Prove that $frac(72!)(36!)^2-1$ is divisible by 73 [closed] The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manara$beginalignh!+!k=p!-!1\rm and p rm primeendalign$ $Rightarrow h!k!!equiv! (-1)^h+1! pmod!p,$ [Wilson Reflection Formula]How can I prove by induction that $9^k - 5^k$ is divisible by 4?Prove that $53^53-33^3$ is divisible by $10$Proving $k$ is divisible by $3$ iff the sum of the digits of $k$ is divisible by 3Fermat's little theoremProve that $ 16^20+29^21+42^22$ is divisible by $13$.$3^n-1$ is divisible by $4 implies n $ is evenProve that integer not divisible by 2 or 3 is not divisible by 6Prove that $n^n-1 - 1$ is divisible by $(n-1)^2$Prove that $4midn$ if and only if the integer formed by the final two digits of $n$ is divisible by 4.Prove without induction that $2×7^n+3×5^n-5$ is divisible by $24$.
Is an up-to-date browser secure on an out-of-date OS?
Using dividends to reduce short term capital gains?
Example of compact Riemannian manifold with only one geodesic.
Does Parliament hold absolute power in the UK?
How do I design a circuit to convert a 100 mV and 50 Hz sine wave to a square wave?
Why doesn't a hydraulic lever violate conservation of energy?
Simulating Exploding Dice
Working through the single responsibility principle (SRP) in Python when calls are expensive
how can a perfect fourth interval be considered either consonant or dissonant?
Is there a writing software that you can sort scenes like slides in PowerPoint?
Mortgage adviser recommends a longer term than necessary combined with overpayments
What do I do when my TA workload is more than expected?
How do you keep chess fun when your opponent constantly beats you?
ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?
What aspect of planet Earth must be changed to prevent the industrial revolution?
What does Linus Torvalds mean when he says that Git "never ever" tracks a file?
Is there a way to generate uniformly distributed points on a sphere from a fixed amount of random real numbers per point?
Does Parliament need to approve the new Brexit delay to 31 October 2019?
Circular reasoning in L'Hopital's rule
Why not take a picture of a closer black hole?
What can I do to 'burn' a journal?
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
Why did Peik Lin say, "I'm not an animal"?
How to determine omitted units in a publication
Prove that $frac(72!)(36!)^2-1$ is divisible by 73 [closed]
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara$beginalignh!+!k=p!-!1\rm and p rm primeendalign$ $Rightarrow h!k!!equiv! (-1)^h+1! pmod!p,$ [Wilson Reflection Formula]How can I prove by induction that $9^k - 5^k$ is divisible by 4?Prove that $53^53-33^3$ is divisible by $10$Proving $k$ is divisible by $3$ iff the sum of the digits of $k$ is divisible by 3Fermat's little theoremProve that $ 16^20+29^21+42^22$ is divisible by $13$.$3^n-1$ is divisible by $4 implies n $ is evenProve that integer not divisible by 2 or 3 is not divisible by 6Prove that $n^n-1 - 1$ is divisible by $(n-1)^2$Prove that $4midn$ if and only if the integer formed by the final two digits of $n$ is divisible by 4.Prove without induction that $2×7^n+3×5^n-5$ is divisible by $24$.
$begingroup$
Prove that $frac(72!)(36!)^2-1$ is divisible by 73.
My approach is as follow $73n=frac(72!)(36!)^2-1$ I tried remainder theorem but could not prove it.
modular-arithmetic divisibility natural-numbers
edited Mar 31 at 9:05
Michael Rozenberg
110k1896201
asked Mar 31 at 6:42
Samar Imam ZaidiSamar Imam Zaidi
1,6221520
$endgroup$
closed as off-topic by user21820, José Carlos Santos, Adrian Keister, K.Power, RRL Apr 4 at 16:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, Adrian Keister, K.Power, RRL
add a comment |
$begingroup$
Prove that $frac(72!)(36!)^2-1$ is divisible by 73.
