08 Aug 2008

# Can you guess the numbers?

I'm thinking of two integer numbers, each of them is more than 1 and their sum is less than 100.

I tell my friend CEan - Ash the sum of these two numbers, and another friend, CEan - MaRo, product of these two numbers.

Then such a dialog took place:

Ash: I can't determine what are these numbers.
MaRo: Ah, i knew you wouldn't be able to do this.
Ash: Oh, then i know what they are!
MaRo: Oh, then i know them too!

Can you determine the numbers? 😁

crazyrajiv

Branch Unspecified
08 Aug 2008
Somehow I feel they should be 97 and 1

because when ash said she(/he) couldnt find and mArO was excited that he(/she knew that ash would not find it), It actually means that marO could find them straightaway. and when ash sees maro excited, she already knows the sum is 98. and now she can guess why maro is so excited(he probably knows the answer,

I feel confused now.
08 Aug 2008
PS: Both Ash & MaRo are 'He' and moderators of CE Forums 😁

13 Aug 2008
... and I used to think we have some smartest brains from all over the world. But I was wrong. We do not have smartest people. We just can't crack a simple puzzle!

PS: Does that hurt your ego? 😁
I'm still waiting for the brainy CEan who can crack this!

anuragh27crony

Branch Unspecified
13 Aug 2008
actually both of them don't know those numbers😉.....just they are bluffing each other 😁

😁........
13 Aug 2008
anuragh27crony
actually both of them don't know those numbers😉.....just they are bluffing each other 😁

😁........
What a ... lame answer. :sleeping:

Sorry !

If you think hard, you'll see that the numbers can be guessed!

anuragh27crony

Branch Unspecified
14 Aug 2008
srry for those lame answer......i think answers are i) 2,2 or ii)2,3

Assumption : Both numbers should be more than 1

since Ash and Mac knew sum and other the product...but they never exchange information abt sum/product....

if numbers are 2,2

Ash knows 4....but basic priciple is both numbers are more than one so
1,3(it differs because...one number is equal to one)
2,2

and Mach know 4...
so it can be
1,4 (it differs because...on number is equal to one)
2,2
............................
if numbers are 2,3

Ash knows 5....but basic priciple is both numbers are more than one so
1,4(it differs because...one number is equal to one)
2,3

and Mach know 6...
so it can be
1,6 (it differs because...on number is equal to one)
2,3
14 Aug 2008
[Update]: 1 < x < y and x+y < 100 [x,y are the numbers].

But, is the answer, right? 😁

Think harder!

[PS: I sincerely & honestly believe that xy cannot be the product of two distinct primes, for then Ash could deduce the numbers.]

Or, am I just wrong here? 😁

anuragh27crony

Branch Unspecified
14 Aug 2008
yeah then 2,3 is the right answer....which can be gussed by both ....even with out transferring peices of information.....

and also satisfies the condition too

1<(X=2)<(Y=3) 😁

[PS: If the both numbers are not disitinct primes then...there's possibility that Mac get's confused with correct pair numbers since it leaves the scope for more than one possibility of xy combination.and yes we can deduce the numbers even they are not distinct primes...if the basic condition is modified as 1<x<=y then answer can be X=Y=2]

ntr_youngtiger

Branch Unspecified
15 Aug 2008
134😁
15 Aug 2008
ntr_youngtiger
134😁
What's that?

anuragh27crony

Branch Unspecified
18 Aug 2008
@The_Big_K ...plzz comment on my answer in the recent post.......
whether my solution is right or not
18 Aug 2008
anuragh27crony
@The_Big_K ...plzz comment on my answer in the recent post.......
whether my solution is right or not
Okay, here's my comment -

If Ash knew the that the sum of numbers is 5. Then he'd have following options in his mind -

(1,4), (2,3)

But, he knew the numbers are both greater than one. Then, he'd not have said the following -

Ash: I can't determine what are these numbers.
That is, Ash could not determine the numbers for sure!

So, unfortunately, we need some hard thinking over here, I think. 😲

anuragh27crony

Branch Unspecified
21 Aug 2008
21 Aug 2008
Sorry sire, the answer will not come. Its open for the non-quitters 😉 . Come on, try harder!

😁

barryn56

Branch Unspecified
24 Aug 2008
Interesting - I'm not sure the problem is correctly posed, since it is obvious given only the sum of the two numbers that you would not be able to derive a solution. Adding that a solution "should" be possible, the assumption would be that you only consider unique values, which for sums gives 4 values: 5, 6, 98 and 99.
Based on a=x+y and b=xy, then x=b/y and y2-ay+b=0, so y=(a+sqrt(a2-4b))/2. The list of integer values of this function for 1<x,y<51 has a number of pairs of results, which makes it diffcult to see how the additional data that there isn't a unique result, helps produce a unique result. Please check the question is accurate.
24 Aug 2008
barryn,

Are you completely ignoring the dialogue between Ash & MaRo? Its there - and it's got all the clues you need to crack the problem.

CEan - anuragh27crony was making a good progress, but looks like he's given up 😉

Think harder.

anuragh27crony

Branch Unspecified
25 Aug 2008
@Big_k....
Ash: I can't determine what are these numbers.
MaRo: Ah, i knew you wouldn't be able to do this.

i can't deduce the significance of these dialouges in solving problem....

barryn56

Branch Unspecified
25 Aug 2008
Are you sure you don't mean their PRODUCT is less than 100?? Otherwise I don't think this is solvable, because there are many solutions...
26 Aug 2008
The conditions have been mentioned and they haven't changed so far. Shall I post the answer or wait for few more days?

sreedu

Branch Unspecified
30 Aug 2008
the product is never a prime number becoz one of the numbers will be 1.so the product will be such a number which has 2 factors other than 1 and the no: itself.so product will be <=10.it can't be 1,2,3,4,5.if product is 6 the numbers are 2&3.so sum is 5.ash may think numbers are (1,4),(2,3).so he needn't have any confusion.so the product is nt 6.its nt 7.if prod is 8 the sum is 6.so numbers can be (2,4),(3,3).here too no confusion.product can't be 9.

so product is 10.sum is 7.numbers can be (1,6),(2,5),(4,3).confusion arises at (2,5),(3,4).but since maro don't hav any confusion,ash can confirm the numbers to be 5,2
is this correct,big_k?😎
30 Aug 2008
is this correct,big_k?
No. It is not.

ummeshh

Branch Unspecified
31 Aug 2008
Aaahh...
I cant get any clue from their dailogues..
If my guess is right., then it would be 4...

