[math-fun] a simple problem, but ?
Here is a simple problem I saw in one math marathon at the University of Orsay for youngsters, it is similar to the PUTNAM. If we have a*b + 7*a*c + 15*c = a*b*c then find all a,b,c that are primes that satisfies the equation. a,b,c are all >=2. By using brute force I can find 3 solution with small primes and 49 others with non-primes, 52 solutions in all. I am stuck with this problem, I do not see how to resolve this without using brute force. Normally, that problem should be feasible with simple means, I don't see how to find the 52 solutions and from that , the 3 triplets of primes. hint, a, b c are relatively small one of the solution is 2, 29, 2 for a, b, c, see below for the complete set (spoiler). here is the original paper : https://www.imo.universite-paris-saclay.fr/marathon/sept20.pdf Best regards, Simon Plouffe 1, 23, 23 1, 24, 12 1, 33, 3 1, 44, 2 2, 15, 30 2, 29, 2 3, 13, 13 3, 14, 7 3, 15, 5 3, 16, 4 3, 18, 3 3, 24, 2 4, 11, 44 5, 11, 11 5, 12, 6 5, 15, 3 5, 20, 2 6, 10, 20 6, 19, 2 8, 9, 72 9, 9, 27 9, 13, 3 10, 9, 18 10, 17, 2 12, 9, 12 12, 11, 4 15, 9, 9 15, 10, 5 15, 12, 3 15, 16, 2 16, 8, 128 17, 8, 68 18, 8, 48 19, 8, 38 20, 8, 32 21, 8, 28 21, 9, 7 23, 8, 23 25, 8, 20 27, 8, 18 30, 8, 16 30, 9, 6 30, 10, 4 30, 15, 2 35, 8, 14 39, 8, 13 45, 8, 12 45, 11, 3 55, 8, 11 75, 8, 10 75, 9, 5 135, 8, 9
hihi - well, clearly c divides a*b, so if it is prime, it divides a or b, and if a and b are prime, then c=a or c=b that should at least make the search much smaller - if c=a, then b*c + 7*c^2 + 15*c = b*c^2, b + 7 * c + 15 = b * c, c * (b - 7) = b + 15, so b > 7, or something along those lines, and somewhat similarly for c=b more later, chris On 2020-10-02 08:48, Simon Plouffe wrote:
Here is a simple problem I saw in one math marathon at the University of Orsay for youngsters, it is similar to the PUTNAM.
If we have a*b + 7*a*c + 15*c = a*b*c then find all a,b,c that are primes that satisfies the equation. a,b,c are all >=2.
By using brute force I can find 3 solution with small primes and 49 others with non-primes, 52 solutions in all.
I am stuck with this problem, I do not see how to resolve this without using brute force.
Normally, that problem should be feasible with simple means, I don't see how to find the 52 solutions and from that , the 3 triplets of primes.
hint, a, b c are relatively small one of the solution is 2, 29, 2 for a, b, c,
see below for the complete set (spoiler). here is the original paper : https://www.imo.universite-paris-saclay.fr/marathon/sept20.pdf
Best regards, Simon Plouffe
1, 23, 23 1, 24, 12 1, 33, 3 1, 44, 2 2, 15, 30 2, 29, 2 3, 13, 13 3, 14, 7 3, 15, 5 3, 16, 4 3, 18, 3 3, 24, 2 4, 11, 44 5, 11, 11 5, 12, 6 5, 15, 3 5, 20, 2 6, 10, 20 6, 19, 2 8, 9, 72 9, 9, 27 9, 13, 3 10, 9, 18 10, 17, 2 12, 9, 12 12, 11, 4 15, 9, 9 15, 10, 5 15, 12, 3 15, 16, 2 16, 8, 128 17, 8, 68 18, 8, 48 19, 8, 38 20, 8, 32 21, 8, 28 21, 9, 7 23, 8, 23 25, 8, 20 27, 8, 18 30, 8, 16 30, 9, 6 30, 10, 4 30, 15, 2 35, 8, 14 39, 8, 13 45, 8, 12 45, 11, 3 55, 8, 11 75, 8, 10 75, 9, 5 135, 8, 9
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For the prime case I'd do this. Rewrite a*b + 7*a*c + 15*c = a*b*c as a b (c - 1) == c (7a + 15) If c==2 then this simplifies to b = 14 + 30/aj which (going through the prime divisors of 30, which number 3) gives us the solution (a=2, b=29, c=2). So other than this solution c>2. The left side must be divisible by c, but c-1 never is divisible by the prime c, so either a=c or b=c. We handle the two cases in turn. If a=c then we have b = 7 + 22 / (c - 1) So for b to be integral c must be 1, 2, 11, or 22. None of these give a new solution. If b=c, then we have: c = 8 + 15/a a can be 3 or 5, and this gives c=13, 11 respectively so this gives the solutions (a=13, b=3, c=3) and (a=11, b=5, c=5). -tom On Fri, Oct 2, 2020 at 8:49 AM Simon Plouffe <simon.plouffe@gmail.com> wrote:
Here is a simple problem I saw in one math marathon at the University of Orsay for youngsters, it is similar to the PUTNAM.
If we have a*b + 7*a*c + 15*c = a*b*c then find all a,b,c that are primes that satisfies the equation. a,b,c are all >=2.
By using brute force I can find 3 solution with small primes and 49 others with non-primes, 52 solutions in all.
I am stuck with this problem, I do not see how to resolve this without using brute force.
