View Full Version : A puzzle for the thinkers
What is the missing number?
31 41 ?9 26 53 58 97
DrMatrix
08-04-2000, 12:59 AM
The missing number is 5.
Going to back that up or have you just plucked a digit from the air? ;)
DarrenS
08-04-2000, 01:16 AM
Come on - that was a piece of pie, erm I mean cake.
DrMatrix
08-04-2000, 01:16 AM
I didn't want to give it away, in case anybody else wanted to play.
You gave the first few digits of Pi.
So...What do I win? :)
aseymayo
08-04-2000, 01:17 AM
That Dr. Matrix - always looking for pi in the sky.
aseymayo
08-04-2000, 01:19 AM
Note to self - learn to type faster.
I just thought we'd throw some puzzles around and giving the answer away be damned!!!
Try this one on:
1) A+B=2
2) 5A+B=5
3) A+5B=7
What do A and B equal?
Prizes? Um... you get to play in the next round. With only one Doper lifeline available. But you have to keep getting the answers right before anyone else.
panamajack
08-04-2000, 01:42 AM
Originally posted by dpr
Try this one on:
1) A+B=2
2) 5A+B=5
3) A+5B=7
What do A and B equal?
Aren't these supposed to be puzzles? This stuff was covered in precalculus. And you don't think someone with a name like Dr.Matrix would miss this, do you?
A and B are ... numbers. And I didn't even have to run MATLAB to figure it out.
panama jack
_______________
Viscera
08-04-2000, 01:42 AM
A = .75
B = 1.25
Next.
Vis :)
DrMatrix
08-04-2000, 01:45 AM
Let's see. A = 3/4, B = 5/4. And I didn't even have to use the third equation :)
What? Well, it's not my fault he screwed up the third equation!
C K Dexter Haven
08-04-2000, 07:47 AM
<< What is the missing number?
31 41 ?9 26 53 58 97 >>
DrMatrix says 5. Sure. But it can be anything you want. Say I want the missing number to be X. Then I take the following formula:
[(31)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)]/(-1)*(-2)*(-3)*(-4)*(-5)*(-6)
+ [(41)(n-1)(n-3)(n-4)(n-5)(n-6)(n-7)]/)]/1*(-1)*(-2)*(-3)*(-4)*(-5)
+ [(X9)(n-1)(n-2)(n-4)(n-5)(n-6)(n-7)]/2*1*(-1)*(-2)*(-3)*(-4)
+ [(26)(n-1)(n-2)(n-3)(n-5)(n-6)(n-7)]/)]/3*2*1(-1)*(-2)*(-3)
+ [(53)(n-1)(n-2)(n-3)(n-4)(n-6)(n-7)]/4*3*2*1*(-1)*(-2)
+ [(58)(n-1)(n-2)(n-3)(n-4)(n-5)(n-7)]/5*4*3*2*1*(-1)
+ [(97)(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)]]/5*4*3*2*1
OK, now I multiply out and simplify the terms and I have a 7th degree polynomial. Ignore that, and look at the pattern above.
When I plug in n = 1 to get the first number in the sequence, it's 31. (The first term in the sum is only one that's non-zero, since all the other terms have (n-1) = 0 as a multiplier and so disappear; the first term is 31.)
When I plug in n = 2 to get the second number in the sequence, it's 41. Again, the second term in the above sum is the only one that's non-zero.
When I plug in n = 3, I get X9 since all the other terms are zero.
When I plug in n = 4 to get the fourth number, it's 26...
Etc.
Thus, I have given you a polynomial in n, and plugging in values of n = 1, 2, etc exactly matches your sequence.
Mathematic Fact: given any finite sequence of numbers, I can invent an infinite number of formulas that answer the question "What's the next number?"
Silly example:
What's next in the sequence: 1, 3, 5, 7 ...
(a) 9
(b) 11
(c) 8
Correct answer: could be any of these.
(a) If this is a sequence of odd numbers, the next one is 9.
(b) If this is a sequence of odd primes, the next one is 11.
(c) If this is a sequence of numbers that have the letter "e" in the English spelling of their names, the next one is 8.
I hate these sequence questions, because they penalize people who think outside the box.
DAVEW0071
08-04-2000, 01:29 PM
Yeah....what Dex said....
sdimbert
08-04-2000, 02:17 PM
What is the sequence?
8 5 4 9 0 7 6 3 2...