Munguti

Branch Unspecified
02 Sep 2008
The numbers range from 2 to 98. Their sum is not unique otherwise ash would have known them(i.e. their sum is anything from 6 to 100, since 1+1=2
2+1=3 and none of the numbers is 1
2+2=4 unique
2+3=5 unique
2+4=3+3=6 start of ambiguity
2+5=3+4=4+3=7
2+6=3+5=..........
then ash has a number between 6 and 100

Then from the information Maro has (product of the two numbers) he knows that their sum is ambiguous and therefore we can say he has a rough idea of the two numbers say maybe two unique pairs at most. The number range Maro has is from 8,9,10,12,15....... up to some number that can be determined from a multiplication table with numbers 2-98 on both sides multiplying by each other.

What i am saying is that by generating a multiplication table and an addition table from the range given above we can get a unique solution based on a simple logic that what should be unique(or having at most two equal solutions for different pairs of numbers) is their product.
i didn't have that much tyme but anyone willing to try it should post the findings
02 Sep 2008
Alright, this should be my last post in this thread; unless of course, no one cracks the puzzle.

Pay attention to the dialog between Ash & MaRo. When I tell them the sum & the product respectively along with the conditions, each one of them has few pairs of numbers ready in their mind before the dialog happens.

Now pay attention: -

Ash also knows possible pairs of numbers MaRo has in his mind. Similarly, MaRo knows Ash has few pairs in his mind. Now starts the elimination game. Ash says "I can't determine what the numbers are". MaRo immediately picks up his clue and understand's Ash's situation (why he could not tell the answer). When MaRo says "I don't know the numbers too", Ash has enough information to eliminate the incorrect pairs - how? he knows if MaRo can't deduce the answer even after knowing that Ash couldn't determine the numbers. So he immediately says "Oh, then I know the numbers".

Now you've enough hints. Think hard.

murali_pclover

Branch Unspecified
04 Sep 2008
Hi Big_K.Can you give the complete question once again.

confounded

Branch Unspecified
04 Sep 2008

Munguti

Branch Unspecified
04 Sep 2008
I think the product given is 12 since it is the first product with more than one option i.e. two 4,3 and 6,2 therefore one can be eliminated. I also think the sum given is seven since it also has two options and one can be eliminated. so i am with confounded

AlanMc

Branch Unspecified
05 Sep 2008
Because MaRo said the he KNEW that Ash couldn't determine the numbers, it means that MaRo was able to deduce the numbers from his product, because the product is unique. Because the sum must be less than 100, the numbers must be between 2 and 97 (inclusive). The larger sets of these numbers have unique products, but not a sum less than 100. Because the sum is less than 100 the unique product must be small. However, because Ash was able to determine the answer from realizing MaRo knew he didn't know, it mean that only one of his numbers has a unique product (which he knew it now must have, because MaRo already knows).

anuragh27crony

Branch Unspecified
06 Sep 2008
@Big_k
Does Maro Knew about the condition of x+y<100 ??

crazyash

Branch Unspecified
06 Sep 2008
any two prime nos whose product is less than 100 should be the answer.
ex: 2,5
3,5
let me know whether its right or not.
07 Sep 2008
Guys, when you post your answer, also post a detailed explanation and the methodology you used to arrive at your [so far incorrect 😁 ] answer(s).

gravz84

Branch Unspecified
07 Sep 2008
My first post as a
😕
not given that x & y are unique?
I'll assume this.
x,y >1
1. Ash says he can't deduce the numbers from the sum
obviously very difficult as sums are not unique.
except for 5(2,3) & 6(2,4)
after 7 its not unique anymore (2,5), (3,4)

2. On hearing this Maro can only deduce that sum is >6
but Maro has the product and products have lesser possibilities, still difficult for him to conclude the numbers.
consider both distinct primes: if this was the case Maro would have been able to find the numbers. Eg 13*11=143 only 13 & 11 and 143 & 1 give this product and since x,y>1 only 13,11 fulfill the condition.

3. On hearing Maro state his failure, Ash now knows that the 2 numbers are not prime & their sum.
But now he has guessed it!

What is the least number for which he could guess this? A sum that has only 1 unique combination (which are not primes), and other combinations are primes since he's just eliminated them at step 2.

7
(2,5) (3,4)
note that if the numbers were 2,5 maro wld have been able to tell the numbers from the product(10)

10
(2,8) (3,7) (4,6)
note that if the numbers were 2,8 maro wld have been able to tell the numbers from the product(16)--same for 21
eliminate first 2 - numbers are 4,6

😕
I think the answer(sum) can be 7, 10 and lots of others but I guess for now I can't do better than taking the least answer as mine.

So I go with 3 & 4 as the numbers, 7 as the sum and 12 as product

Raj_rising_engi

Branch Unspecified
08 Sep 2008
let 2 nos be a & b,thn Ash has x=a+b n Maro:y=a*b
so y=a*b=(x-b)*b=xb-(b*b)
Satisfying this condition s obviously a=4 and b=2...

Please lemme know if there s any correction....
08 Sep 2008
:sleeping:

Wake me up when you get the right answer.

anuragh27crony

Branch Unspecified
08 Sep 2008
@Big_K....

This is my guess.. is the right answer (49,50) ???

If it's right i'll post the explination
08 Sep 2008
anuragh27crony
@Big_K....

This is my guess.. is the right answer (49,50) ???

