Maths at Nine
Let’s start with the problem but it has to be said that the reasoning behind the solution is complex, unless you are a mathematician:
Al Turing and Bertie Einstein, two layabout mathematicians from West Brom, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. Turing is given the value x+y and Einstein is given the value xy. They then have the following conversation.
Einstein: I cannot determine the two numbers.
Turing: I knew that.
Einstein: Now I can determine them.
Turing: So can I.Given that the above statements are true, what are the two numbers?
For all you non-mathematicians out there, some hints:
1. First of all, xy cannot be prime. It also cannot be the square of a prime, for that would imply x = y. For example, xy cannot be the product of two distinct primes, for then P could deduce the numbers. Likewise, xy cannot be the cube of a prime, such as 33 = 27, for then 3×9 would be a unique factorization; or the fourth power of a prime.
2. Einstein’s statement implies that xy cannot have exactly two distinct proper factors whose sum is less than 100. Call such a pair of factors eligible.
3. Eligible sums: 11, 17, 23, 27, 29, 35, 37, 41, 47, 53.
4. Goldbach’s Conjecture states that every even integer greater than 2 can be expressed as the sum of two primes, to deduce that the above list can contain only odd numbers. Although the conjecture remains unproven, it has been verified empirically up to 1018.
Answer shall be given once the speculation reaches a wall.
Filed under: Diversions, Technology & ideas
Well, I’m there.
I dunno if there is only one answer, but I’d guess 3 and 4.
Before either party says anything E has the number 12 (so he knows it’s either 2+6 or 3+4) and T has the number 7 (so he knows it’s either 2+5 or 3+4).
E admits he can’t tell.
T then knows that it can’t be 2+5, because if E had a 10, he’d know it was 2+5, so he can rule that out, which leaves 3+4.
I’m there meaning that you have the answer, Lord T?
Mark – not yet.
Sorry for posting OT, didn’t know when an eligible post would come.
Things are getting kinda urgent.
Read This post,
And understand that Gold is turning parabolic, via a “Swiss Stair” which signifies planned, controlled NATIONAL buying, desperately.
Be warned!
This could go on the next post, which is relevant.
What-ever you choose, James
Germany will shortly announce they will be buying gold.
The Day the Dollar Died Part 2.
OK, had a go at this but am struggling….(too much vino is my excuse!)
Einstein can’t work it out therefore the product (xy) cannot be semiprime – as in one prime multiplied by another (it can’t be prime either as X and Y are greater than 1).
Thus at least one of x and y is non prime.
Turing knows this as the sum he has (X+Y) is an odd number, and therefore cannot be made up of two prime numbers. The number 2 is a bit of an exception, however Turing knows that X+Y-2 is not a prime number either (prob looked them up on Google like I have lol).
So we have x * y = non prime or non semiprime
X+Y = odd number where X+Y-2 is not prime and < 100
Candidates:
3 and 8
4 and 7
5 and 6?
gawd help!
James, if you want to move the above posts, and this post, to your next subject post, please do so.
Given warnings by several ratings agencies to the UK, and the OECD, IMF, and This, and also, Download the Scribd and read and also This speech,
maybe AA should petition the Queen for the dissolution of Parliament, and an election sometime in the holidays, between Xmas and the new year.
This could all move out of control before the planned con, ermmm, election.
Just joking, maybe!
This was bleddy cruel and I’m beaten and off to peel the veg. Mutter, mutter.
Actually, I withdraw 3+4 from the race. If that’s the combination, then T can’t have known that E wouldn’t know either, because there is a possibility that E has 10 written on his piece of paper.
So let’s try 3 + 8, the first one on Don’s list.
T has 11, so he knows that E has either 18 (which might be 2+9 or 3+6, from E’s point of view) or 30 (which might be 3+10 or 5+6 from E’s point of view).
So T knows that E can’t know.
E admits he doesn’t know, so T confirms he knew that.
In which case knows that T has either 11 or 9 or 13 or 11, none of which helps E.
To apply logic rather than maths, neither number is prime and they add up to an odd number. Rule out 1, 2 and 3.
Try 4+9. T has 13, but this might be 2+11, and if E had 22, E would know that E knows, rule out.
Try 4+15. T has 19. He knows that E has either 48, 60, 70, 78, 72, 88 or 90, none of which has a unique solution.
That’s my final offer, the numbers are 4+15.
I did what Don did and worked out all the candidates. Then I reread the post and realised that these guys had been given the answers to X+Y and XY yet they could not work it out which means they must be large enough to have several candidates.
