Can you guess the numbers?

[The_Big_K]

Originally Posted by gravz84

Well alrite..
oh no I didn't mean that...
I meant that if the numbers are (3,4) they can guess the numbers in the fashion stated..
Similarly there are other numbers for which they can solve in the same fashion ie those 4 statements..

...and that does not lead us to a unique solution, right?

[gravz84]

Originally Posted by The_Big_K

...and that does not lead us to a unique solution, right?

It doesn't lead us to a unique solution but it does lead them to a unique solution!!
"
We need a list of values of x+y that are never the sum of precisely two elitible factors.
"
that's what gives multiples..

[The_Big_K]

Originally Posted by gravz84

It doesn't lead us to a unique solution but it does lead them to a unique solution!!
"
We need a list of values of x+y that are never the sum of precisely two elitible factors.
"
that's what gives multiples..

Boy, that was a hint! Of course then the answer needs to be picked up from that list!

[wtf]

ok if we go by proper mathematical cum logical concept and not only by logic...we get 4 and 13 as the answer.coz 4 and 13 is the only pair that can come in the list...in fact i think even 4 and 61 can come but i m not very sure abt it.

[wtf]

big k please give the answer..i have like spent 3 days on this question...even bunked my campus connect lecture for it...just tell us wht u have deduced.

[wtf]

and yeah while in my explaination post i had written that we need to prepare a list of values of x+y that are never the sum of precisely two eligible factors,i also mentioned that we have to select only tht pair of sum which doesnt repeat in tht list and so the answer.

[gravz84]

Originally Posted by wtf

and yeah while in my explaination post i had written that we need to prepare a list of values of x+y that are never the sum of precisely two eligible factors,i also mentioned that we have to select only tht pair of sum which doesnt repeat in tht list and so the answer.

I think we'll have to wait for the answer. I am not convinced yet that (3,4) is not a possible answer..
but big_k says the scheme/logic should work out to give you a unique answer so our logic might be incorrect.

[wtf]

yup lets wait for the final thump.

[The_Big_K]

Guys, I like your enthusiasm & attitude. I believe you're almost there. However, you have different answers. All I know is, there's a unique solution. I'm not posting the answer unless someone solves it with detailed & correct explanation.

Hint #2 -

We need a list of values of x+y that are never the sum of precisely two elitible factors.

--- and this list can contain only odd numbers.

[PS: Am I complicating the things for you guys?]

[wtf]

ok i m posting the list of all such possible values in no particular order ...but once a particular 'sum' has been mentioned atleast twice i will not write it down again(which automatically implies that it cant be the answer)...so here it goes:

13+4=17
16+7=23
13+10=23
27+2=29
25+4=29
30+5=35
17+10=27
32+5=37
8+3=11
26+9=35
29+24=53
7+4=11
38+3=41
27+10=37
26+27=53
22+19=41
41+6=47
31+16=47
18+9=27

i think these are the only sums possible with the conditions on two numbers.
we can easily sort out that 17 is the only number which has occurred only once...
hope this ends up the forum discussion.