If it's right i'll post the explination
:sleepy: :sleeping: :sleepy:

Sorry?

gravz84

Branch Unspecified
08 Sep 2008
gravz84
My first post as a
😕
not given that x & y are unique?
I'll assume this.
x,y >1
1. Ash says he can't deduce the numbers from the sum
obviously very difficult as sums are not unique.
except for 5(2,3) & 6(2,4)
after 7 its not unique anymore (2,5), (3,4)

2. On hearing this Maro can only deduce that sum is >6
but Maro has the product and products have lesser possibilities, still difficult for him to conclude the numbers.
consider both distinct primes: if this was the case Maro would have been able to find the numbers. Eg 13*11=143 only 13 & 11 and 143 & 1 give this product and since x,y>1 only 13,11 fulfill the condition.

3. On hearing Maro state his failure, Ash now knows that the 2 numbers are not prime & their sum.
But now he has guessed it!

What is the least number for which he could guess this? A sum that has only 1 unique combination (which are not primes), and other combinations are primes since he's just eliminated them at step 2.

7
(2,5) (3,4)
note that if the numbers were 2,5 maro wld have been able to tell the numbers from the product(10)

10
(2,8) (3,7) (4,6)
note that if the numbers were 2,8 maro wld have been able to tell the numbers from the product(16)--same for 21
eliminate first 2 - numbers are 4,6

😕
I think the answer(sum) can be 7, 10 and lots of others but I guess for now I can't do better than taking the least answer as mine.

So I go with 3 & 4 as the numbers, 7 as the sum and 12 as product
lol..such a long post...
My answer was that there are no unique numbers that the two have thought of and the least possible combination is (3,4)
At least write if the line of thought/ approach is right...
and what is the mistake..
simply posting the solution is of no use, i agree; so I would appreciate if some discussion could ensue as to each person's solution and approach!
anyone??

gravz84

Branch Unspecified
08 Sep 2008
Raj_rising_engi
let 2 nos be a & b,thn Ash has x=a+b n Maro:y=a*b
so y=a*b=(x-b)*b=xb-(b*b)
Satisfying this condition s obviously a=4 and b=2...

Please lemme know if there s any correction....
To advance what I just posted, I'll analyze some solutions suggested by others.
Raj,
the two are not communicating the results ie sum and product that they have been given. They only express whether they have independently concluded the 2 numbers.
What you have stated, requires the knowledge of both sum and product. Plus you can substitute any 2 given numbers in that equation and get the same result. obviously all numbers fulfill this:
y=a*b=(x-b)*b=xb-(b*b)

In statement 1, Ash(guy given the sum of the 2 numbers) says he can't find out.
In statement 2, Maro(guy given the product of the 2 numbers) says he can't find out.
In statement 3, Ash now says he has found out the solution.
In statement 4, Maro now says he has found out the solution.
there must have been extra information for Ash between statement 1 & 3 which is what you should be looking at-since there is no external information the extra information that has helped him solve the puzzle is statement 2.
Similarly for Maro.

gravz84

Branch Unspecified
08 Sep 2008
crazyash
any two prime nos whose product is less than 100 should be the answer.
ex: 2,5
3,5
let me know whether its right or not.
😒
If the 2 numbers were both primes then the product is unique, right?
ie there is only one way to factorize 15 3*5, only 1 way to factorize 55 11*5, so the person who knows the product would instantly know the 2 numbers (both numbers, it is assumed, are distinct and greater than 1) but it is not so so the solution based on both primes is incorrect in my opinion.

gravz84

Branch Unspecified
08 Sep 2008
Munguti
I think the product given is 12 since it is the first product with more than one option i.e. two 4,3 and 6,2 therefore one can be eliminated. I also think the sum given is seven since it also has two options and one can be eliminated. so i am with confounded
it is indeed the first product which satisfies but there are others that have 2 possibilities that are solved by Ash only after the 1st 2 statements are made.
So in my opinion there are many such solutions but 3,4 is the least one.

tanuj

Branch Unspecified
10 Sep 2008
nos r 4 and 3
sum - 7(Ash)
product - 12 (MaRo)

himanshu_cl

Branch Unspecified
11 Sep 2008
i believe that the product of the numbers should be 16
and the sum of the numbers is 8

so maro has the product (16)
he has got 2 possibilities of numbers (2,8) and (4,4)
the respective sums are 10 and 8
for both the sums he is sure that Ash wouldnt know the correct answer

but when ash realizes that maro knows that she cant guess the nos
and she knows the sum is 8
she thinks of the possibilities
(2,6) (3,5) (4,4)
for (6,2) product is12
for (3,5) product is 15
for (4,4) product is 16

15 cant be the product as in that case maro would have known the nos
12 cant be the product as in that case maro had possibilities (3,4) (2,6)
giving the sums as 7 and 8. in that case maro couldnt have said that he knew that ash wouldnt know the answer

so ash realizes that the nos are (4,4)

when maro comes to know that ash knows the answer he rethinks over his possibilities (2,8) and (4,4,)
for (2,8) the sum was 10 for which ash couldnt have told the correct answer
so maro also realizes that the nos are (4,4)
11 Sep 2008

This is oh-so-interesting!

mayurpathak

Branch Unspecified
11 Sep 2008
I have got the answer and the solution too... But I'm going to wait till people try their level best.
11 Sep 2008
mayurpathak
I have got the answer and the solution too... But I'm going to wait till people try their level best.
I *HOPE* you don't have the *WRONG* answer

Yeah, I believe we should wait till someone from other galaxy cracks this puzzle for us and tells everyone why his/her answer is THE RIGHT ANSWER.

mayurpathak

Branch Unspecified
11 Sep 2008
No I'm sure of the answer now. 😀

But let me wait a while

gravz84

Branch Unspecified
11 Sep 2008
mayurpathak
No I'm sure of the answer now. 😀

But let me wait a while
I still think given the constraints(which are quite wide) and assuming that the 2 numbers are unique, there can still be multiple answers and everyone's giving 1 of those multiple solutions.
Still 3,4 is the least possible solution in my opinion.
😕
Hope you guys give out the answer and a satisfactory solution shortly.