The bit that totally confused me was when they said they couldn’t work it out then they could when they found out the other couldn’t. That means that when they work out X+Y and XY then only one pair meets the criteria. However I have no idea how to do that without knowing the results. Clearly there is a Mathematical trick that does it and not being a maths guys I’m stuck. Head firmly against wall as I meant by my comment yesterday.
Sorry, I explained that wrong. Ignore previous comment.
Actually, I withdraw 3+4 from the race. If that’s the combination, then T can’t have known that E wouldn’t know either, because there is a possibility that E has 10 written on his piece of paper.
So let’s try 3 + 8, the first one on Don’s list.
a. T has 11, so he knows that E has either 18 (which might be 2+9 or 3+6, from E’s point of view) or 24 (which might be 3+8 or 4+6) or 28 (which might be 4+7 or 2+14) or 30 (which might be 3+10 or 5+6 from E’s point of view).
b. So T knows that E can’t know.
c. However, if E has 24, then E knows that it’s either 3+8 or 4+6, so T might have 10 on his piece of paper. If T had 10, he knows that E might have 21, which is semi-prime, so if T had 10, he couldn’t be sure that E doesn’t know.
So rule that out.
To apply logic rather than maths, neither number is prime and they add up to an odd number. Rule out 1, 2 and 3 (all prime) so the smallest even number is 4…
Try 4+9. T has 13, but this might be 2+11, and if E had 22, that has only one solution.
Try 4+15. T has 19, but this might be 2+17. He knows that E might have 34, which has a unique solution.
Try 4+21. T has 25, but this might be 2+23. He knows that E might have 46, which has a unique solution.
OK, next ad hoc rule is starting to emerge. The total of the two numbers minus 2 can’t be prime. The smallest two odd numbers that are not prime numbers that are two apart are 25 and 27.
Therefore the smallest odd number it can include is 25.
Try 4+25. T has 29, this might be 2+27, but even if E had 54, there is no unique solution to this. T knows that the other numbers that E might have are 78, 100,
So my final, final offer is 4+25.
Right, I think I’ve got it (head has now cleared after the vino!)
Einstein’s number can’t be a semiprime (otherwise it would have two unique factors only)
Turing knows this because the sum he has cannot be represented by adding two prime numbers together. Thus the sum
is not an even number, nor is the sum-2 equal to a prime number.
So x and y are made up of one even number and one odd number.
Once Einstein knows that the sum of the numbers cannot be represented by 2 prime numbers, he can solve it.
Reason for this is the number (the product) Einstein has is so small, it can only have one unique set of factors which are an even and odd pair.
Ditto for Turing. Once he knows Einstein can solve it, he knows that the number must be so small that even-odd pair must make up unique
factors in that number.
Thus the numbers must be the smallest they can be (to be unique factors) which satisfy the rules.
x and y are both greater than 1, and they are not equal.
Try x=2
y must be an odd non prime number, therefore the lowest value of y is 9
Unfortunately the product does not have a unique set of odd-even factors (could be 2×9 or 3×6).
No point in increasing the value of y (i.e. to 15) as if the lowest value doesn’t have unique factors, anything higher can’t.
Try x=3
y must be an even number
With y=4, the sum is 7. This is unacceptable as it can be represented by 2 primes and therefore Turing couldn’t know for sure that Einstein couldn’t solve it.
y=6
The sum is 9 which again violates our rule that it can be represented by 2 primes (2 and 7)
y=8
The sum is 11 which means Turing’s rules are satisfied (i.e. cannot be made up of two primes)
The product is 24 which only has one set of unique even-odd factors, namely 8 and 3.
By my reckoning, the numbers are 8 and 3.
Very interesting….perhaps someone can placate my Mrs. as to why the decorating has been going slowly this
morning!
Well, I hope one of you is right.
OK, time for the explanation Nick gave. In the original, Turing was S and Einstein was P:
Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.
P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.
Given that the above statements are true, what are the two numbers?
First of all, trivially, xy cannot be prime. It also cannot be the square of a prime, for that would imply x = y.
We now deduce as much as possible from each of the logicians’ statements. We have only public information: the problem statement, the logicians’ statements, and the knowledge that the logicians, being perfect, will always make correct and complete deductions. Each logician has, in addition, one piece of private information: sum or product.
P: I cannot determine the two numbers.
P’s statement implies that xy cannot have exactly two distinct proper factors whose sum is less than 100. Call such a pair of factors eligible.
