How to prove that $$\frac{1000!}{(500!)^2}$$ is not divisible by 7?
I reduced the above fraction to: $$\frac{\prod_{k=501}^{1000}k}{500!} $$
But I don't know how to proceed. Please help.
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$\begingroup$ Hint: consider prime factorisations. Focus on $7$'s. $\endgroup$ – Arthur Jan 24 '17 at 15:29
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It is a general formula that the number of times a prime $p$ occurs in the prime factorisation of $n!$ is given by
$$ \left[\frac{n}{p}\right] + \left[\frac{n}{p^2}\right] + \left[\frac{n}{p^3}\right] + \cdots $$
where $[x]$ denotes the greatest integer $≤x$.
In $1000!$ you'd have
$$ \left[\frac{1000}{7}\right] + \left[\frac{1000}{7^2}\right] + \left[\frac{1000}{7^3}\right] + \cdots = 164 $$
$7$'s, and in $500!$ you have
$$ \left[\frac{500}{7}\right] + \left[\frac{500}{7^2}\right] + \left[\frac{500}{7^3}\right] + \cdots = 82 $$
$7$'s. So you'll have a total of $164$ 7's in $(500!)^2$.
The rest follows.
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$\begingroup$ I am sorry but would you kindly help in the conclusion? I scratched my head for a while but did not understand how that proves that it is not divisible by 7. $\endgroup$ – Macindows Nov 14 '18 at 14:59
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This is overkill, but by Lucas's Theorem$$\frac{1000!}{(500!)^2}=\binom{1000}{500} \equiv \binom{2}{1}\binom{6}{3}\binom{2}{1}\binom{6}{3} \equiv 4 \pmod{7}$$ As $1000=2626_{7}$ and $500=1313_{7}$.
Here is another , more elementary argument, from Legendre's formula, $$\nu_7(n!) = \frac{n - s_7(n)}{6}$$ So $\nu_7(100!)=164, \nu_7 (50!)=82$. So the $7$ s in the denominator and the numerator cancel out. So we have the desired result.
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$\begingroup$ Got it! But you told that applying Lucas's Theorem is overkill, so, is there any other solution? $\endgroup$ – ami_ba Jan 24 '17 at 15:37
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$\begingroup$ @AmitayasBanerjee I added one right now. This one is a more elementary result. $\endgroup$ – S.C.B. Jan 24 '17 at 15:38
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3$\begingroup$ (+1) Two beautiful proofs in a single answer make a well-deserved upvote. $\endgroup$ – Jack D'Aurizio Jan 24 '17 at 15:43
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$\begingroup$ [+1] I love Lucas' theorem, and a good application of it. $\endgroup$ – Andreas Caranti Jan 24 '17 at 16:14
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$\begingroup$ @AmitayasBanerjee Could you tell me why you unaccepted my answer? $\endgroup$ – S.C.B. Apr 29 '17 at 4:01
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You can also overkill it by Kummer's theorem:
Notice that $500$ is $1313$ in base $7$. Since every digit is smaller than $\frac{7}{2}$ there are no carries when adding $1313+1313$ in base $7$.
Therefore $\binom{1000}{500}$ is not a multiple of $7$.
In general we have that a prime $p$ divides $\binom{2n}{n}$ if and only if $n$ has at least one digit larger than or equal to $p/2$ in its base $p$ expansion.
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Consider prime factorisations:
$$\lfloor\frac{1000}{7}\rfloor+\lfloor\frac{1000}{7^2}\rfloor+\lfloor\frac{1000}{7^3}\rfloor+...=142+20+2=164$$ Then $$7^{164}|1000!$$
Similarly:
$$\lfloor\frac{500}{7}\rfloor+\lfloor\frac{500}{7^2}\rfloor+\lfloor\frac{500}{7^3}\rfloor+...=71+10+1=82$$ Then $$7^{164}=7^{82\cdot2}=(7^{82})^2|(500!)^2$$
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1$\begingroup$ The fact that $7^{164}$ divides $1000!$ and $(500!)^2$ does not suffice to decide if $7$ divides their ratio or not. Please be more careful. $\endgroup$ – Did Jan 24 '17 at 15:42
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$\begingroup$ So the point is not that $7^{164}$ divides $1000!$, but that $7^{165}$ does not. $\endgroup$ – Callus - Reinstate Monica Jan 24 '17 at 15:49
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In $\{1,\dots, 500\}$, there are 71 multiples of 7, 10 multiples of $7^2$ and one multiple of $7^3$.
How many multiples of $7$, $7^2$, $7^3$ do you have in $\{501,\dots, 1000\}$? Same amount.
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1$\begingroup$ 504 is first multiple of 7 (with multiplicity of seven is 1 in factorization), and $504 + 7 \cdot 70 = 994$. Also $7^2 \cdot 11 = 539$ (multiplicity is 2) and $7^2 \cdot 20 = 980$. At last $7^3 \cdot 2 = 686$ (fits only once). $\endgroup$ – johnnycrab Jan 24 '17 at 16:18
