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Prove that $3cdot 5^2n+1 +2^3n+1$ is divisible by $17$ for all $n ∈ mathbbN$
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up vote
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Use mathematical induction to prove that $3cdot 5^2n+1 +2^3n+1$
is divisible by $17$ for all $n ∈ mathbbN$.
I've tried to do it as follow.
If $n = 1$ then $392/17 = 23$.
Assume it is true when $n = p$. Therefore $3cdot 5^2p+1 +2^3p+1 = 17k $ where $k ∈ mathbbN $. Consider now $n=p+1$. Then
beginalign
&3cdot 5^2(p+1)+1 +2^3(p+1)+1=\
&3cdot 5^2p+1+2 + 2^3p+1+3=\
&3cdot5^2p+1cdot 5^2 + 2^3p+1cdot 2^3.
endalign
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.
induction divisibility
edited 2 mins ago
TheSimpliFire
11.5k62256
asked 44 mins ago
Shehan Tearz
874
add a comment |Â
up vote
4
down vote
favorite
Use mathematical induction to prove that $3cdot 5^2n+1 +2^3n+1$
is divisible by $17$ for all $n ∈ mathbbN$.
I've tried to do it as follow.
If $n = 1$ then $392/17 = 23$.
Assume it is true when $n = p$. Therefore $3cdot 5^2p+1 +2^3p+1 = 17k $ where $k ∈ mathbbN $. Consider now $n=p+1$. Then
beginalign
&3cdot 5^2(p+1)+1 +2^3(p+1)+1=\
&3cdot 5^2p+1+2 + 2^3p+1+3=\
&3cdot5^2p+1cdot 5^2 + 2^3p+1cdot 2^3.
endalign
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.
induction divisibility
edited 2 mins ago
TheSimpliFire
11.5k62256
asked 44 mins ago
Shehan Tearz
874
2
When $n=p$ you assume that $3cdot 5^2p+1+2^3p+1 = 17k$, don't you?
– Gibbs
38 mins ago
You did. I just corrected the notation for products.
– Gibbs
32 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Use mathematical induction to prove that $3cdot 5^2n+1 +2^3n+1$
is divisible by $17$ for all $n ∈ mathbbN$.
I've tried to do it as follow.
If $n = 1$ then $392/17 = 23$.
Assume it is true when $n = p$. Therefore $3cdot 5^2p+1 +2^3p+1 = 17k $ where $k ∈ mathbbN $. Consider now $n=p+1$. Then
beginalign
&3cdot 5^2(p+1)+1 +2^3(p+1)+1=\
&3cdot 5^2p+1+2 + 2^3p+1+3=\
&3cdot5^2p+1cdot 5^2 + 2^3p+1cdot 2^3.
endalign
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.
induction divisibility
edited 2 mins ago
TheSimpliFire
11.5k62256
asked 44 mins ago
Shehan Tearz
874
Use mathematical induction to prove that $3cdot 5^2n+1 +2^3n+1$
is divisible by $17$ for all $n ∈ mathbbN$.
I've tried to do it as follow.
If $n = 1$ then $392/17 = 23$.
Assume it is true when $n = p$. Therefore $3cdot 5^2p+1 +2^3p+1 = 17k $ where $k ∈ mathbbN $. Consider now $n=p+1$. Then
beginalign
&3cdot 5^2(p+1)+1 +2^3(p+1)+1=\
&3cdot 5^2p+1+2 + 2^3p+1+3=\
&3cdot5^2p+1cdot 5^2 + 2^3p+1cdot 2^3.
endalign
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.
induction divisibility
induction divisibility
edited 2 mins ago
TheSimpliFire
11.5k62256
asked 44 mins ago
Shehan Tearz
874
edited 2 mins ago
TheSimpliFire
11.5k62256
asked 44 mins ago
Shehan Tearz
874
edited 2 mins ago
TheSimpliFire
11.5k62256
edited 2 mins ago
TheSimpliFire
11.5k62256
edited 2 mins ago
TheSimpliFire
11.5k62256
11.5k62256
asked 44 mins ago
Shehan Tearz
874
asked 44 mins ago
Shehan Tearz
874
asked 44 mins ago
Shehan Tearz
874
874
2
When $n=p$ you assume that $3cdot 5^2p+1+2^3p+1 = 17k$, don't you?