Normally, that problem should be feasible with simple means, I don't see how to find the 52 solutions and from that , the 3 triplets of primes.
hint, a, b c are relatively small one of the solution is 2, 29, 2 for a, b, c,
see below for the complete set (spoiler). here is the original paper : https://www.imo.universite-paris-saclay.fr/marathon/sept20.pdf
Best regards, Simon Plouffe
1, 23, 23 1, 24, 12 1, 33, 3 1, 44, 2 2, 15, 30 2, 29, 2 3, 13, 13 3, 14, 7 3, 15, 5 3, 16, 4 3, 18, 3 3, 24, 2 4, 11, 44 5, 11, 11 5, 12, 6 5, 15, 3 5, 20, 2 6, 10, 20 6, 19, 2 8, 9, 72 9, 9, 27 9, 13, 3 10, 9, 18 10, 17, 2 12, 9, 12 12, 11, 4 15, 9, 9 15, 10, 5 15, 12, 3 15, 16, 2 16, 8, 128 17, 8, 68 18, 8, 48 19, 8, 38 20, 8, 32 21, 8, 28 21, 9, 7 23, 8, 23 25, 8, 20 27, 8, 18 30, 8, 16 30, 9, 6 30, 10, 4 30, 15, 2 35, 8, 14 39, 8, 13 45, 8, 12 45, 11, 3 55, 8, 11 75, 8, 10 75, 9, 5 135, 8, 9
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Ah yes, thanks to all of you, of course, c divides a*b, and from there your reasoning, quite clear, it matches the solutions found, 2, 29, 2 3, 13, 13, 5, 11, 11 best regards and thanks. simon plouffe Le 2020-10-02 à 18:45, Tomas Rokicki a écrit :
For the prime case I'd do this.
Rewrite
a*b + 7*a*c + 15*c = a*b*c
as
a b (c - 1) == c (7a + 15)
If c==2 then this simplifies to
b = 14 + 30/aj
which (going through the prime divisors of 30, which number 3) gives us the solution (a=2, b=29, c=2).
So other than this solution c>2. The left side must be divisible by c, but c-1 never is divisible by the prime c, so either a=c or b=c. We handle the two cases in turn.
If a=c then we have
b = 7 + 22 / (c - 1)
So for b to be integral c must be 1, 2, 11, or 22. None of these give a new solution.
If b=c, then we have:
c = 8 + 15/a
a can be 3 or 5, and this gives c=13, 11 respectively so this gives the solutions (a=13, b=3, c=3) and (a=11, b=5, c=5).
-tom
On Fri, Oct 2, 2020 at 8:49 AM Simon Plouffe <simon.plouffe@gmail.com> wrote:
Here is a simple problem I saw in one math marathon at the University of Orsay for youngsters, it is similar to the PUTNAM.
If we have a*b + 7*a*c + 15*c = a*b*c then find all a,b,c that are primes that satisfies the equation. a,b,c are all >=2.
By using brute force I can find 3 solution with small primes and 49 others with non-primes, 52 solutions in all.
I am stuck with this problem, I do not see how to resolve this without using brute force.
Normally, that problem should be feasible with simple means, I don't see how to find the 52 solutions and from that , the 3 triplets of primes.
hint, a, b c are relatively small one of the solution is 2, 29, 2 for a, b, c,
see below for the complete set (spoiler). here is the original paper : https://www.imo.universite-paris-saclay.fr/marathon/sept20.pdf
Best regards, Simon Plouffe
1, 23, 23 1, 24, 12 1, 33, 3 1, 44, 2 2, 15, 30 2, 29, 2 3, 13, 13 3, 14, 7 3, 15, 5 3, 16, 4 3, 18, 3 3, 24, 2 4, 11, 44 5, 11, 11 5, 12, 6 5, 15, 3 5, 20, 2 6, 10, 20 6, 19, 2 8, 9, 72 9, 9, 27 9, 13, 3 10, 9, 18 10, 17, 2 12, 9, 12 12, 11, 4 15, 9, 9 15, 10, 5 15, 12, 3 15, 16, 2 16, 8, 128 17, 8, 68 18, 8, 48 19, 8, 38 20, 8, 32 21, 8, 28 21, 9, 7 23, 8, 23 25, 8, 20 27, 8, 18 30, 8, 16 30, 9, 6 30, 10, 4 30, 15, 2 35, 8, 14 39, 8, 13 45, 8, 12 45, 11, 3 55, 8, 11 75, 8, 10 75, 9, 5 135, 8, 9
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I went through a similar but slightly different process when solving it, obtaining three solutions which I've verified (they're also on Simon's list): a, b, c = 2, 29, 2 a, b, c = 3, 13, 13 a, b, c = 5, 11, 11 The values for these are 116, 507, and 605 (for both the left and right sides of course). You basically got it, but the final substitutions were wrong for the last two, where you reversed a vs. (b and c): a, b, c = 13, 3, 3 (doesn't work) a, b, c = 11, 5, 5 (doesn't work) neither of which is correct. Tom Tomas Rokicki writes:
For the prime case I'd do this.
Rewrite
a*b + 7*a*c + 15*c = a*b*c
as
a b (c - 1) == c (7a + 15)
If c==2 then this simplifies to
b = 14 + 30/aj
which (going through the prime divisors of 30, which number 3) gives us the solution (a=2, b=29, c=2).
So other than this solution c>2. The left side must be divisible by c, but c-1 never is divisible by the prime c, so either a=c or b=c. We handle the two cases in turn.
If a=c then we have
b = 7 + 22 / (c - 1)
So for b to be integral c must be 1, 2, 11, or 22. None of these give a new solution.
If b=c, then we have:
c = 8 + 15/a
a can be 3 or 5, and this gives c=13, 11 respectively so this gives the solutions (a=13, b=3, c=3) and (a=11, b=5, c=5).
-tom
participants (4)
-
christopher landauer -
Simon Plouffe -
Tom Karzes -
Tomas Rokicki