wtf

Branch Unspecified
12 Sep 2008
I Have Guessed. 😀

the answer is 13 and 4.
sum=17
prod.=52
when first friend said initially tht he cant fine the numbers,it implied that the product didnot have
CASE (1)-any absolute factors(factors except 1 and the number itself are called absolute factors,e.g. in case of 10: 1 and 10 are not absolute factors,but 2 and 5 are) or
CASE (2) any power of prime i.e 8(2x2x2) or 27(3x3x3) or
CASE (3) product of 2 primes.
now when 2nd one admitted his failure to find out the numbers it meant that none of the possible values of the sum can be such that their product has exactly 2 numbers whose product gives the given value.
now in "higher algebra-silva" an empirical rule is given which helps us find such numbers.we can restrict our list to small numbers since we know the sum of the two numbers is less than 100.

after this both can get the answer by the given method

make a list of all left possible combinations of numbers and just check which number can b expressed in one form only.
we get only 2 possible values:13 and 4.

mayurpathak

Branch Unspecified
12 Sep 2008
*thinking* Hmmm... well. Any other explanations?

wtf

Branch Unspecified
12 Sep 2008
Is there a shorter and simpler explanation for this one?

gravz84

Branch Unspecified
12 Sep 2008
Re: I Have Guessed. 😀

wtf
the answer is 13 and 4.
sum=17
prod.=52
when first friend said initially tht he cant fine the numbers,it implied that the product didnot have
CASE (1)-any absolute factors(factors except 1 and the number itself are called absolute factors,e.g. in case of 10: 1 and 10 are not absolute factors,but 2 and 5 are) or
CASE (2) any power of prime i.e 8(2x2x2) or 27(3x3x3) or
CASE (3) product of 2 primes.
now when 2nd one admitted his failure to find out the numbers it meant that none of the possible values of the sum can be such that their product has exactly 2 numbers whose product gives the given value.
now in "higher algebra-silva" an empirical rule is given which helps us find such numbers.we can restrict our list to small numbers since we know the sum of the two numbers is less than 100.

after this both can get the answer by the given method

make a list of all left possible combinations of numbers and just check which number can b expressed in one form only.
we get only 2 possible values:13 and 4.
I am not sure if you have the right answer or not but you definitely have one mistake in your solution. Might not change the answer though:
The first guy, Ash is given the sum and the second one the product. You took them as the opposite..
12 Sep 2008
So many engineers, so many answers.

I'm thinking if crazyengineers ever decide to build a skyscraper, we'll never be able to finalize the design.

{Turns on the music player and watches everyone scratch head :music: }

wtf

Branch Unspecified
12 Sep 2008
the order does matter but when the 'sum' guy said that he cant find the numbers he obviously had ruled out the clause that the sum he has is a number with exactly one possible absolute factor combination(else the other guy would have known the answer) and his assumption was based on the fact that the 'product' guy is not dumb and he had considered the fact that the number can't be product of 2 distinct primes or a power of some prime number.
So in this question the order does not hamper the conditions for solution,assuming that both friends are genius.

gravz84

Branch Unspecified
12 Sep 2008
wtf
the order does matter but when the 'sum' guy said that he cant find the numbers he obviously had ruled out the clause that the sum he has is a number with exactly one possible absolute factor combination(else the other guy would have known the answer) and his assumption was based on the fact that the 'product' guy is not dumb and he had considered the fact that the number can't be product of 2 distinct primes or a power of some prime number.
So in this question the order does not hamper the conditions for solution,assuming that both friends are genius.
Hope you take all this in a sporting spirit. 😉
"
when the 'sum' guy said that he cant find the numbers he obviously had ruled out the clause that the sum he has is a number with exactly one possible absolute factor combination(else the other guy would have known the answer)
"
you have made an extra assumption.. ie the "sum guy" Ash already knows that Maro has stated that he doesn't know the answer.
Of course since this is a theoretical situation, information must flow distinctly as dialogues... maybe at the time ash is making his statement maro already knows the answer but this can be established only when maro makes his first statement.
Also one more thing in your explanation that I take exception to is:
"the number can't be product of 2 distinct primes or a power of some prime number."
the number(s) can definitely be powers of prime numbers!!
8=2^3 and 4=2^2 bt from the product 32 he still can only deduce that the numbers are either (2,16) or (8,4) so they can definitely be powers of prime numbers!!
Also how you get 13 and 4 as the only possible solution escapes me..
😕
clarify--for eg why can't the solution be (3,4), (23,4) and (17,4) just for the sake of elimination by samples..

wtf

Branch Unspecified
12 Sep 2008
y isnt no one replying back? 😔
is this answer wrong?😒..plz big k jus temme if this is right,i have my campus connect classes from 5:30...gotta go for that....wont b able to concentrate there if this thing dusnt get out of head.
proud 2 b a

gravz84

Branch Unspecified
12 Sep 2008
The_Big_K
So many engineers, so many answers.

I'm thinking if crazyengineers ever decide to build a skyscraper, we'll never be able to finalize the design.

{Turns on the music player and watches everyone scratch head :music: }
And you can play the role of "skeptic" (in management speak) shooting down answers!!
lol, I kid I kid...
seriously itching for the solution now

wtf

Branch Unspecified
12 Sep 2008
ok..i'll explain y my assumption makes sense and y the answer still holds true ...gimme sum tym,i'll type and explain everything,even though it myt take sum time as there are many numbers which comes in the 'not possible list'

gravz84

Branch Unspecified
12 Sep 2008
just do the ones i mentioned..
3,4
13,4
maybe 17, 4 as well

wtf

Branch Unspecified
12 Sep 2008
@ gaurav

first of all let us assume that u know the sum and i know the product.
case1:u know your number is 7(4+3),u dont know my number,u say u don't knw the two numbers,i say i dnt knw,then u say u knw and i say even i knw.
now 7=3+4 or 5+2(chuck the '1+something' part in this as well as all future cases as thts out of point) and 12=6*2 or 4*3. when u say tht u know the answer, concluding it from the fact that i dint know the answer initially,u made a mistake at that point because u cant pinpoint the two numbers exactly to be 4 and 3 because u think i have the number 12 with me but i could have even had 10(5*2 or 5+2 for ur case sake)...so if u guess the numbers to b 3 and 4 its just a guess and we cant provide a valid reason for that,which in the end comes out to be some kind of logical fallacy.
similarly for the case 2 and 3.