For example, xy cannot be the product of two distinct primes, for then P could deduce the numbers. Likewise, xy cannot be the cube of a prime, such as 33 = 27, for then 3×9 would be a unique factorization; or the fourth power of a prime.
Other combinations are ruled out by the fact that the sum of the two factors must be less than 100. For example, xy cannot be 242 = 2×112, since 11×22 is the unique eligible factorization; 2×121 being ineligible. Similarly for xy = 318 = 2×3×53.
S: I knew that.
If S was sure that P could not deduce the numbers, then none of the possible summands of x+y can be such that their product has exactly one pair of eligible factors. For example, x+y could not be 51, since summands 17 and 34 produce xy = 578, which would permit P to deduce the numbers.
We can generate a list of values of x+y that are never the sum of precisely two eligible factors. The following list is generated by JavaScript; the function may be inspected by viewing JavaScript: function genSum (plain text.)
Eligible sums: 11, 17, 23, 27, 29, 35, 37, 41, 47, 53.
(We can use Goldbach’s Conjecture, which states that every even integer greater than 2 can be expressed as the sum of two primes, to deduce that the above list can contain only odd numbers. Although the conjecture remains unproven, it has been verified empirically up to 1018.)
P: Now I can determine them.
P now knows that x+y is one of the values listed above. If this enables P to deduce x and y, then, of the eligible factorizations of xy, there must be precisely one for which the sum of the factors is in the list. The table below, generated by JavaScript (view plain text JavaScript: function genProd), shows all such xy, together with the corresponding x, y, and x+y. The table is sorted by sum and then product.
Note that a product may be absent from the table for one of two reasons. Either none of its eligible factorizations appears in the above list of eligible sums (example: 12 = 2×6 and 3×4; sums 8 and 7), or more than one such factorization appears (example: 30 = 2×15 and 5×6; sums 17 and 11.)
S: So can I.
If S can deduce the numbers from the table below, there must be a sum that appears exactly once in the table. Checking the table, we find just one such sum: 17.
Therefore, we are able to deduce that the numbers are x = 4 and y = 13.
Eligible products and sums
Product x y Sum
18 2 9 11
24 3 8 11
28 4 7 11
52 4 13 17
76 4 19 23
112 7 16 23
130 10 13 23
50 2 25 27
92 4 23 27
110 5 22 27
140 7 20 27
152 8 19 27
162 9 18 27
170 10 17 27
176 11 16 27
182 13 14 27
54 2 27 29
100 4 25 29
138 6 23 29
154 7 22 29
168 8 21 29
190 10 19 29
198 11 18 29
204 12 17 29
208 13 16 29
96 3 32 35
124 4 31 35
150 5 30 35
174 6 29 35
196 7 28 35
216 8 27 35
234 9 26 35
250 10 25 35
276 12 23 35
294 14 21 35
304 16 19 35
306 17 18 35
160 5 32 37
186 6 31 37
232 8 29 37
252 9 28 37
270 10 27 37
322 14 23 37
336 16 21 37
340 17 20 37
180 5 36 41
114 3 38 41
148 4 37 41
238 7 34 41
288 9 32 41
310 10 31 41
348 12 29 41
364 13 28 41
378 14 27 41
390 15 26 41
400 16 25 41
408 17 24 41
414 18 23 41
418 19 22 41
132 3 44 47
172 4 43 47
246 6 41 47
280 7 40 47
370 10 37 47
396 11 36 47
442 13 34 47
462 14 33 47
480 15 32 47
496 16 31 47
510 17 30 47
522 18 29 47
532 19 28 47
540 20 27 47
546 21 26 47
550 22 25 47
552 23 24 47
Mmmm. Bit of a swizz there James. Another of these maths puzzles that are only good for theoretical Maths people.
For example if Einstein had the number 20 as his result it meant that 1+19, 2+18 .. 9+11, etc. could meet his requirements and he doesn’t know the answer to x and y. If Turing had 99 as his answer he could have 1×99, 3×33 … and so on. He doesn’t know either. They both say they don’t know and it doesn’t help one jot unless they give each other the information they hold.
Have I mentioned I now hate maths.
Nice one!
Should have checked my hypothesis further. Would have fallen down at 4 and 7 lol!
Thanks – a fun challenge!
The masochist in me pleads for more well posed teasers, James.
They’re in the pipeline. Well done.
“Let’s start with the problem but it has to be said that the reasoning behind the solution is complex, unless you are a mathematician:”
Not fair, you’re not excluding physicists, computer scientists and engineers — ok, we know they suck, but this problem should be easy enough even for them.
*sniff*