One term of (2π+5)^n = 288000π^8, what's n?
Without using calculator what's the value of n?
Using binomial expansion I get:
nCp * 2n-p * πn-p * 5p = 288000π8
Easily I know that n-p=8, by the π's power
Then the power of 2 is also 8, so I can divide both sides
nCp * 5p = 1125
Well I know that
- n-p=8
- n-8=p
- n=p+8
And with "pascal triangle's" rule(nCp = nCn-p)
nC8 * 5p = 1125
nC8 * 5n-8 = 1125
Well there are definitively many ways to equate this, but how can I get the value of n without calculator or guessing(Trial and error)?
Or is there any other approach?
The answer is n=10 and p=2...
Sorry for no LaTeX(I don't know how to work with it
combinatorics binomial-coefficients binomial-theorem
add a comment |
Without using calculator what's the value of n?
Using binomial expansion I get:
nCp * 2n-p * πn-p * 5p = 288000π8
Easily I know that n-p=8, by the π's power
Then the power of 2 is also 8, so I can divide both sides
nCp * 5p = 1125
Well I know that
- n-p=8
- n-8=p
- n=p+8
And with "pascal triangle's" rule(nCp = nCn-p)
nC8 * 5p = 1125
nC8 * 5n-8 = 1125
Well there are definitively many ways to equate this, but how can I get the value of n without calculator or guessing(Trial and error)?
Or is there any other approach?
The answer is n=10 and p=2...
Sorry for no LaTeX(I don't know how to work with it
combinatorics binomial-coefficients binomial-theorem
1
math.meta.stackexchange.com/questions/5020/…
– saulspatz
Nov 29 '18 at 22:12
1
The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
– platty
Nov 29 '18 at 22:19
It was my bad sorry, corrected it
– Nuno Mateus
Nov 29 '18 at 22:23
add a comment |
Without using calculator what's the value of n?
Using binomial expansion I get:
nCp * 2n-p * πn-p * 5p = 288000π8
Easily I know that n-p=8, by the π's power
Then the power of 2 is also 8, so I can divide both sides
nCp * 5p = 1125
Well I know that
- n-p=8
- n-8=p
- n=p+8
And with "pascal triangle's" rule(nCp = nCn-p)
nC8 * 5p = 1125
nC8 * 5n-8 = 1125
Well there are definitively many ways to equate this, but how can I get the value of n without calculator or guessing(Trial and error)?
Or is there any other approach?
The answer is n=10 and p=2...
Sorry for no LaTeX(I don't know how to work with it
combinatorics binomial-coefficients binomial-theorem
Without using calculator what's the value of n?
Using binomial expansion I get:
nCp * 2n-p * πn-p * 5p = 288000π8
Easily I know that n-p=8, by the π's power
Then the power of 2 is also 8, so I can divide both sides
nCp * 5p = 1125
Well I know that
- n-p=8
- n-8=p
- n=p+8
And with "pascal triangle's" rule(nCp = nCn-p)
nC8 * 5p = 1125
nC8 * 5n-8 = 1125
Well there are definitively many ways to equate this, but how can I get the value of n without calculator or guessing(Trial and error)?
Or is there any other approach?
The answer is n=10 and p=2...
Sorry for no LaTeX(I don't know how to work with it
combinatorics binomial-coefficients binomial-theorem
combinatorics binomial-coefficients binomial-theorem
edited Nov 29 '18 at 22:23
Nuno Mateus
asked Nov 29 '18 at 22:07
Nuno MateusNuno Mateus
494
494
1
math.meta.stackexchange.com/questions/5020/…
– saulspatz
Nov 29 '18 at 22:12
1
The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
– platty
Nov 29 '18 at 22:19
It was my bad sorry, corrected it
– Nuno Mateus
Nov 29 '18 at 22:23
add a comment |
1
math.meta.stackexchange.com/questions/5020/…
– saulspatz
Nov 29 '18 at 22:12
1
The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
– platty
Nov 29 '18 at 22:19
It was my bad sorry, corrected it
– Nuno Mateus
Nov 29 '18 at 22:23
1
1
math.meta.stackexchange.com/questions/5020/…
– saulspatz
Nov 29 '18 at 22:12
math.meta.stackexchange.com/questions/5020/…
– saulspatz
Nov 29 '18 at 22:12
1
1
The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
– platty
Nov 29 '18 at 22:19
The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
– platty
Nov 29 '18 at 22:19
It was my bad sorry, corrected it
– Nuno Mateus
Nov 29 '18 at 22:23
It was my bad sorry, corrected it
– Nuno Mateus
Nov 29 '18 at 22:23
add a comment |
2 Answers
2
active
oldest
votes
The term of degree $8$ in the expansion of $(x+a)^n$ is
$$
binom{n}{8}a^{n-8}x^8
$$
so you want to solve
$$
binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
$$
hence
$$
binom{n}{8}5^{n-11}=9
$$
We can exclude $n>11$, because the right hand side is not divisible by $5$.
