very very tough math puzzle 2

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LAexpress12

theres only 100 possible answers...

00,01,02,03,04,05,06,07,08,09,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30.....

jerry2468

7, and 1?

__vxD_mAte

whats the answer please?

__vxD_mAte

[editing]

__vxD_mAte

ok here is my theory on the matter (specially rephrased)

ok ... (this is rephrased)

 

the sum of the numbers in the first circle is 63.

The first circle contains the pairs {3,3}, {8,8}, {9,9}, {6,6}

the first circle contains the non-matched numbers {4,7}

 

The sum of the numbers in the second circle is 57

The second circle contains the pairs {5,5}, {7,7}, {9,9}

The second circle contains the non-matched numbers {2,3,4,6}

 

The sum of the numbers shown in the third circle is 39

The third circle contains {6,6}, {3,3}

with non-matched numbers {2,4,8}

 

and overall ... the following pairs have appeared

{3,3} {8,8}, {9,9}, {6,6} {5,5} {7,7} {9,9}, {6,6}, {3,3}

 

{3,3} has appeared twice

{9,9} has appeared twice

{6,6} has appeared twice

 

{5,5}, {7,7}, {8,8} have appeared once

 

in sequence (and set form)

{3,3}, {5,5}, {6,6}, {7,7}, {8,8}, {9,9}

missing 2 and 4

 

the following numbers have appeared without pairs

{4,7,2,3,4,6,2,4,8}

 

in sequence (set form)

{2,3,4,6,7,8}

missing 5 and 9

 

observations

each circle has no number more than twice

the first circle adds up to a bigger number than the 2nd circle

if the sum of numbers in the 3rd circle is less then it could be a descending sequence ...

at most it must be a number less than or equal to the 2nd because 39 + 9+9 = 57

this would mean the pair of 9's will have appeared 3 times

so ... if the sequence subtracts 6 for the first number, 63-57=6

and doubles 57-12 = 45 

then we have multiples of 3

 

45 = 3*15

57 = 3*19

63 = 3*21

 

and the differences are...

19-15 = 4

21-19 = 2

 

the numbers 4&2

 

they also happen to finish the set of pairs so that every number will have been in a pair.

 

5+9 = 14 and 39+14 = 53, this number isn't a mulitple of 3.

 

also, 57-53 = 4, and the sequence 63, 57, 53 has the differences -6,-4. 

 

So perhaps 5+9 is a bad fit.

 

also if the sequence of sumes descended by 6 each time, i.e. 63, 57, 51

then the number must add up to 12,

this leaves the numbers 5&7 and 8&4 ... but neither of these numbers complete the pairs or the lack of pairs, i.e the numbers

without pairs still miss number 9 and for the 8& 4 case the numbers with pairs stil miss number 2.

 

also 8&4 already appear adjacent to each other in the third circle.

So this attempts to rule out the other possibilities and leaves us with the sequence of sums

 

-6, -12

 

where 12=6*2

 

and 57-12 = 45 and the numbers fitting are 2&4 where 4=2*2

 

then these match up the missing pairs and also make a pretty number ...

the digits in 45,57,63 make up a set ... 3,4,5,6,7 with 5 repeated

 

as the sum of the 2nd circle has the numbers 5&7, I looked at the idea that 5&7 fit into the third circle and they do,

but in the 2nd circle 6&3 the digits from the number of the sum of the first circle - these two numbers don't have pairs

while the number 7 does in the third circle. Also 5&7 contains the digits of the middle number...

 

I am guessing 4 & 2 because they look nicer with the mulitple of 3 sums, however the 5&7 arguement is tempting so I'll hold that in reserve

shequan
mhs84 wrote:
AndyClifton wrote:

What do IQ & math puzzles have to do with a chess course?


Chess improves intelligence, and who has high intelligence will be good in chess. so IQ puzzles could help us to improve our calculations and imagination, and we need these things to be a good in chess.


chess mainly involves one kind of intelligence in my opinion, spatial/visual intelligence. all other kinds of intelligence that are involved in chess are necessarily filtered through this when you play. there are many different kinds of intelligence. someone can be very intelligent and also not really excel all that much in chess (by not excel I mean they wouldn't ever become really strong masters, but I do think most people who are highly intelligent in some area or another will at the least be able to become average players if they applied themselves). I just think it's probably true that some highly intelligent people's "spatial/visual analysis filter", for lack of a better term, just isn't that super strong, that their intelligence mainly resides in a different area.

 

I do think your visual/spatial questions probably would help with chess.