read the part where i have mentioned the absolute factors thing.for the time being do one thing...make a list of such products(x*y) which cannot have exactly two distinct absolute factors(such that the sum of those factors is less than 100).Now make a list of values of sum(i.e. x+y) that can never be the sum of exactly two absolute factors(call them absolute sum).and try to sort out the relation betwen the sum and the product list.only one pair exist in the list which has exactly one absolute factor and one absolute sum.

wtf

Branch Unspecified
12 Sep 2008
big k just verify if (4,61) can also be a possible answer??

gravz84

Branch Unspecified
12 Sep 2008
Re: @ gaurav

wtf
first of all let us assume that u know the sum and i know the product.
case1:u know your number is 7(4+3),u dont know my number,u say u don't knw the two numbers,i say i dnt knw,then u say u knw and i say even i knw.
now 7=3+4 or 5+2(chuck the '1+something' part in this as well as all future cases as thts out of point) and 12=6*2 or 4*3. when u say tht u know the answer, concluding it from the fact that i dint know the answer initially,u made a mistake at that point because u cant pinpoint the two numbers exactly to be 4 and 3 because u think i have the number 12 with me but i could have even had 10(5*2 or 5+2 for ur case sake)...so if u guess the numbers to b 3 and 4 its just a guess and we cant provide a valid reason for that,which in the end comes out to be some kind of logical fallacy.
similarly for the case 2 and 3.

read the part where i have mentioned the absolute factors thing.for the time being do one thing...make a list of such products(x*y) which cannot have exactly two distinct absolute factors(such that the sum of those factors is less than 100).Now make a list of values of sum(i.e. x+y) that can never be the sum of exactly two absolute factors(call them absolute sum).and try to sort out the relation betwen the sum and the product list.only one pair exist in the list which has exactly one absolute factor and one absolute sum.
Here's where you went wrong..
Maybe you got muddled up because you are not explicitly thinking of the sequence of events,
Assume I have the number 7!
Statement 1: I don't know the number
I think the number can be 3,4 or 2,5
ie you have the products 12 or 10
Statement 2(you): I don't know either
HERE'S WHERE YOU WENT WRONG. IF YOU HAD 10, you would have said in your 1st statement itself- I have the Numbers 2,5!!
which means you have 12!! but can't say for sure what I have
You know its either (3,4) or (2,6) but can't eliminate 2,6 yet, its still a possibility because I could have sum as 8 (meaning 3,5 or 2,6)!!
(maybe 2,6 maybe 3,4)

Statement 3(Me): Oh then I know the number.
So from your statement I have concluded that you don't have 10 ie you have 12..I DECLARE "I have found the numbers"

Statement 4(You): Then I know as well.
Now you have the product 12...
You knew its either (3,4) or (2,6)
Why did you eliminate 2,6??
cos if you had 2,6 sum would be 8!! From 8 as sum I concluded 3,5 or 2,6 but if 3,5 you WOULD HAVE SAID THAT YOU HAD 3,5 since they are primes and their product unique
So Finally you eliminate 2,6 and DECLARE THAT YOU ALSO KNOW THE ANSWER NOW.

wtf

Branch Unspecified
12 Sep 2008
from sum 8,we can even have 4 and 4??..that will perhaps give more possible further clauses?...i dont think its given in the question that the two integers cant be same?

wtf

Branch Unspecified
12 Sep 2008
just clarify one thing...if i knew it ws either 3,4 or 2,6 how can i reject 2,6?...cos i had the product which was 12...so u either had 7 or 8(from my point of view)if u had 7 it could be either 4,3 or 5,2 ...if u had 8 it cud b 2,6 or 3,5...so according to your logic we need one more round of conversation because u wrote
"From 8 as sum I concluded 3,5 or 2,6 but if 3,5 you WOULD HAVE SAID THAT YOU HAD 3,5 since they are primes and their product unique"
how cud i have said tht i had 3,5 when i had actually either 3,4 or 2,6?

gravz84

Branch Unspecified
12 Sep 2008
wtf
from sum 8,we can even have 4 and 4??..that will perhaps give more possible further clauses?...i dont think its given in the question that the two integers cant be same?
actually I did consider 4,4 ie same numbers as you'll see in the following explanation:
which is what helps 2nd person ie you eliminate 2,6 as an option!! read on..

gravz84

Branch Unspecified
12 Sep 2008
wtf
just clarify one thing...if i knew it ws either 3,4 or 2,6 how can i reject 2,6?...cos i had the product which was 12...so u either had 7 or 8(from my point of view)if u had 7 it could be either 4,3 or 5,2 ...if u had 8 it cud b 2,6 or 3,5...so according to your logic we need one more round of conversation because u wrote
"From 8 as sum I concluded 3,5 or 2,6 but if 3,5 you WOULD HAVE SAID THAT YOU HAD 3,5 since they are primes and their product unique"
how cud i have said tht i had 3,5 when i had actually either 3,4 or 2,6?
Assume numbers are 3,4
i have 7 as sum, u have 12 as product
Statement 1:
Possibilities: (2,5) (3,4)
I don't know the answer 😔
Statement 2: you:
Possibilities: (2,6) (3,4)
Me neither!! 😔
Statement 3:
Possibilities: (2,5) (3,4)
2,5 eliminated since that would give 10 and you would have guessed right away!!
Statement 4:
you think --hmm he knows the answer so, my statement must have helped him eliminate 1 option and he must have had only 2 options to begin with.
that gives me 3,4!!!
because if it was 2,6 you would think sum is 8 which can be (2,6) (3,5) or (4,4) which is 3 possibilities!!!
SO YOU ALSO SAY
YES NOW I ALSO KNOW THE ANSWER.

I am still not ruling out the possibilities of having even more solutions, but 3,4 is the least one possible.

gravz84

Branch Unspecified
12 Sep 2008
The_Big_K
I'm thinking of two integer numbers, each of them is more than 1 and their sum is less than 100.