– Gibbs
38 mins ago
You did. I just corrected the notation for products.
– Gibbs
32 mins ago
add a comment |Â
2
When $n=p$ you assume that $3cdot 5^2p+1+2^3p+1 = 17k$, don't you?
– Gibbs
38 mins ago
You did. I just corrected the notation for products.
– Gibbs
32 mins ago
2
2
When $n=p$ you assume that $3cdot 5^2p+1+2^3p+1 = 17k$, don't you?
– Gibbs
38 mins ago
When $n=p$ you assume that $3cdot 5^2p+1+2^3p+1 = 17k$, don't you?
– Gibbs
38 mins ago
You did. I just corrected the notation for products.
– Gibbs
32 mins ago
You did. I just corrected the notation for products.
– Gibbs
32 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
Having shown that
$$3cdot5^2p+1+2^3p+1=17k$$
we have (continuing from the last line)
$$3cdot5^2p+15^2+2^3p+12^3=17(3cdot5^2p+1)+8(3cdot5^2p+1+2^3p+1)=17(3cdot5^2p+1+8k)$$
so the next expression is also divisible by 17, as required.
answered 38 mins ago
Parcly Taxel
35.5k136991
Can you please help me understand how you got into the second step because thats the step i cant understand. Thanks
– Shehan Tearz
25 mins ago
@ShehanTearz $3cdot5^2p+1cdot5^2=3cdot5^2p+1cdot17+3cdot5^2p+1cdot8$. Then I rearrange and collect the two terms with 8.
– Parcly Taxel
23 mins ago
add a comment |Â
up vote
2
down vote
Step $n+1:$
$3(5^2)5^2p+1+(2^3)2^3p+1=$
$3(17+8)5^2p+1 +8 cdot 2^3p+1=$
$8[3 cdot 5^2p+1+2^3p+1]+ 17cdot 3 cdot 5^2p+1.$
The first term in the above sum is divisible by $17$ (hypothesis), the second term is a multiple of $17$.
answered 22 mins ago
Peter Szilas
9,2362720
add a comment |Â
up vote
1
down vote
I'll start with the inductive step
$$3cdot5^2(n+1)+1+2^3(n+1)+1$$
$$3cdot5^2n+1+2+2^3n+1+3$$
$$3cdot25cdot5^2(n+1)+8cdot2^3n+1$$
We know that $$3cdot5^2n+1+2^3n+1=17k$$
Therefore $$3cdot5^2n+1=17k-2^3n+1$$
Plug this in:
$$25cdot(17k-2^3n+1)+8cdot2^3n+1$$
$$425kcdot-25cdot2^3n+1+8cdot2^3n+1$$
$$425kcdot-17cdot2^3n+1$$
$$17(25-2^3n+1)$$
answered 19 mins ago
Ethan Chan
716324
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Having shown that
$$3cdot5^2p+1+2^3p+1=17k$$
we have (continuing from the last line)
$$3cdot5^2p+15^2+2^3p+12^3=17(3cdot5^2p+1)+8(3cdot5^2p+1+2^3p+1)=17(3cdot5^2p+1+8k)$$
so the next expression is also divisible by 17, as required.
answered 38 mins ago
Parcly Taxel
35.5k136991
Can you please help me understand how you got into the second step because thats the step i cant understand. Thanks
– Shehan Tearz
25 mins ago
@ShehanTearz $3cdot5^2p+1cdot5^2=3cdot5^2p+1cdot17+3cdot5^2p+1cdot8$. Then I rearrange and collect the two terms with 8.
– Parcly Taxel
23 mins ago
add a comment |Â
up vote
6
down vote
accepted
Having shown that
$$3cdot5^2p+1+2^3p+1=17k$$
we have (continuing from the last line)
$$3cdot5^2p+15^2+2^3p+12^3=17(3cdot5^2p+1)+8(3cdot5^2p+1+2^3p+1)=17(3cdot5^2p+1+8k)$$
so the next expression is also divisible by 17, as required.
answered 38 mins ago
Parcly Taxel
35.5k136991
Can you please help me understand how you got into the second step because thats the step i cant understand. Thanks
– Shehan Tearz
25 mins ago
@ShehanTearz $3cdot5^2p+1cdot5^2=3cdot5^2p+1cdot17+3cdot5^2p+1cdot8$. Then I rearrange and collect the two terms with 8.
