to see that my theory was valid so far, so I continued
608 = 32×19
2116 = 92×23
377 = 13×29
2263 = 73×31
518 = 14×37
1394 = 34×41
3182 = 74×43
1645 = 35×47
Following the above reasoning I though that the 13th number has to be a multiple of 53 => 13th = Zx53. Then I thought that 944 should be a multiple of Z inverted.
While trying to think of a solution for this number, I noticed that 7, 11, 19, 23, …, 43, 47, 53 are the prime numbers in ascending order, so 944 should be a multiple of 59 (944 = 16×59).
Knowing that the 13th number is a multiple of 53 and that 944 is a multiple of 59 => 13th number is 5035 = 95×53.
Thank you for the puzzle. It was a nice way to start my day
.
Comment by Alex Rotaru — December 7, 2010 @ 1:51 am