I tell my friend CEan - Ash the sum of these two numbers, and another friend, CEan - MaRo, product of these two numbers.

Then such a dialog took place:

Ash: I can't determine what are these numbers.
MaRo: Ah, i knew you wouldn't be able to do this.
Ash: Oh, then i know what they are!
MaRo: Oh, then i know them too!

Can you determine the numbers? 😁
HEY HAVE THE STATEMENTS BEEN CHANGED???
I thought statement 2 ie by Maro was
"I can't determine them either"
but now is
"i knew you wouldn't be able to do this."
which has different connotations!!!
can we assume the 2 to mean the same??

and BIG K could you at least tell everyone, ie those who haven't been replied to(me, wtf, others) whether their answers are right..
I think you can end the suspense now and tell the answers..
😒
PLEEEEEAAAASEEE
12 Sep 2008
gravz84
HEY HAVE THE STATEMENTS BEEN CHANGED???
I thought statement 2 ie by Maro was
"I can't determine them either"
but now is
"i knew you wouldn't be able to do this."
which has different connotations!!!
can we assume the 2 to mean the same??

and BIG K could you at least tell everyone, ie those who haven't been replied to(me, wtf, others) whether their answers are right..
I think you can end the suspense now and tell the answers..
😒
PLEEEEEAAAASEEE
Nothing has changed. The statements appear in the first post. Gravz - from your post -

So in my opinion there are many such solutions but 3,4 is the least one.
As you said, there are multiple answers. That indicates neither Ash, nor MaRo would have been able to determine the numbers.

But they did guess the correct numbers.
12 Sep 2008
H I N T
We need a list of values of x+y that are never the sum of precisely two elitible factors.
Hint #2: Drink more coffee. It activates brain. Muwa Muwahahaha 😁

gravz84

Branch Unspecified
12 Sep 2008
The_Big_K
Nothing has changed. The statements appear in the first post. Gravz - from your post -

As you said, there are multiple answers. That indicates neither Ash, nor MaRo would have been able to determine the numbers.

But they did guess the correct numbers.
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..
12 Sep 2008
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

Branch Unspecified
12 Sep 2008
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..
12 Sep 2008
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!

:sshhh:

wtf

Branch Unspecified
12 Sep 2008
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

Branch Unspecified
12 Sep 2008
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

Branch Unspecified
12 Sep 2008
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

Branch Unspecified
12 Sep 2008
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

Branch Unspecified
12 Sep 2008
yup lets wait for the final thump.
12 Sep 2008
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

Branch Unspecified
12 Sep 2008
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.

gravz84

Branch Unspecified
13 Sep 2008
"hope this ends up the forum discussion"
I guess Bigk doesn't think so 😔
I don't understand how Ash and Maro would solve for (13,4) in the sequence mentioned..
Ash won't be able to eliminate 6,11 9,8 3,14 5,12 7,10 as all of them have multiples that can be arrived at using different numbers.
Can you tell the sequence for 13,4 according to you?

mayurpathak

Branch Unspecified
15 Sep 2008
Alright boys, let me ease your pressure a bit. The list wtf generated in absolutely on target. I must say wtf has done a great job.

Now for the hint, if Maro was sure Ash couldn't do it, it means none of the possible summations of the two number can be such that they have exactly one pair of eligible factors. So the list is correct. Now since Ash knows the summation, he knows the number as well (refer list)

Now for how will Maro guess the numbers?
i also mentioned that we have to select only tht pair of sum which doesnt repeat in tht list and so the answer.
Ponder on this

SAYLI P

Branch Unspecified
22 Sep 2008
here it is given that there are 2 nos. let it be x, y
where,
x > 1 , y > 1 , x!=y
also x+y < 100
the nos. less than 100 are (2,3,4,5,6,.......99)
let's take 1st no. = 2
we can't get the no. 2 as the sum of two nos. which is greater than 1
next is 3 here also same condition can't satisfied i.e.
we can't get the no. 3 as the sum of two nos. which is greater than 1
next is 4
1+3 = 4 ----------not accepted
2+2 = 4 ----------- both x , y are same. not accepted.
next is 5
1 + 4 = 5------------not accepted
2+ 3 = 5 ------------yes it satisfies condition
other nos such as 7,8,9.....99 can get by many possible pair .
also 6 has unique solution i.e. 2+4 = 6 but
as 6<100 also 5 <100 & 5 < 6 hence the sum is 5 and product is 6

mayurpathak

Branch Unspecified
22 Sep 2008
@SAYLI P: Nice try. But unfortunately the answer is wrong. Read the entire post. You will get your clues there.

shah pooja

Branch Unspecified
01 Oct 2008
a i think dis no's are
2,2
bcz when ash says her first sentance den maro supports the same,& next time as ash says other sen. that she knows den again maro suppots in d same manner.
i.e.
X+Y=X*Y;
& it can be possible only when both are 2
as 2+2=2*2
(may be u can laugh at its answer but what my mind is saying i'm saying so,just like as what ash was saying maro was saying so)

shah pooja

Branch Unspecified
01 Oct 2008
2,2
bcz here x+y=x*y
dat can be posiible only in case of 2 & 2
2+2=2*2

mayurpathak

Branch Unspecified
01 Oct 2008
The answer is wrong Pooja. Read the post again and also go through last 10-15 posts. you will get clues to solve it.

Don't worry. No one is laughing. We all suck at solving puzzles :-D
01 Oct 2008
Fellas,

Please pay attention to the dialog between MaRo & Ash!

Hussanal Faroke

Branch Unspecified
02 Oct 2008
I think the answer is 4 & 3

Hussanal Faroke

Branch Unspecified
02 Oct 2008
assume a+b=7 then the only possibility for a&b are(5,2) and (4,3).
but if the solution is (5,2) the second one can easily said the result is (5,2).because no more factors for 10 other than 5 and 2. ie a*b=12.
and get answer as (4,3) for fist person.
for sln (4,3), a*b=12.ie possible solutions for second person are (4,3)&(6,2)
for solns (4,3) as well as (6,2), a+b=7, or 8,and first person never get the result for first sight.
But the first one got the answer from second one's dialog. Due to this, second person got the result as(4,3).