Therefore you just have to check $n=8,9,10,11$:
begin{align}
binom{8}{8}&=1 \
binom{9}{8}&=binom{9}{1}=9 \
binom{10}{8}&=binom{10}{2}=cdots\
binom{11}{8}&=binom{11}{3}=cdots
end{align}
add a comment |
${nchoose 8} * 5^{n-8} = 1125$
But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$
And $5^{n-11}{nchoose 8} = 9$.
$9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.
Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.
So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so
so $n =10$.
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The term of degree $8$ in the expansion of $(x+a)^n$ is
$$
binom{n}{8}a^{n-8}x^8
$$
so you want to solve
$$
binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
$$
hence
$$
binom{n}{8}5^{n-11}=9
$$
We can exclude $n>11$, because the right hand side is not divisible by $5$.
Therefore you just have to check $n=8,9,10,11$:
begin{align}
binom{8}{8}&=1 \
binom{9}{8}&=binom{9}{1}=9 \
binom{10}{8}&=binom{10}{2}=cdots\
binom{11}{8}&=binom{11}{3}=cdots
end{align}
add a comment |
The term of degree $8$ in the expansion of $(x+a)^n$ is
$$
binom{n}{8}a^{n-8}x^8
$$
so you want to solve
$$
binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
$$
hence
$$
binom{n}{8}5^{n-11}=9
$$
We can exclude $n>11$, because the right hand side is not divisible by $5$.
Therefore you just have to check $n=8,9,10,11$:
begin{align}
binom{8}{8}&=1 \
binom{9}{8}&=binom{9}{1}=9 \
binom{10}{8}&=binom{10}{2}=cdots\
binom{11}{8}&=binom{11}{3}=cdots
end{align}
add a comment |
The term of degree $8$ in the expansion of $(x+a)^n$ is
$$
binom{n}{8}a^{n-8}x^8
$$
so you want to solve
$$
binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
$$
hence
$$
binom{n}{8}5^{n-11}=9
$$
We can exclude $n>11$, because the right hand side is not divisible by $5$.
Therefore you just have to check $n=8,9,10,11$:
begin{align}
binom{8}{8}&=1 \
binom{9}{8}&=binom{9}{1}=9 \
binom{10}{8}&=binom{10}{2}=cdots\
binom{11}{8}&=binom{11}{3}=cdots
end{align}
The term of degree $8$ in the expansion of $(x+a)^n$ is
$$
binom{n}{8}a^{n-8}x^8
$$
so you want to solve
$$
binom{n}{8}5^{n-8}2^8=288000=2^83^25^3
$$
hence
$$
binom{n}{8}5^{n-11}=9
$$
We can exclude $n>11$, because the right hand side is not divisible by $5$.
Therefore you just have to check $n=8,9,10,11$:
begin{align}
binom{8}{8}&=1 \
binom{9}{8}&=binom{9}{1}=9 \
binom{10}{8}&=binom{10}{2}=cdots\
binom{11}{8}&=binom{11}{3}=cdots
end{align}
answered Nov 29 '18 at 22:39
egregegreg
179k1485202
179k1485202
add a comment |
add a comment |
${nchoose 8} * 5^{n-8} = 1125$
But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$
And $5^{n-11}{nchoose 8} = 9$.
$9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.
Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.
So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so
so $n =10$.
add a comment |
${nchoose 8} * 5^{n-8} = 1125$
But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$
And $5^{n-11}{nchoose 8} = 9$.
$9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.
Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.
So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so
so $n =10$.
add a comment |
${nchoose 8} * 5^{n-8} = 1125$
But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$
And $5^{n-11}{nchoose 8} = 9$.
$9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.
Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.
So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so
so $n =10$.
${nchoose 8} * 5^{n-8} = 1125$
But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8le n le 11$
And $5^{n-11}{nchoose 8} = 9$.
$9not mid 5$ so either $n-11=0$ or $5^{11-n}|{nchoose 8}$.
Now as $8le nle 11$ and ${nchoose 8}=frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.
So $n = 10$ or $11$. Now as no prime but $3,5$ divide $frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so
so $n =10$.
edited Nov 30 '18 at 0:14
answered Nov 30 '18 at 0:04
fleabloodfleablood
68.7k22685
68.7k22685
add a comment |
add a comment |
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1
math.meta.stackexchange.com/questions/5020/…
– saulspatz
Nov 29 '18 at 22:12
1
The initial use of binomial theorem doesn't look quite right -- you'd want $pi^p$ instead of $pi^{(n-p) * p}$. So either $p = 8$ or $n-p = 8$ and you would have $2^p$ instead of $2^{n-p}$.
– platty
Nov 29 '18 at 22:19
It was my bad sorry, corrected it
– Nuno Mateus
Nov 29 '18 at 22:23