The sum is 19 the product 34 and the two numbers are 17 and 2, the only way that the sum of 19 can produce a product under 40 is with 17+2. and 34 only has two other numbers it is divisible by except 1 and itself and these are 17 and 2.
Puzzles,conundrums, riddles and thoughts?
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The sum is 19 the product 34 and the two numbers are 17 and 2, the only way that the sum of 19 can produce a product under 40 is with 17+2. and 34 only has two other numbers it is divisible by except 1 and itself and these are 17 and 2.
Wrong. a and b are in [2,40]. There was no specification for the product and the sum to be under 40.
Of course, the sum is in [4,80], and the product in [4, 1600].
Ah sorry I misinterpreted the rules. In that case I'm going to pass, as it's trial and error and I have better things to do. I'm sure some maths whizz know immediately a number that has only a prime as a sum and is only divisable between 2 and some other number and itself. and probably is actually the same number multiplied by itself, but I don't have all day so pass
Typically British to withdraw in the safety of their island and wait for the others to cope with the difficulty when they face a decent opponent.
Do you hope to find again some Prussian to solve the French problem?
Do you hope to find again some Prussian to solve the French problem?
is the sum 11?
No, it isn't 11.
grrrrr, it works with 11 til the last step though
Say, is there a non trial and error approach to the question?
(don't want to what it is, just if it exists)
No as said unless it's got some solution that isn't trial and error which I doubt, why should I spend the next 4 hours trying to stumble blindly across a number between 40 and 1600? No offence but I've got better things to do.
What'll happen is someone with some sort of qualification in maths will remember that some perfect square number has these features and then after spending 4 hours going from 40 to 1600 I'll feel like a total idiot for even bothering.
Anyway that aside it's a good problem for those who like farting around with 28 gajillion piece all white puzzles and so on. But not my cup of tea.Good luck.
There are non trial and error way to help solve the problem.grrrrr, it works with 11 til the last step though
Say, is there a non trial and error approach to the question?
(don't want to what it is, just if it exists)
pboily
fingerlickinmathematickin
EDIT: hint removed, I've found a better way.
pboily
fingerlickinmathematickin
Here is a hint:
what is Mr. S implying when he says that even if Mr. P knew the sum, he wouldn't be able to find the numbers?
It means that {a,b} are not the only integers greater than 2 which give you a product of ab and a sum of a+b.
Call that other set {c,d}.
Then ab=cd and a+b=c+d.
Square the second equation:
(a+b)^2 = (c+d)^2
a^2+2ab+b^2=c^2+2cd+d^2
a^2+b^2=c^2+d^2, because ab=cd.
EDIT: Wait, I can't find any appropriate Pythagorean triples.
Back to the drawing board.
what is Mr. S implying when he says that even if Mr. P knew the sum, he wouldn't be able to find the numbers?
It means that {a,b} are not the only integers greater than 2 which give you a product of ab and a sum of a+b.
Call that other set {c,d}.
Then ab=cd and a+b=c+d.
Square the second equation:
(a+b)^2 = (c+d)^2
a^2+2ab+b^2=c^2+2cd+d^2
a^2+b^2=c^2+d^2, because ab=cd.
EDIT: Wait, I can't find any appropriate Pythagorean triples.
Back to the drawing board.
pboily
fingerlickinmathematickin
Aha!
4 and 13.
EDIT:
Mr P knows Mr S knows the sum, but Mr P doesn't know the value of the sum
means that ab CANNOT be the product of exactly two prime integers.
Mr S knows Mr P knows the product, but Mr S doesn't know the value of the product.
means that a+b CANNOT be the sum of exactly two prime integers.
How many possibilities are there?
The sum a+b has to be one of:
11
17
23
27
29
35
37
41
43
...
(there are 32 in total, the last one is 79).
Similarly, the product ab cannot be one of
4
6
9
10
14
15
21
22
25
26
33
34
35
38
39
...
(there are 78 in total, the last one is 1369).
For each of the numbers in the sum list, write the possible two term summations a+b, together with the product ab.
11=2+9 ---> 2(9)=18
11=3+8 ---> 3(8)=24
11=4+7 ---> 4(9)=36
11=5+6 ---> 5(6)=30
etc...
A bit later, Mr P called Mr S back and says "I've found a and b"
means that there is only one number a+b on the sum list for which only one product ab is not in the product list
In the above example, every product is not on the list, so the sum cannot be 11.
Eliminating all the ones that are left, (a little computer program) there is only the sum 17 left.
Eventually, Mr S called Mr P and says "Me too".
means that there is only one of the product ab for which the sum a+b has this property
That product is 52=4(13).
4 and 13.
