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How to prove divisibility of a number using the binomial expansion?
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I have the following problem:
Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.
How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.
algebra-precalculus binomial-coefficients binomial-theorem
asked 1 hour ago
Adam Grey
294
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up vote
1
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favorite
I have the following problem:
Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.
How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.
algebra-precalculus binomial-coefficients binomial-theorem
asked 1 hour ago
Adam Grey
294
See here for a better example and also a presentation using modular arithmetic.
– Bill Dubuque
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the following problem:
Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.
How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.
algebra-precalculus binomial-coefficients binomial-theorem
asked 1 hour ago
Adam Grey
294
I have the following problem:
Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.
How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.
algebra-precalculus binomial-coefficients binomial-theorem
algebra-precalculus binomial-coefficients binomial-theorem
asked 1 hour ago
Adam Grey
294
asked 1 hour ago
Adam Grey
294
asked 1 hour ago
Adam Grey
294
asked 1 hour ago
Adam Grey
294
asked 1 hour ago
Adam Grey
294
294
See here for a better example and also a presentation using modular arithmetic.
– Bill Dubuque
1 hour ago
add a comment |Â
See here for a better example and also a presentation using modular arithmetic.
– Bill Dubuque
1 hour ago
See here for a better example and also a presentation using modular arithmetic.
– Bill Dubuque
1 hour ago
See here for a better example and also a presentation using modular arithmetic.
– Bill Dubuque
1 hour ago
add a comment |Â
4 Answers
4
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oldest
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up vote
3
down vote
accepted
$$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$
answered 1 hour ago
greedoid
30.7k94184
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up vote
2
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HINT:
$$(5+1)^n=sum_k=0^n binomnk5^k$$
Which terms of this sum are divisible by $5$?
answered 1 hour ago
Frpzzd
18.1k63695
(+1) I would have written an answer similar to this.
– robjohn♦
46 mins ago
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up vote
0
down vote
Guide:
Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.
Then try to factor $5$ out from the remaining term.
answered 1 hour ago
Siong Thye Goh
85.1k1457107
add a comment |Â
up vote
0
down vote
$$6^n-1 = (5+1)^n-1$$
$$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
$$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
$n choose n1 = 1$ so $1$ and $-1$ cancel out.
$$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.
answered 1 hour ago
KM101
5427
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$
answered 1 hour ago
greedoid
30.7k94184
add a comment |Â
up vote
3
down vote
accepted
$$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$
answered 1 hour ago
greedoid
30.7k94184
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$
answered 1 hour ago
greedoid
30.7k94184
$$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$
answered 1 hour ago
greedoid
30.7k94184
answered 1 hour ago
greedoid
30.7k94184
answered 1 hour ago
greedoid
30.7k94184
answered 1 hour ago
greedoid
30.7k94184
30.7k94184
add a comment |Â
add a comment |Â
up vote
2
down vote
HINT:
$$(5+1)^n=sum_k=0^n binomnk5^k$$
Which terms of this sum are divisible by $5$?
answered 1 hour ago
Frpzzd
18.1k63695
(+1) I would have written an answer similar to this.
– robjohn♦
46 mins ago
add a comment |Â
up vote
2
down vote
HINT:
$$(5+1)^n=sum_k=0^n binomnk5^k$$
Which terms of this sum are divisible by $5$?
answered 1 hour ago
Frpzzd
18.1k63695
(+1) I would have written an answer similar to this.
– robjohn♦
46 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
HINT:
$$(5+1)^n=sum_k=0^n binomnk5^k$$
Which terms of this sum are divisible by $5$?
answered 1 hour ago
Frpzzd
18.1k63695
HINT:
$$(5+1)^n=sum_k=0^n binomnk5^k$$
Which terms of this sum are divisible by $5$?
answered 1 hour ago
Frpzzd
18.1k63695
answered 1 hour ago
Frpzzd
18.1k63695
answered 1 hour ago
Frpzzd
18.1k63695
answered 1 hour ago
Frpzzd
18.1k63695
18.1k63695
(+1) I would have written an answer similar to this.
– robjohn♦
46 mins ago
add a comment |Â
(+1) I would have written an answer similar to this.
– robjohn♦
46 mins ago
(+1) I would have written an answer similar to this.
– robjohn♦
46 mins ago
(+1) I would have written an answer similar to this.
– robjohn♦
46 mins ago
add a comment |Â
up vote
0
down vote
Guide:
Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.
Then try to factor $5$ out from the remaining term.
answered 1 hour ago
Siong Thye Goh
85.1k1457107
add a comment |Â
up vote
0
down vote
Guide:
Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.
Then try to factor $5$ out from the remaining term.
answered 1 hour ago
Siong Thye Goh
85.1k1457107
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Guide:
Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.
Then try to factor $5$ out from the remaining term.
answered 1 hour ago
Siong Thye Goh
85.1k1457107
Guide:
Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.
Then try to factor $5$ out from the remaining term.
answered 1 hour ago
Siong Thye Goh
85.1k1457107
answered 1 hour ago
Siong Thye Goh
85.1k1457107
answered 1 hour ago
Siong Thye Goh
85.1k1457107
answered 1 hour ago
Siong Thye Goh
85.1k1457107
85.1k1457107
add a comment |Â
add a comment |Â
up vote
0
down vote
$$6^n-1 = (5+1)^n-1$$
$$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
$$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
$n choose n1 = 1$ so $1$ and $-1$ cancel out.
$$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.
answered 1 hour ago
KM101
5427
add a comment |Â
up vote
0
down vote
$$6^n-1 = (5+1)^n-1$$
$$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
$$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
$n choose n1 = 1$ so $1$ and $-1$ cancel out.
$$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.
answered 1 hour ago
KM101
5427
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$6^n-1 = (5+1)^n-1$$
$$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
$$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
$n choose n1 = 1$ so $1$ and $-1$ cancel out.
$$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.
answered 1 hour ago
KM101
5427
$$6^n-1 = (5+1)^n-1$$
$$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
$$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
$n choose n1 = 1$ so $1$ and $-1$ cancel out.
$$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.
answered 1 hour ago
KM101
5427
edited 1 hour ago
answered 1 hour ago
KM101
5427
answered 1 hour ago
KM101
5427
answered 1 hour ago
KM101
5427
5427
add a comment |Â
add a comment |Â
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See here for a better example and also a presentation using modular arithmetic.
– Bill Dubuque
1 hour ago