– Parcly Taxel
23 mins ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Having shown that
$$3cdot5^2p+1+2^3p+1=17k$$
we have (continuing from the last line)
$$3cdot5^2p+15^2+2^3p+12^3=17(3cdot5^2p+1)+8(3cdot5^2p+1+2^3p+1)=17(3cdot5^2p+1+8k)$$
so the next expression is also divisible by 17, as required.
answered 38 mins ago
Parcly Taxel
35.5k136991
Having shown that
$$3cdot5^2p+1+2^3p+1=17k$$
we have (continuing from the last line)
$$3cdot5^2p+15^2+2^3p+12^3=17(3cdot5^2p+1)+8(3cdot5^2p+1+2^3p+1)=17(3cdot5^2p+1+8k)$$
so the next expression is also divisible by 17, as required.
answered 38 mins ago
Parcly Taxel
35.5k136991
edited 28 mins ago
answered 38 mins ago
Parcly Taxel
35.5k136991
answered 38 mins ago
Parcly Taxel
35.5k136991
answered 38 mins ago
Parcly Taxel
35.5k136991
35.5k136991
Can you please help me understand how you got into the second step because thats the step i cant understand. Thanks
– Shehan Tearz
25 mins ago
@ShehanTearz $3cdot5^2p+1cdot5^2=3cdot5^2p+1cdot17+3cdot5^2p+1cdot8$. Then I rearrange and collect the two terms with 8.
– Parcly Taxel
23 mins ago
add a comment |Â
Can you please help me understand how you got into the second step because thats the step i cant understand. Thanks
– Shehan Tearz
25 mins ago
@ShehanTearz $3cdot5^2p+1cdot5^2=3cdot5^2p+1cdot17+3cdot5^2p+1cdot8$. Then I rearrange and collect the two terms with 8.
– Parcly Taxel
23 mins ago
Can you please help me understand how you got into the second step because thats the step i cant understand. Thanks
– Shehan Tearz
25 mins ago
Can you please help me understand how you got into the second step because thats the step i cant understand. Thanks
– Shehan Tearz
25 mins ago
@ShehanTearz $3cdot5^2p+1cdot5^2=3cdot5^2p+1cdot17+3cdot5^2p+1cdot8$. Then I rearrange and collect the two terms with 8.
– Parcly Taxel
23 mins ago
@ShehanTearz $3cdot5^2p+1cdot5^2=3cdot5^2p+1cdot17+3cdot5^2p+1cdot8$. Then I rearrange and collect the two terms with 8.
– Parcly Taxel
23 mins ago
add a comment |Â
up vote
2
down vote
Step $n+1:$
$3(5^2)5^2p+1+(2^3)2^3p+1=$
$3(17+8)5^2p+1 +8 cdot 2^3p+1=$
$8[3 cdot 5^2p+1+2^3p+1]+ 17cdot 3 cdot 5^2p+1.$
The first term in the above sum is divisible by $17$ (hypothesis), the second term is a multiple of $17$.
answered 22 mins ago
Peter Szilas
9,2362720
add a comment |Â
up vote
2
down vote
Step $n+1:$
$3(5^2)5^2p+1+(2^3)2^3p+1=$
$3(17+8)5^2p+1 +8 cdot 2^3p+1=$
$8[3 cdot 5^2p+1+2^3p+1]+ 17cdot 3 cdot 5^2p+1.$
The first term in the above sum is divisible by $17$ (hypothesis), the second term is a multiple of $17$.
answered 22 mins ago
Peter Szilas
9,2362720
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Step $n+1:$
$3(5^2)5^2p+1+(2^3)2^3p+1=$
$3(17+8)5^2p+1 +8 cdot 2^3p+1=$
$8[3 cdot 5^2p+1+2^3p+1]+ 17cdot 3 cdot 5^2p+1.$
The first term in the above sum is divisible by $17$ (hypothesis), the second term is a multiple of $17$.
