Prove that $forall n in Bbb N, ~40^n n! mid (5n)!$
$begingroup$
I'm having trouble trying to solve this problem:
Prove that $forall n in Bbb N,~ 40^n n! mid (5n)!$
I must be overlooking something simple because I can't go through it with induction.
Base case $P(1)$ works.
I want to see that it holds for $P(n+1)$.
So I'm looking at $40^{n+1} (n+1)! | (5n+5)!$ with the hypothesis that what I want to prove is right.
But operating I end up with: $dfrac{(5n+5)!}{40^{n+1} (n+1)!}$
If I expand the factorial in the numerator I see my inductive hypothesis but I have an 8 in the denominator and some other factors above that I don't know how to deal with: $$dfrac{(5n+4)(5n+3)(5n+2)(5n+1)5n!}{40^{n} (n)! cdot 8}$$
Perhaps this is not the way to do it. I've also tried starting with my hypothesis and then multiplying by $dfrac{40}{40}$ and $dfrac{(n+1)}{(n+1)}$ but it's almost the same thing.
I'll be thankful with any suggestion!
elementary-number-theory proof-verification proof-writing divisibility factorial
edited Apr 25 '17 at 23:54
Jaideep Khare
17.7k32569
asked Apr 25 '17 at 23:38
Joaquin RomeraJoaquin Romera
883520
$endgroup$
add a comment |
$begingroup$
I'm having trouble trying to solve this problem:
Prove that $forall n in Bbb N,~ 40^n n! mid (5n)!$
I must be overlooking something simple because I can't go through it with induction.
Base case $P(1)$ works.
I want to see that it holds for $P(n+1)$.
So I'm looking at $40^{n+1} (n+1)! | (5n+5)!$ with the hypothesis that what I want to prove is right.
But operating I end up with: $dfrac{(5n+5)!}{40^{n+1} (n+1)!}$
If I expand the factorial in the numerator I see my inductive hypothesis but I have an 8 in the denominator and some other factors above that I don't know how to deal with: $$dfrac{(5n+4)(5n+3)(5n+2)(5n+1)5n!}{40^{n} (n)! cdot 8}$$
Perhaps this is not the way to do it. I've also tried starting with my hypothesis and then multiplying by $dfrac{40}{40}$ and $dfrac{(n+1)}{(n+1)}$ but it's almost the same thing.
I'll be thankful with any suggestion!
elementary-number-theory proof-verification proof-writing divisibility factorial
edited Apr 25 '17 at 23:54
Jaideep Khare
17.7k32569
asked Apr 25 '17 at 23:38
Joaquin RomeraJoaquin Romera
883520
$endgroup$
2
$begingroup$
So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:42
1
$begingroup$
You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
$endgroup$
– Marcelo Fornet
Apr 25 '17 at 23:44
1
$begingroup$
The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:46
add a comment |
$begingroup$
I'm having trouble trying to solve this problem:
Prove that $forall n in Bbb N,~ 40^n n! mid (5n)!$
I must be overlooking something simple because I can't go through it with induction.
Base case $P(1)$ works.
I want to see that it holds for $P(n+1)$.
So I'm looking at $40^{n+1} (n+1)! | (5n+5)!$ with the hypothesis that what I want to prove is right.
But operating I end up with: $dfrac{(5n+5)!}{40^{n+1} (n+1)!}$
If I expand the factorial in the numerator I see my inductive hypothesis but I have an 8 in the denominator and some other factors above that I don't know how to deal with: $$dfrac{(5n+4)(5n+3)(5n+2)(5n+1)5n!}{40^{n} (n)! cdot 8}$$
Perhaps this is not the way to do it. I've also tried starting with my hypothesis and then multiplying by $dfrac{40}{40}$ and $dfrac{(n+1)}{(n+1)}$ but it's almost the same thing.
I'll be thankful with any suggestion!
elementary-number-theory proof-verification proof-writing divisibility factorial
edited Apr 25 '17 at 23:54
Jaideep Khare
17.7k32569
asked Apr 25 '17 at 23:38
Joaquin RomeraJoaquin Romera
883520
$endgroup$
I'm having trouble trying to solve this problem:
Prove that $forall n in Bbb N,~ 40^n n! mid (5n)!$
I must be overlooking something simple because I can't go through it with induction.
Base case $P(1)$ works.
I want to see that it holds for $P(n+1)$.
So I'm looking at $40^{n+1} (n+1)! | (5n+5)!$ with the hypothesis that what I want to prove is right.
But operating I end up with: $dfrac{(5n+5)!}{40^{n+1} (n+1)!}$
If I expand the factorial in the numerator I see my inductive hypothesis but I have an 8 in the denominator and some other factors above that I don't know how to deal with: $$dfrac{(5n+4)(5n+3)(5n+2)(5n+1)5n!}{40^{n} (n)! cdot 8}$$
Perhaps this is not the way to do it. I've also tried starting with my hypothesis and then multiplying by $dfrac{40}{40}$ and $dfrac{(n+1)}{(n+1)}$ but it's almost the same thing.