EDIT:
Mr P knows Mr S knows the sum, but Mr P doesn't know the value of the sum
means that ab CANNOT be the product of exactly two prime integers.
Mr S knows Mr P knows the product, but Mr S doesn't know the value of the product.
means that a+b CANNOT be the sum of exactly two prime integers.
How many possibilities are there?
The sum a+b has to be one of:
11
17
23
27
29
35
37
41
43
...
(there are 32 in total, the last one is 79).
Similarly, the product ab cannot be one of
4
6
9
10
14
15
21
22
25
26
33
34
35
38
39
...
(there are 78 in total, the last one is 1369).
For each of the numbers in the sum list, write the possible two term summations a+b, together with the product ab.
11=2+9 ---> 2(9)=18
11=3+8 ---> 3(8)=24
11=4+7 ---> 4(9)=36
11=5+6 ---> 5(6)=30
etc...
A bit later, Mr P called Mr S back and says "I've found a and b"
means that there is only one number a+b on the sum list for which only one product ab is not in the product list
In the above example, every product is not on the list, so the sum cannot be 11.
Eliminating all the ones that are left, (a little computer program) there is only the sum 17 left.
Eventually, Mr S called Mr P and says "Me too".
means that there is only one of the product ab for which the sum a+b has this property
That product is 52=4(13).
Correct!
That is 4 and 13.
How did you reach this conclusion?
That is 4 and 13.
How did you reach this conclusion?
Syterion
Voodoo Economist
No 13 is 2*6.5 and he meant a is 4 and b is 13, not that the sum and products are 4 and 13.
Oh I know now thanks I got it after pboily's edit. And sadly I know it well, regardless of my stupidity pboily is up.
Glad someone got it because with my provisos I'd of been farting around for years.
pboily, the PhD wasn't wasted, if you want to put one up, feel free.
pboily
fingerlickinmathematickin
I'm having a tough time with the edits, sorry people.
I'm almost done.
EDIT: Done. Well, I wouldn't have been able to do it without a computer to compute the sums and products. What a time-consuming puzzle. Is there an easier way, Steph?
I'm almost done.
EDIT: Done. Well, I wouldn't have been able to do it without a computer to compute the sums and products. What a time-consuming puzzle. Is there an easier way, Steph?
On a side non, they run the algorithm used by PBoy, increasing the range, from [2,40] to [2, several millions], and apparently, 4,13 is the ONLY solution
pboily
fingerlickinmathematickin
Answer each of the 8 questions with a letter from A to D.
The word "answer" in the test refers to YOUR answer, not some hypothetical "best" answer.
After choosing the 8 answers score the test by comparing each question with your answers.
Score 1 point for each question answered correctly, 0 otherwise.
Keep re-taking the test, trying to get the highest possible score.
(Of course you must know what the highest possible score is in order to correctly score
the last question!)
(1) The next question with the same answer as this one is:
(A) 2 (B) 3 (C) 4 (D) 5
(2) The first question with answer C is:
(A) 1 (B) 2 (C) 3 (D) 4
(3) The last question with answer A is:
(A) 5 (B) 6 (C) 7 (D) 8
(4) The number of questions with answer D is:
(A) 1 (B) 2 (C) 3 (D) 4
(5) The answer occuring the most is: (if tied, first alphabetically)
(A) A (B) B (C) C (D) D
(6) The first question with the same answer as the question following it is:
(A) 2 (B) 3 (C) 4 (D) 5
(7) The answer occuring the least is: (if tied, last alphabetically)
(A) A (B) B (C) C (D) D
(8) The highest possible score on this test is:
(A) 5 (B) 7 (C) 6 (D) 8
The word "answer" in the test refers to YOUR answer, not some hypothetical "best" answer.
After choosing the 8 answers score the test by comparing each question with your answers.
Score 1 point for each question answered correctly, 0 otherwise.
Keep re-taking the test, trying to get the highest possible score.
(Of course you must know what the highest possible score is in order to correctly score
the last question!)
(1) The next question with the same answer as this one is:
(A) 2 (B) 3 (C) 4 (D) 5
(2) The first question with answer C is:
(A) 1 (B) 2 (C) 3 (D) 4
(3) The last question with answer A is:
(A) 5 (B) 6 (C) 7 (D) 8
(4) The number of questions with answer D is:
(A) 1 (B) 2 (C) 3 (D) 4
(5) The answer occuring the most is: (if tied, first alphabetically)
(A) A (B) B (C) C (D) D
(6) The first question with the same answer as the question following it is:
(A) 2 (B) 3 (C) 4 (D) 5
(7) The answer occuring the least is: (if tied, last alphabetically)
(A) A (B) B (C) C (D) D
(8) The highest possible score on this test is:
(A) 5 (B) 7 (C) 6 (D) 8
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