Find the Factorial From the Given Prime Relationship
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Find the Factorial From the Given Prime Relationship
$begingroup$
$Given$:
$P$, $Q$, $R$ are three distinct Prime Numbers
$P!$ = $Q$ x $R^Q$ x $P$
Find P, Q, R.
mathematics no-computers
asked 9 hours ago
UvcUvc
1,264119
$endgroup$
add a comment |
$begingroup$
$Given$:
$P$, $Q$, $R$ are three distinct Prime Numbers
$P!$ = $Q$ x $R^Q$ x $P$
Find P, Q, R.
mathematics no-computers
asked 9 hours ago
UvcUvc
1,264119
$endgroup$
add a comment |
$begingroup$
$Given$:
$P$, $Q$, $R$ are three distinct Prime Numbers
$P!$ = $Q$ x $R^Q$ x $P$
Find P, Q, R.
mathematics no-computers
asked 9 hours ago
UvcUvc
1,264119
$endgroup$
$Given$:
$P$, $Q$, $R$ are three distinct Prime Numbers
$P!$ = $Q$ x $R^Q$ x $P$
Find P, Q, R.
mathematics no-computers
mathematics no-computers
asked 9 hours ago
UvcUvc
1,264119
asked 9 hours ago
UvcUvc
1,264119
asked 9 hours ago
UvcUvc
1,264119
asked 9 hours ago
UvcUvc
1,264119
asked 9 hours ago
UvcUvc
1,264119
1,264119
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think it’s
P = 5, Q = 3, and R = 2.
This gives
$5! = 120 = 3 times 2^3 times 5$.
We note that
$P! = P times (P-1)!$, so $(P-1)! = R^Q times Q$. Noting that one of P, Q, and R had to be at least 5, I noted that $4 = 2^2$ and so $4 times 2 = 2^3$ was probably a convenient way to include these factors of $P!$.
answered 9 hours ago
El-GuestEl-Guest
23.4k35395
$endgroup$
add a comment |
$begingroup$
It's:
$5!=3cdot2^3cdot5$
Because:
$P=5$ as there are exactly $3$ prime factors in the factorial.
Therefore $24=Qcdot R^Q=3cdot2^3$, and so $Q=3$ and $R=2$.
answered 9 hours ago
JonMark PerryJonMark Perry
22.2k643103
$endgroup$
$begingroup$
Great answer and explanation!
$endgroup$
– El-Guest
8 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think it’s
P = 5, Q = 3, and R = 2.
This gives
$5! = 120 = 3 times 2^3 times 5$.
We note that
$P! = P times (P-1)!$, so $(P-1)! = R^Q times Q$. Noting that one of P, Q, and R had to be at least 5, I noted that $4 = 2^2$ and so $4 times 2 = 2^3$ was probably a convenient way to include these factors of $P!$.
answered 9 hours ago
El-GuestEl-Guest
23.4k35395
$endgroup$
add a comment |
$begingroup$
I think it’s
P = 5, Q = 3, and R = 2.
This gives
$5! = 120 = 3 times 2^3 times 5$.
We note that
$P! = P times (P-1)!$, so $(P-1)! = R^Q times Q$. Noting that one of P, Q, and R had to be at least 5, I noted that $4 = 2^2$ and so $4 times 2 = 2^3$ was probably a convenient way to include these factors of $P!$.
answered 9 hours ago
El-GuestEl-Guest
23.4k35395
$endgroup$
add a comment |
$begingroup$
I think it’s
P = 5, Q = 3, and R = 2.
This gives
$5! = 120 = 3 times 2^3 times 5$.
We note that
$P! = P times (P-1)!$, so $(P-1)! = R^Q times Q$. Noting that one of P, Q, and R had to be at least 5, I noted that $4 = 2^2$ and so $4 times 2 = 2^3$ was probably a convenient way to include these factors of $P!$.
answered 9 hours ago
El-GuestEl-Guest
23.4k35395
$endgroup$
I think it’s
P = 5, Q = 3, and R = 2.
This gives
$5! = 120 = 3 times 2^3 times 5$.
We note that
$P! = P times (P-1)!$, so $(P-1)! = R^Q times Q$. Noting that one of P, Q, and R had to be at least 5, I noted that $4 = 2^2$ and so $4 times 2 = 2^3$ was probably a convenient way to include these factors of $P!$.
answered 9 hours ago
El-GuestEl-Guest
23.4k35395
answered 9 hours ago
El-GuestEl-Guest
23.4k35395
answered 9 hours ago
El-GuestEl-Guest
23.4k35395
answered 9 hours ago
El-GuestEl-Guest
23.4k35395
23.4k35395
add a comment |
add a comment |
$begingroup$
It's:
$5!=3cdot2^3cdot5$
Because:
$P=5$ as there are exactly $3$ prime factors in the factorial.
Therefore $24=Qcdot R^Q=3cdot2^3$, and so $Q=3$ and $R=2$.
answered 9 hours ago
JonMark PerryJonMark Perry
22.2k643103
$endgroup$
$begingroup$
Great answer and explanation!
$endgroup$
– El-Guest
8 hours ago
add a comment |
$begingroup$
It's:
$5!=3cdot2^3cdot5$
Because:
$P=5$ as there are exactly $3$ prime factors in the factorial.
Therefore $24=Qcdot R^Q=3cdot2^3$, and so $Q=3$ and $R=2$.
answered 9 hours ago
JonMark PerryJonMark Perry
22.2k643103
$endgroup$
$begingroup$
Great answer and explanation!
$endgroup$
– El-Guest
8 hours ago
add a comment |
$begingroup$
It's:
$5!=3cdot2^3cdot5$
Because:
$P=5$ as there are exactly $3$ prime factors in the factorial.
Therefore $24=Qcdot R^Q=3cdot2^3$, and so $Q=3$ and $R=2$.
answered 9 hours ago
JonMark PerryJonMark Perry
22.2k643103
$endgroup$
It's:
$5!=3cdot2^3cdot5$
Because:
$P=5$ as there are exactly $3$ prime factors in the factorial.
Therefore $24=Qcdot R^Q=3cdot2^3$, and so $Q=3$ and $R=2$.
answered 9 hours ago
JonMark PerryJonMark Perry
22.2k643103
answered 9 hours ago
JonMark PerryJonMark Perry
22.2k643103
answered 9 hours ago
JonMark PerryJonMark Perry
22.2k643103
answered 9 hours ago
JonMark PerryJonMark Perry
22.2k643103
22.2k643103
$begingroup$
Great answer and explanation!
$endgroup$
– El-Guest
8 hours ago
add a comment |
$begingroup$
Great answer and explanation!
$endgroup$
– El-Guest
8 hours ago
$begingroup$
Great answer and explanation!
$endgroup$
– El-Guest
8 hours ago
$begingroup$
Great answer and explanation!
$endgroup$
– El-Guest
8 hours ago
add a comment |
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