I'll be thankful with any suggestion!
elementary-number-theory proof-verification proof-writing divisibility factorial
elementary-number-theory proof-verification proof-writing divisibility factorial
edited Apr 25 '17 at 23:54
Jaideep Khare
17.7k32569
asked Apr 25 '17 at 23:38
Joaquin RomeraJoaquin Romera
883520
edited Apr 25 '17 at 23:54
Jaideep Khare
17.7k32569
asked Apr 25 '17 at 23:38
Joaquin RomeraJoaquin Romera
883520
edited Apr 25 '17 at 23:54
Jaideep Khare
17.7k32569
edited Apr 25 '17 at 23:54
Jaideep Khare
17.7k32569
edited Apr 25 '17 at 23:54
Jaideep Khare
17.7k32569
17.7k32569
asked Apr 25 '17 at 23:38
Joaquin RomeraJoaquin Romera
883520
asked Apr 25 '17 at 23:38
Joaquin RomeraJoaquin Romera
883520
asked Apr 25 '17 at 23:38
Joaquin RomeraJoaquin Romera
883520
883520
2
$begingroup$
So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:42
1
$begingroup$
You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
$endgroup$
– Marcelo Fornet
Apr 25 '17 at 23:44
1
$begingroup$
The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:46
add a comment |
2
$begingroup$
So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:42
1
$begingroup$
You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
$endgroup$
– Marcelo Fornet
Apr 25 '17 at 23:44
1
$begingroup$
The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:46
2
2
$begingroup$
So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:42
$begingroup$
So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:42
1
1
$begingroup$
You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
$endgroup$
– Marcelo Fornet
Apr 25 '17 at 23:44
$begingroup$
You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
$endgroup$
– Marcelo Fornet
Apr 25 '17 at 23:44
1
1
$begingroup$
The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:46
$begingroup$
The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are actually done:
$$
frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
$$
The first is an integer, by the inductive hypothesis.
To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).
Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.
Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.
answered Apr 25 '17 at 23:45
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.1k33376
$endgroup$
add a comment |
$begingroup$
You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.
Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
answered Apr 25 '17 at 23:45
Jaideep KhareJaideep Khare
17.7k32569
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You are actually done:
$$
frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
$$
The first is an integer, by the inductive hypothesis.
To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).
Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.
Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.
answered Apr 25 '17 at 23:45
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.1k33376
$endgroup$
add a comment |
$begingroup$
You are actually done:
$$
frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
$$
The first is an integer, by the inductive hypothesis.
To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).
Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.
Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.
answered Apr 25 '17 at 23:45
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.1k33376
$endgroup$
add a comment |
$begingroup$
You are actually done:
$$
frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
$$
The first is an integer, by the inductive hypothesis.
To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).
Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.
Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.
answered Apr 25 '17 at 23:45
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.1k33376
$endgroup$
You are actually done:
$$
frac{(5n+5)!}{40^{n+1}(n+1)!} = frac{(5n)!}{40^n (n)!} times frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{5 times 8}
$$
The first is an integer, by the inductive hypothesis.
To see that the second is an integer, note that $5$ divides $5n+5$. Of the remaining $4$ consecutive numbers $5n+1 ... 5n+4$, at least two are even. Furthermore, one of them is a multiple of $4$ ($4$ consecutive numbers must leave all possible remainders mod $4$, including zero).
Hence, the product of these is a multiple of $4 times 2 = 8$, so it follows that the other expression is also an integer. Hence, the LHS is an integer, and the induction is complete.
Example : $n=7$, then $5n+1 ... 5n+4 = 36,37,38,39$, where $36$ is a multiple of $4$ and $38$ is a multiple of $2$. Similarly, $n=10$, then $52 = 5n+2$ is a multiple of $4$, and $54$ is a multiple of $2$.
answered Apr 25 '17 at 23:45
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.1k33376
edited Jan 3 at 17:51
answered Apr 25 '17 at 23:45
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.1k33376
answered Apr 25 '17 at 23:45
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.1k33376
answered Apr 25 '17 at 23:45
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.1k33376
38.1k33376
add a comment |
add a comment |
$begingroup$
You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.
Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
answered Apr 25 '17 at 23:45
Jaideep KhareJaideep Khare
17.7k32569
$endgroup$
add a comment |
$begingroup$
You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.
Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
answered Apr 25 '17 at 23:45
Jaideep KhareJaideep Khare
17.7k32569
$endgroup$
add a comment |
$begingroup$
You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.
Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
answered Apr 25 '17 at 23:45
Jaideep KhareJaideep Khare
17.7k32569
$endgroup$
You only need to prove that : $$ 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
Since they are four consecutive numbers, at least one of them is divisible by $4$, and out of remaining three, at least one is even. Hence you get a factor of $4$ from one of them and a factor of $2$ from remaining three.
Therefore : $$4 times 2 mid(5n+4)(5n+3)(5n+2)(5n+1) implies 8 mid(5n+4)(5n+3)(5n+2)(5n+1)$$
answered Apr 25 '17 at 23:45
Jaideep KhareJaideep Khare
17.7k32569
answered Apr 25 '17 at 23:45
Jaideep KhareJaideep Khare
17.7k32569
answered Apr 25 '17 at 23:45
Jaideep KhareJaideep Khare
17.7k32569
answered Apr 25 '17 at 23:45
Jaideep KhareJaideep Khare
17.7k32569
17.7k32569
add a comment |
add a comment |
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$begingroup$
So you need $(5n+4)(5n+3)(5n+2)(5n+1)$ divisible by $8$...
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:42
1
$begingroup$
You are on the track. Try to prove that $8$ divides $(5n+4)(5n+3)(5n+2)(5n+1)$
$endgroup$
– Marcelo Fornet
Apr 25 '17 at 23:44
1
$begingroup$
The combinatorial proof would be to show that $frac{(5n)!}{n!(5!)^n}$ counts the number of ways to partition a set of $5n$ elements into $n$ sets of $5$ elements each.
$endgroup$
– Thomas Andrews
Apr 25 '17 at 23:46