Hussanal Faroke

Branch Unspecified
02 Oct 2008
02 Oct 2008
Nope 😔

Hussanal Faroke

Branch Unspecified
02 Oct 2008
Pls tell me,what is wrong with the soln
02 Oct 2008

Hussanal Faroke

Branch Unspecified
02 Oct 2008
Ash get 7 as sum. So the possible slns are (5,2) and (4,3). so he said "I can't determine what are these numbers".
MaRo get 12 as product.so the possible slns are (6,2) and (4,3). so the possible sum for a&b are 8 or 7. Even the sum is 7 or 8, we can generate that sum in two or more ways. So with confidence he said that "Ah, i knew you wouldn't be able to do this".
Ash already know the sum is 7.If the soln is (5,2) maro can easily tell the soln.Because there is no more feasible sln to generate aproduct of 10.
so he got the soln as (4,3).
if the sum is eight ash can't get the correct soln. because the feasible sln for
sum=8 are (4,4),(3,5),(2,6). He can never reach a sln.
From this Maro get the soln.ie(4,3).

jejt

Branch Unspecified
02 Oct 2008
First time poster..

Using the rules and set theory (and a database of the possible answers)

1 < x
1 < y

x + y < 100

and that one person knows x+y and the other knows x*y, and that each knows the other can't work out the solution, gives us the following:

There are 2352 possible number combinations. Of these, we can ignore 2 because they allow the person who knows the sum to work it out straight away (i.e. 4 and 5) and 606 more that the product guy would be able to work out straight away (multiples of two primes e.g. 17*17 can only be 17*17).

Of the sets of solutions which are in a group of two, there are just two from the point of view of the sum - those that create the product answer of 6 and 7. (for the product, theres a large group of 2 answers).

If the answer had been 6, then the pairs would have been 2,4 or 3,3. 2,4 is the product of 8, which only has one option (so is instantly solvable), as well as 3,3 (being the product of two primes) also removing it as an option.

If the answer had been 7, then the pairs would have been 2,5 or 3,4. 2,5 are again multiplications of prime numbers which have been removed, so the only possibility (in retrospect again) is that the pair was 3,4.

1 < x < y then the only possible solution (using set theory) is that the pair is 3,4 and that the sum was 7 product was 12.

As long as neither of the two people have lied, or made any mistakes, then it seems that 3,4 is the only option:

3,4 (giving the Dilemma for the Sum as being 7 - either 2,5 or 3,4 - when he finds out the product does not give the solution, he knows its not 2,5 and :. 3,4).

3,4 product 12 - for which he has two answers -2,6 or 3,4. He knows that if the answer was 2,6 then the sum would have been 8, for which the sum guy has 3 possible choices, so thats not solvable), forcing him to select the 3,4 option.

Did I miss something from my summation of the rules? (if there was a rule which said 1 < x < y then it also adds more sets of answers into this set)

Also, I think I have the same solution as the last guy to post. but different workings out.

jejt

Branch Unspecified
02 Oct 2008
The solutions which match the additionl rule

1 < x < y

actually are where sum = 7 or 8 Having the sum as 7 was discussed in my previous post, but having the sum of 8 was not.

If this rule is in place, therea are two possiblities: 2/5 and 3/4. Because 2 and 5 are primes, their product (10) is only atainable by 2*5, so the product would have lead straight to their solution. There is no other dilemma solution for the Sum.

There are of course potentially other solutions involving simultatious solutions, but 7 is still the smallest.

mspeed

Branch Unspecified
04 Oct 2008
Big K,😉
if all d engineers in this forum r giving it there best shot and u still haven't gotten an answer or probably waving it because of inappropriate deductions, then i can say, the question is about teamwork if both Ash and Maro talk about what they know about the question then an answer would arise.

secondly,if u told us the numbers u gave them we can come up wit something.

ishak.hussainul

Branch Unspecified
04 Oct 2008
gravz84
it is indeed the first product which satisfies but there are others that have 2 possibilities that are solved by Ash only after the 1st 2 statements are made.
So in my opinion there are many such solutions but 3,4 is the least one.
i also thout the same.
but it is not correct....
more than 2 possibilities may occur.
but only one possibility must be confusing..i mean all other possibility may give the one (who knows the product) the correct answer.....

nishank_eureka

Branch Unspecified
07 Oct 2008
This is a very tricky question. I think some programming is required. I will post the programme as soon as i am done. Logic will be pretty much clear from the programme itself.
14 Oct 2008
Nishank - I guess you are on the right path 😉

Btw, we have 100 replies over here - and NOT A SINGLE CORRECT SOLUTION! DAMN!

Anyway, Keep Trying

Want more hints, guys?

nishank_eureka

Branch Unspecified
15 Oct 2008
Sorry I give up. Could not get to something concrete. But , i will keep trying till the correct answer is posted.

mayurpathak

Branch Unspecified
16 Oct 2008
The_Big_K
Btw, we have 100 replies over here - and NOT A SINGLE CORRECT SOLUTION! DAMN!
I doubt it. One guy arrived at the answer but the explanation he gave was incorrect.

Is the hint big enough?
26 Oct 2008
La la la :music:

No one no one yet!
CEans can't solve it,
I bet!

😁
26 Oct 2008
The_Big_K
I'm thinking of two integer numbers, each of them is more than 1 and their sum is less than 100.

I tell my friend CEan - Ash the sum of these two numbers, and another friend, CEan - MaRo, product of these two numbers.

....

Can you determine the numbers? 😁
Probably 4 and 13....
I came to CE after a long time.. and seems it has some appetizers ready for me.

😁
27 Oct 2008
crazyboy
Probably 4 and 13....
I came to CE after a long time.. and seems it has some appetizers ready for me.

😁
This is utter slogging... 😁

Look for the dialogs.... then make pairs... then start looking for combination... and your night will flew like anything.. but believe me it would be worth.