I also think geometry/math problems would probably be best for chess improvement.

ringwraith10
__vxD_mAte wrote:

ok here is my theory on the matter (specially rephrased)

 

ok ... (this is rephrased)

 

the sum of the numbers in the first circle is 63.

The first circle contains the pairs {3,3}, {8,8}, {9,9}, {6,6}

the first circle contains the non-matched numbers {4,7}

 

The sum of the numbers in the second circle is 57

The second circle contains the pairs {5,5}, {7,7}, {9,9}

The second circle contains the non-matched numbers {2,3,4,6}

 

The sum of the numbers shown in the third circle is 39

The third circle contains {6,6}, {3,3}

with non-matched numbers {2,4,8}

 

and overall ... the following pairs have appeared

{3,3} {8,8}, {9,9}, {6,6} {5,5} {7,7} {9,9}, {6,6}, {3,3}

 

{3,3} has appeared twice

{9,9} has appeared twice

{6,6} has appeared twice

 

{5,5}, {7,7}, {8,8} have appeared once

 

in sequence (and set form)

{3,3}, {5,5}, {6,6}, {7,7}, {8,8}, {9,9}

missing 2 and 4

 

the following numbers have appeared without pairs

{4,7,2,3,4,6,2,4,8}

 

in sequence (set form)

{2,3,4,6,7,8}

missing 5 and 9

 

observations

each circle has no number more than twice

the first circle adds up to a bigger number than the 2nd circle

if the sum of numbers in the 3rd circle is less then it could be a descending sequence ...

at most it must be a number less than or equal to the 2nd because 39 + 9+9 = 57

this would mean the pair of 9's will have appeared 3 times

so ... if the sequence subtracts 6 for the first number, 63-57=6

and doubles 57-12 = 45 

then we have multiples of 3

 

45 = 3*15

57 = 3*19

63 = 3*21

 

and the differences are...

19-15 = 4

21-19 = 2

 

the numbers 4&2

 

they also happen to finish the set of pairs so that every number will have been in a pair.

 

5+9 = 14 and 39+14 = 53, this number isn't a mulitple of 3.

 

also, 57-53 = 4, and the sequence 63, 57, 53 has the differences -6,-4. 

 

So perhaps 5+9 is a bad fit.

 

also if the sequence of sumes descended by 6 each time, i.e. 63, 57, 51

then the number must add up to 12,

this leaves the numbers 5&7 and 8&4 ... but neither of these numbers complete the pairs or the lack of pairs, i.e the numbers

without pairs still miss number 9 and for the 8& 4 case the numbers with pairs stil miss number 2.

 

also 8&4 already appear adjacent to each other in the third circle.

So this attempts to rule out the other possibilities and leaves us with the sequence of sums

 

-6, -12

 

where 12=6*2

 

and 57-12 = 45 and the numbers fitting are 2&4 where 4=2*2

 

then these match up the missing pairs and also make a pretty number ...

the digits in 45,57,63 make up a set ... 3,4,5,6,7 with 5 repeated

 

as the sum of the 2nd circle has the numbers 5&7, I looked at the idea that 5&7 fit into the third circle and they do,

but in the 2nd circle 6&3 the digits from the number of the sum of the first circle - these two numbers don't have pairs

while the number 7 does in the third circle. Also 5&7 contains the digits of the middle number...

 

I am guessing 4 & 2 because they look nicer with the mulitple of 3 sums, however the 5&7 arguement is tempting so I'll hold that in reserve


hats off to you for having the time to post such a long explanation on a logic problem on chess.com.

__vxD_mAte

hmm, thats not a very long answer ... its supposed to be a "very hard" math problem so I decided to take a look. And yes before someone asks I am trying to improve my IQ.

I was probably wrong anyway after all that.

fireballz

there is a question-mark for each eye in the human head.

if we switch the lights on, we can see two question-marks.

if we switch the  light off, then we can remember seeing it on our computer, even if it go into  power-saving mode...

we are now entering a  dreamstate where we can start imaginating answers, where random can either be true or false...

jwol
mhs84 wrote:
jwol wrote:

2 and 4.


 Why ??


The number in the top half of the decagon is double that of the bottom half.  The simplest solution is often the correct one!

Wrinn
X_PLANET wrote:

4 and 5 just look in diagonal. Diagonal for 7 is 5  and for 2 is 4. I got it. very easy.


None of the other numbers in the circle follow that rule concretely.  It was a nice try though.

__vxD_mAte
jwol wrote:
mhs84 wrote:
jwol wrote:

2 and 4.