Anyone?
panamajack
08-04-2000, 02:36 PM
well, if I were British, I'd venture to guess that that's the alphabetical ordering of the arabic numerals by pronunciation, i.e. :
Ait Five Four Nine Nought Seven Six Three Too ...
and the next digit is of course Won.
But I'm not British, I'm not even Australian, so I'll reveal the correct answer : It is the first digits of the phone number of the palace of the ruler of Uqbar (using that country's internal codes), and the next digit is 1. This number will connect with the ruler's cellular phone in the event that she is vacationing on Tlon.
panama jack
______________
"Mirrors and copulation are abominable, for they multiply the number of mankind." - Heresiarch of Uqbar
Lance Turbo
08-04-2000, 02:43 PM
Originally posted by dpr
1) A+B=2
2) 5A+B=5
3) A+5B=7
What do A and B equal?
The first equation is A+B=2, therefore A and B equal 2.
I only had to use one equation.
RM Mentock
08-04-2000, 03:43 PM
Originally posted by DrMatrix
Let's see. A = 3/4, B = 5/4. And I didn't even have to use the third equation :)
What? Well, it's not my fault he screwed up the third equation!
Of course he didn't. I got the same answer using only the last two equations.
OK, here's one.
Without division, find the next digits of 1/89 = .011235...
aseymayo
08-04-2000, 05:04 PM
um...81321?
panamajack
08-04-2000, 06:02 PM
well, it's certainly possible for calculators/computers to perform the calculation with reasonable accuracy without division. So one could do that.
Another way to do that would be using logarithms. So I take 89, and look it up on the log table. Or I grab my slide rule (the slide rule is so I can be sure I'm looking at the '89' line of the log table. Why, what else would it be used for?) I get some number (like 1.4218) which I negate, then look around on the chart to find out what number that is the log of, and that's the answer. (What? Oh, that's because you're not using base 23.5 log tables like I am. You really ought to switch.)
My favorite method, since it's most easily extended to any precision, is to multiply whatever you have by 89, and see how far off from 1 you are. Your original number (.011235) yields .999915, so I know the next digit has to be a 9 (89 * 9 is almost 90 * 9, which is 810, and it can't get bigger than that). If I use 9, I can keep adding more by multiplication until I get a satisfactorily rational answer.
panama jack
Okay, what's the next number in this sequence?
1,2,720!,...
RM Mentock
08-05-2000, 01:50 PM
Originally posted by aseymayo
um...81321?
Good guess! It is the fibonacci sequence
It goes like this though:
.0
+.01
+.001
+.0002
+.00003
+.000005
+.0000008
+.00000013
+.000000021
+.0000000034
+.00000000055
On out forever, so the next digits are "95". They
start repeating after 44 digits I think.
Speaker for the Dead
08-05-2000, 03:37 PM
I just got out of grade eight, and I'm scared to be doing this stuff, EVER.
Bill H.
08-06-2000, 07:07 PM
Some of you may enjoy this site:
http://www.research.att.com/~njas/sequences/eisonline.html
And I just thought this would be a bit of fun. No offense was intended I assure you.
SO let's move away from mathematical:
What are the next four letters in this series.
Y Y H L Y E Y T ...
Another one for any drive by solver...
The Pope has it but he does not use it.
Your father has it but your mother uses it.
Nuns do not need it.
Your lady friend's husband has it and she uses it.
What is it?
Lance Turbo
08-08-2000, 08:46 PM
Originally posted by dpr
Another one for any drive by solver...
The Pope has it but he does not use it.
Your father has it but your mother uses it.
Nuns do not need it.
Your lady friend's husband has it and she uses it.
What is it?
A last name.
I beg you, please stop asking questions that you already know the answer to.
dylan_73
08-09-2000, 06:31 AM
Originally posted by dpr
And I just thought this would be a bit of fun. No offense was intended I assure you.
SO let's move away from mathematical:
What are the next four letters in this series.
Y Y H L Y E Y T ...
R R R R
I think.....and the total number of letters in the sequence should give the explanation away.
dylan_73
panamajack
08-09-2000, 01:56 PM
dpr, don't take any offense at comments about what you've posted. It's just that this particular forum is designed to answer questions that people want answered, not as a place to pose riddles.
Some people do object to the posing of riddles in general, since the answers are often well-known and not quite the sort of questions expected by many on the board.
i don't think anyone would object to you posting to MPSIMS. You might not get as quick a response as here, but you won't upset anybody.
panama jack
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