Between.. Biggi confirm is this te right answer...?
27 Oct 2008
The_Big_K
La la la :music:

No one no one yet!
CEans can't solve it,
I bet!

😁
Wait wait wait biggi.....
I can bet CEAN's can solve any puzzle anybody post of the web. After all we all are crazy...

pooja sehgal

Branch Unspecified
04 Nov 2008
are the numbers be-(3,4)
as 1<3<4
ash knew 7 as the sum, wa onfuse in 2,5 & 3,4
when mach told he wnt guess he thought if it wud 2,5 .............
then mach would have easily guessed because 2*5= 10
so mach knew 12 i.e he was confused between
6,2 & 3,4
mach came to know that if it would 2,6 ash would have guessed
so ultimalety it is- 3 & 4

Yosippavar

Branch Unspecified
10 Nov 2008
Are U sure, that you narrated the question in a RIGHT way? I found only lose threads.😒

mayurpathak

Branch Unspecified
10 Nov 2008
The question is absolutely right. And the answer has been given in 'one' of the threads.

:sshhh:Oops! Did I say some thing?:sshhh:

Yosippavar

Branch Unspecified
17 Nov 2008
The numbers are 3 and 4. Ash has 7 and Maro has 12.
I'll try to explain to my best.
Both Ash and Maro would have some sets of numbers from the clue number they have got. Lets call Ash's sets of numbers as sum sets and Maro's sets as product sets.
So Sum sets starts from the number
4 - (2,2)
5 - (2,3)
6 - (2,4)(3,3)
7 - (2,5)(3,4)
8 - (2,6)(3,5)(4,4)
.
.
.

And Product sets start from the number
4 - (2*2)
6 - (2*3)
8 - (2*4)
9 - (3*3)
10 - (2*5)
12 - (2*2*3) - (2*6)(4*3)
14 - (2*7)
15 - (3*5)
16 - (2*2*2*2) - (2*8)(4*4)
18 - (2*3*3) - (2*9)(3,6)
.
.
.
Its obvious that both Ash and Maro are having more than a single set in their mind. Or else they can guess the number at very beginning itself. So we can omit all single set numbers.

Maro's first statement is "Ah, i knew you wouldn't be able to do this.". This means Maro also cant find the answer and he is sure that Ash also couldnt make it.

So Maro has a number with all its product sets tends Ash also have more than one set.

Maro's minimum number which can have two sets is 12. So we consider that Maro is having 12.

It have two sets - (2*6)&(4*3). These two sets if summed will come - 8 and 7. Maro is considering the possibilities of Ash. For both 8 and 7, Ash would have more than one set of numbers. So he is sure that Ash couldnt make the right numbers. But still Maro is not sure about which product set is right one. So he makes his first statement.

Now consider if Ash is having 8. which leads to the sets (2,6)(3,5)(4,4). He will consider the product sets of these sets
(2*6=12)(3*5=15)(4*4=16). All these product sets have multiple sets of Maro. SO Ash cant guess the number at this point if he have 8.

But after hearing Maro's first statement, Ash is able to find the numbers.
So lets now consider the other option that Ash is having 7, which leads the sum sets (2,5)(3,4). The product sets of these numbers are (2*5=10)(3*4=12). No Ash would think in his way that, if Maro has 10 which has only one product set(2,5), he would have guessed the number before making his first statement. But he hadnt. So he has the number 12.

And so, now Ash can determine that numbers were 3 and 4. And he says so.

Already, we assumed Maro is having 12 and so, now he will think in this way. As Ash had found the numbers, they could not be (2,6), because their sum is 8 and Ash knows sum sets of 8 will tends multiple options for Maro, SO Ash could not be sure about the numbers after Maro's first statement. But if Ash got 7, he can guess the numbers. So Maro decides Ash got 7. And so the numbers were 3 and 4.

I think this is right answer and enough explanation.😎

mayurpathak

Branch Unspecified
17 Nov 2008
Good explanation but wrong answer. 😀
17 Nov 2008
mayurpathak
Good explanation but wrong answer. 😀

Mayur - You have not confirmed that 4 and 13 are wrong answer... ? 😁

Yosippavar

Branch Unspecified
17 Nov 2008
Could you explain what's wrong with my answer?

shalini_goel14

Branch Unspecified
22 Nov 2008
Is the answer 3 and 4?

Branch Unspecified
04 Dec 2008
The two numbers should be prime numbers... becoz the product known is directly the numbers multiplication...whereas sum can have any number so ash cant guess.whereas maro can know.

shiwa436

Branch Unspecified
04 Feb 2009
The numbers are 4 and 13.

shankycet

Branch Unspecified
22 Feb 2009
i think the answer is(2,2) or(3,3) or(5,5) or(7,7).

shalini_goel14

Branch Unspecified
22 Feb 2009
It has been a long time that its answer is not posted here. I guess now its time that the answer should be told to all curious minds here.

Saandeep Sreerambatla

Branch Unspecified
02 Mar 2009
the answer is 4 and 13.. 😎😎

a_chinese_guy

Branch Unspecified
03 May 2009
<removed>
...so the the numbers must be 64 and 73

Hussanal Faroke

Branch Unspecified
16 Aug 2012
I'm thinking of two integer numbers, each of them is more than 1 and their sum is less than 100. How sum of 64 and 73 is less than 100?

a_chinese_guy
<removed>
...so the the numbers must be 64 and 73

Anoop Kumar

Branch Unspecified
16 Aug 2012
The_Big_K
I'm thinking of two integer numbers, each of them is more than 1 and their sum is less than 100.

I tell my friend CEan - Ash the sum of these two numbers, and another friend, CEan - MaRo, product of these two numbers.

Then such a dialog took place:

Ash: I can't determine what are these numbers.
MaRo: Ah, i knew you wouldn't be able to do this.
Ash: Oh, then i know what they are!
MaRo: Oh, then i know them too!

Can you determine the numbers? 😁
This is known as Impossible Puzzle https://en.wikipedia.org/wiki/Impossible_Puzzle
Here is solution of this puzzle https://people.sc.fsu.edu/~jburkardt/fun/puzzles/impossible_solution.html