 Why ??


The number in the top half of the decagon is double that of the bottom half.  The simplest solution is often the correct one!


+1

planterns

It's 2 and 4.

Bottom equals top 5 divided by 2.

87396/2 = 43698

59534/2 = 29767

48672/2 = 24336

PrawnEatsPrawn
planterns wrote:

It's 2 and 4.

Bottom equals top 5 divided by 2.

87396/2 = 43698

59534/2 = 29767

48672/2 = 24336


Well spotted.

Fabian0Marijuana

Its 2 and 4!

Almost everyone is guessing those numbers in last 2 pages so it should be the good answer

planterns
Yeah it seems jwol got it first. Kudos to you!
__vxD_mAte

saying "the simplest solution is always the best one" to the question of "very hard math problem 2" isn't really correct ... but the fact of the pairs and the slightly worse logic of the other patterns yeilds the solution by the process of elimination, the point about the sum of numbers in the top half decagon being twice the sum of numbers in the bottom half really nails the point. Still waiting for the answer from the author/poster ... he said "why" to the question 2&4 and this suggests giving the wrong reason "why" might show how the question is mistitled... 

druggedbug
__vxD_mAte wrote:

hmm, thats not a very long answer ... its supposed to be a "very hard" math problem so I decided to take a look. And yes before someone asks I am trying to improve my IQ.

I was probably wrong anyway after all that.


I read this and I felt that it deserved a response. It is not possible to 'increase your IQ,' but it is possible to increase your intelligence. I've noticed over the years that there is an increasing number of people who don't really understand what an IQ implies. An IQ is a measure of one's ability to learn something. A low IQ score would not necessarily imply that you are stupid, rather it would imply that you pick up things slower than others. You can know more than someone and have a lower IQ, after all, what you do with the brain you have is what really matters. An example of an IQ being used in the real world is assessing a student for special needs education.

Doing these math problems would be beneficial to you I agree, but not because of IQ as you suggested, but because it would help your ability to recognize patterns. Chess is patterns, math is patterns, chess == math, therefore knowing patterns is very beneficial to math practice. Pattern recognition is how computer chess players work as they analyze every move.

Maybe, this will come as a shock, but your IQ is something that never changes no matter how hard you try. The good news is you will be a much smarter person for all of your effort to increase your IQ!

I mean no harm or disrespect in my post! My hats off to you for your attempt at the math problem.Wink

froghollow

 2  and  4  ,  then numbers 2 3 4 5 6 7 8 9  are represented  ( twice ) at least once ? if right /  addition was applied to the three , no pattern . pairs were obvious until the 3rd / attempted before reading other posts - 1 was similar ; but i did"nt comprehend the reserve answer .

__vxD_mAte
druggedbug wrote:
__vxD_mAte wrote:

hmm, thats not a very long answer ... its supposed to be a "very hard" math problem so I decided to take a look. And yes before someone asks I am trying to improve my IQ.

I was probably wrong anyway after all that.


I read this and I felt that it deserved a response. It is not possible to 'increase your IQ,' but it is possible to increase your intelligence. I've noticed over the years that there is an increasing number of people who don't really understand what an IQ implies. An IQ is a measure of one's ability to learn something. A low IQ score would not necessarily imply that you are stupid, rather it would imply that you pick up things slower than others. You can know more than someone and have a lower IQ, after all, what you do with the brain you have is what really matters. An example of an IQ being used in the real world is assessing a student for special needs education.

Doing these math problems would be beneficial to you I agree, but not because of IQ as you suggested, but because it would help your ability to recognize patterns. Chess is patterns, math is patterns, chess == math, therefore knowing patterns is very beneficial to math practice. Pattern recognition is how computer chess players work as they analyze every move.

Maybe, this will come as a shock, but your IQ is something that never changes no matter how hard you try. The good news is you will be a much smarter person for all of your effort to increase your IQ!

I mean no harm or disrespect in my post! My hats off to you for your attempt at the math problem.


I would disagree, if you take the right vitamins your temporal IQ should increase, you have different physical states, the state when you are malnourished and the state where you have all the vitamins are the most obvious examples, (drunken state, drugged state?). So there is no test that can really measure your IQ if your Quotient is dependent on nutrition and (lack of) drug exposure. 

So for bevity "temporal IQ" in this case means "Your IQ at the moment" or "how well you can potentially solve problems in your current state of fitness".

I am always good in tests however my chess skill isn't as great as I would like it to be, perhaps I would need much more knowledge, or perhaps it is simply too hardcore.