My approach is as follow $73n=frac(72!)(36!)^2-1$ I tried remainder theorem but could not prove it.
modular-arithmetic divisibility natural-numbers
edited Mar 31 at 9:05
Michael Rozenberg
110k1896201
asked Mar 31 at 6:42
Samar Imam ZaidiSamar Imam Zaidi
1,6221520
$endgroup$
closed as off-topic by user21820, José Carlos Santos, Adrian Keister, K.Power, RRL Apr 4 at 16:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, Adrian Keister, K.Power, RRL
$begingroup$
Use $73-aequiv-apmod73$.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:45
$begingroup$
$binom72kequiv(-1)^kpmod73$
$endgroup$
– robjohn♦
Mar 31 at 9:26
$begingroup$
Special case of this.
$endgroup$
– Bill Dubuque
Apr 1 at 3:04
add a comment |
$begingroup$
Prove that $frac(72!)(36!)^2-1$ is divisible by 73.
My approach is as follow $73n=frac(72!)(36!)^2-1$ I tried remainder theorem but could not prove it.
modular-arithmetic divisibility natural-numbers
edited Mar 31 at 9:05
Michael Rozenberg
110k1896201
asked Mar 31 at 6:42
Samar Imam ZaidiSamar Imam Zaidi
1,6221520
$endgroup$
Prove that $frac(72!)(36!)^2-1$ is divisible by 73.
My approach is as follow $73n=frac(72!)(36!)^2-1$ I tried remainder theorem but could not prove it.
modular-arithmetic divisibility natural-numbers
modular-arithmetic divisibility natural-numbers
edited Mar 31 at 9:05
Michael Rozenberg
110k1896201
asked Mar 31 at 6:42
Samar Imam ZaidiSamar Imam Zaidi
1,6221520
edited Mar 31 at 9:05
Michael Rozenberg
110k1896201
asked Mar 31 at 6:42
Samar Imam ZaidiSamar Imam Zaidi
1,6221520
edited Mar 31 at 9:05
Michael Rozenberg
110k1896201
edited Mar 31 at 9:05
Michael Rozenberg
110k1896201
edited Mar 31 at 9:05
Michael Rozenberg
110k1896201
110k1896201
asked Mar 31 at 6:42
Samar Imam ZaidiSamar Imam Zaidi
1,6221520
asked Mar 31 at 6:42
Samar Imam ZaidiSamar Imam Zaidi
1,6221520
asked Mar 31 at 6:42
Samar Imam ZaidiSamar Imam Zaidi
1,6221520
1,6221520
closed as off-topic by user21820, José Carlos Santos, Adrian Keister, K.Power, RRL Apr 4 at 16:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, Adrian Keister, K.Power, RRL
closed as off-topic by user21820, José Carlos Santos, Adrian Keister, K.Power, RRL Apr 4 at 16:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, Adrian Keister, K.Power, RRL
$begingroup$
Use $73-aequiv-apmod73$.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:45
$begingroup$
$binom72kequiv(-1)^kpmod73$
$endgroup$
– robjohn♦
Mar 31 at 9:26
$begingroup$
Special case of this.
$endgroup$
– Bill Dubuque
Apr 1 at 3:04
add a comment |
$begingroup$
Use $73-aequiv-apmod73$.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:45
$begingroup$
$binom72kequiv(-1)^kpmod73$
$endgroup$
– robjohn♦
Mar 31 at 9:26
$begingroup$
Special case of this.
$endgroup$
– Bill Dubuque
Apr 1 at 3:04
$begingroup$
Use $73-aequiv-apmod73$.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:45
$begingroup$
Use $73-aequiv-apmod73$.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:45
$begingroup$
$binom72kequiv(-1)^kpmod73$
$endgroup$
– robjohn♦
Mar 31 at 9:26
$begingroup$
$binom72kequiv(-1)^kpmod73$
$endgroup$
– robjohn♦
Mar 31 at 9:26
$begingroup$
Special case of this.
$endgroup$
– Bill Dubuque
Apr 1 at 3:04
$begingroup$
Special case of this.