answered 22 mins ago
Peter Szilas
9,2362720
Step $n+1:$
$3(5^2)5^2p+1+(2^3)2^3p+1=$
$3(17+8)5^2p+1 +8 cdot 2^3p+1=$
$8[3 cdot 5^2p+1+2^3p+1]+ 17cdot 3 cdot 5^2p+1.$
The first term in the above sum is divisible by $17$ (hypothesis), the second term is a multiple of $17$.
answered 22 mins ago
Peter Szilas
9,2362720
answered 22 mins ago
Peter Szilas
9,2362720
answered 22 mins ago
Peter Szilas
9,2362720
answered 22 mins ago
Peter Szilas
9,2362720
9,2362720
add a comment |Â
add a comment |Â
up vote
1
down vote
I'll start with the inductive step
$$3cdot5^2(n+1)+1+2^3(n+1)+1$$
$$3cdot5^2n+1+2+2^3n+1+3$$
$$3cdot25cdot5^2(n+1)+8cdot2^3n+1$$
We know that $$3cdot5^2n+1+2^3n+1=17k$$
Therefore $$3cdot5^2n+1=17k-2^3n+1$$
Plug this in:
$$25cdot(17k-2^3n+1)+8cdot2^3n+1$$
$$425kcdot-25cdot2^3n+1+8cdot2^3n+1$$
$$425kcdot-17cdot2^3n+1$$
$$17(25-2^3n+1)$$
answered 19 mins ago
Ethan Chan
716324
add a comment |Â
up vote
1
down vote
I'll start with the inductive step
$$3cdot5^2(n+1)+1+2^3(n+1)+1$$
$$3cdot5^2n+1+2+2^3n+1+3$$
$$3cdot25cdot5^2(n+1)+8cdot2^3n+1$$
We know that $$3cdot5^2n+1+2^3n+1=17k$$
Therefore $$3cdot5^2n+1=17k-2^3n+1$$
Plug this in:
$$25cdot(17k-2^3n+1)+8cdot2^3n+1$$
$$425kcdot-25cdot2^3n+1+8cdot2^3n+1$$
$$425kcdot-17cdot2^3n+1$$
$$17(25-2^3n+1)$$
answered 19 mins ago
Ethan Chan
716324
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'll start with the inductive step
$$3cdot5^2(n+1)+1+2^3(n+1)+1$$
$$3cdot5^2n+1+2+2^3n+1+3$$
$$3cdot25cdot5^2(n+1)+8cdot2^3n+1$$
We know that $$3cdot5^2n+1+2^3n+1=17k$$
Therefore $$3cdot5^2n+1=17k-2^3n+1$$
Plug this in:
$$25cdot(17k-2^3n+1)+8cdot2^3n+1$$
$$425kcdot-25cdot2^3n+1+8cdot2^3n+1$$
$$425kcdot-17cdot2^3n+1$$
$$17(25-2^3n+1)$$
answered 19 mins ago
Ethan Chan
716324
I'll start with the inductive step
$$3cdot5^2(n+1)+1+2^3(n+1)+1$$
$$3cdot5^2n+1+2+2^3n+1+3$$
$$3cdot25cdot5^2(n+1)+8cdot2^3n+1$$
We know that $$3cdot5^2n+1+2^3n+1=17k$$
Therefore $$3cdot5^2n+1=17k-2^3n+1$$
Plug this in:
$$25cdot(17k-2^3n+1)+8cdot2^3n+1$$
$$425kcdot-25cdot2^3n+1+8cdot2^3n+1$$
$$425kcdot-17cdot2^3n+1$$
$$17(25-2^3n+1)$$
answered 19 mins ago
Ethan Chan
716324
answered 19 mins ago
Ethan Chan
716324
answered 19 mins ago
Ethan Chan
716324
answered 19 mins ago
Ethan Chan
716324
716324
add a comment |Â
add a comment |Â
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2
When $n=p$ you assume that $3cdot 5^2p+1+2^3p+1 = 17k$, don't you?
– Gibbs
38 mins ago
You did. I just corrected the notation for products.
– Gibbs
32 mins ago