$endgroup$
– Bill Dubuque
Apr 1 at 3:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because $$frac72!(36!)^2=frac72cdot71cdot...cdot371cdot2cdot...cdot36equivfrac-1cdot(-2)cdot...cdot(-36)1cdot2cdot...cdot36=(-1)^36=1.$$
answered Mar 31 at 8:03
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Hint: Prove that $$72! equiv (36!)^2 pmod73$$
Hint 2: Deduce from above that
$$73 | left(frac(72!)(36!)^2-1right) cdot (36!)^2$$
Since 73 is prime....
answered Mar 31 at 6:50
N. S.N. S.
105k7115210
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because $$frac72!(36!)^2=frac72cdot71cdot...cdot371cdot2cdot...cdot36equivfrac-1cdot(-2)cdot...cdot(-36)1cdot2cdot...cdot36=(-1)^36=1.$$
answered Mar 31 at 8:03
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Because $$frac72!(36!)^2=frac72cdot71cdot...cdot371cdot2cdot...cdot36equivfrac-1cdot(-2)cdot...cdot(-36)1cdot2cdot...cdot36=(-1)^36=1.$$
answered Mar 31 at 8:03
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Because $$frac72!(36!)^2=frac72cdot71cdot...cdot371cdot2cdot...cdot36equivfrac-1cdot(-2)cdot...cdot(-36)1cdot2cdot...cdot36=(-1)^36=1.$$
answered Mar 31 at 8:03
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
Because $$frac72!(36!)^2=frac72cdot71cdot...cdot371cdot2cdot...cdot36equivfrac-1cdot(-2)cdot...cdot(-36)1cdot2cdot...cdot36=(-1)^36=1.$$
answered Mar 31 at 8:03
Michael RozenbergMichael Rozenberg
110k1896201
answered Mar 31 at 8:03
Michael RozenbergMichael Rozenberg
110k1896201
answered Mar 31 at 8:03
Michael RozenbergMichael Rozenberg
110k1896201
answered Mar 31 at 8:03
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
add a comment |
add a comment |
$begingroup$
Hint: Prove that $$72! equiv (36!)^2 pmod73$$
Hint 2: Deduce from above that
$$73 | left(frac(72!)(36!)^2-1right) cdot (36!)^2$$
Since 73 is prime....
answered Mar 31 at 6:50
N. S.N. S.
105k7115210
$endgroup$
add a comment |
$begingroup$
Hint: Prove that $$72! equiv (36!)^2 pmod73$$
Hint 2: Deduce from above that
$$73 | left(frac(72!)(36!)^2-1right) cdot (36!)^2$$
Since 73 is prime....
answered Mar 31 at 6:50
N. S.N. S.
105k7115210
$endgroup$
add a comment |
$begingroup$
Hint: Prove that $$72! equiv (36!)^2 pmod73$$
Hint 2: Deduce from above that
$$73 | left(frac(72!)(36!)^2-1right) cdot (36!)^2$$
Since 73 is prime....
answered Mar 31 at 6:50
N. S.N. S.
105k7115210
$endgroup$
Hint: Prove that $$72! equiv (36!)^2 pmod73$$
Hint 2: Deduce from above that
$$73 | left(frac(72!)(36!)^2-1right) cdot (36!)^2$$
Since 73 is prime....
answered Mar 31 at 6:50
N. S.N. S.
105k7115210
answered Mar 31 at 6:50
N. S.N. S.
105k7115210
answered Mar 31 at 6:50
N. S.N. S.
105k7115210
answered Mar 31 at 6:50
N. S.N. S.
105k7115210
105k7115210
add a comment |
add a comment |
$begingroup$
Use $73-aequiv-apmod73$.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:45
$begingroup$
$binom72kequiv(-1)^kpmod73$
$endgroup$
– robjohn♦
Mar 31 at 9:26
$begingroup$
Special case of this.
$endgroup$
– Bill Dubuque
Apr 1 at 3:04