What is the last digit of $n^5-5n^3+4n+7, forall n in mathbb{N}$? [closed]

2

$begingroup$

If $n$ is a natural number larger than two, what is the last digit of the expression $n^5-5n^3+4n+7$? How do you prove it ?

number-theory

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edited Feb 3 at 12:18

jvdhooft

5,65961641

asked Feb 3 at 12:05

dburanszkidburanszki

183

$endgroup$

closed as off-topic by YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin Feb 12 at 11:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts? Have you tried several $n$ at least?
  $endgroup$
  – lulu
  Feb 3 at 12:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The last digit is $7$,maybe this helps you to come up with a proof.
  $endgroup$
  – Alexdanut
  Feb 3 at 12:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes i have already figure it out that the last digit is seven, however i cannot prove it.
  $endgroup$
  – dburanszki
  Feb 3 at 12:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
  $endgroup$
  – Yanko
  Feb 3 at 12:20
  

  
  
  
  

add a comment | 

2

$begingroup$

If $n$ is a natural number larger than two, what is the last digit of the expression $n^5-5n^3+4n+7$? How do you prove it ?

number-theory

share|cite|improve this question

edited Feb 3 at 12:18

jvdhooft

5,65961641

asked Feb 3 at 12:05

dburanszkidburanszki

183

$endgroup$

closed as off-topic by YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin Feb 12 at 11:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts? Have you tried several $n$ at least?
  $endgroup$
  – lulu
  Feb 3 at 12:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The last digit is $7$,maybe this helps you to come up with a proof.
  $endgroup$
  – Alexdanut
  Feb 3 at 12:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes i have already figure it out that the last digit is seven, however i cannot prove it.
  $endgroup$
  – dburanszki
  Feb 3 at 12:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
  $endgroup$
  – Yanko
  Feb 3 at 12:20
  

  
  
  
  

add a comment | 

2

2

2

2

$begingroup$

If $n$ is a natural number larger than two, what is the last digit of the expression $n^5-5n^3+4n+7$? How do you prove it ?

number-theory

share|cite|improve this question

edited Feb 3 at 12:18

jvdhooft

5,65961641

asked Feb 3 at 12:05

dburanszkidburanszki

183

$endgroup$

If $n$ is a natural number larger than two, what is the last digit of the expression $n^5-5n^3+4n+7$? How do you prove it ?

number-theory

number-theory

share|cite|improve this question

edited Feb 3 at 12:18

jvdhooft

5,65961641

asked Feb 3 at 12:05

dburanszkidburanszki

183

share|cite|improve this question

edited Feb 3 at 12:18

jvdhooft

5,65961641

asked Feb 3 at 12:05

dburanszkidburanszki

183

share|cite|improve this question

share|cite|improve this question

edited Feb 3 at 12:18

jvdhooft

5,65961641

edited Feb 3 at 12:18

jvdhooft

5,65961641

edited Feb 3 at 12:18

jvdhooft

5,65961641

5,65961641

asked Feb 3 at 12:05

dburanszkidburanszki

183

asked Feb 3 at 12:05

dburanszkidburanszki

183

asked Feb 3 at 12:05

dburanszkidburanszki

183

183

closed as off-topic by YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin Feb 12 at 11:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin Feb 12 at 11:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, Cesareo, José Carlos Santos, Kemono Chen, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts? Have you tried several $n$ at least?
  $endgroup$
  – lulu
  Feb 3 at 12:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The last digit is $7$,maybe this helps you to come up with a proof.
  $endgroup$
  – Alexdanut
  Feb 3 at 12:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes i have already figure it out that the last digit is seven, however i cannot prove it.
  $endgroup$
  – dburanszki
  Feb 3 at 12:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
  $endgroup$
  – Yanko
  Feb 3 at 12:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts? Have you tried several $n$ at least?
  $endgroup$
  – lulu
  Feb 3 at 12:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The last digit is $7$,maybe this helps you to come up with a proof.
  $endgroup$
  – Alexdanut
  Feb 3 at 12:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes i have already figure it out that the last digit is seven, however i cannot prove it.
  $endgroup$
  – dburanszki
  Feb 3 at 12:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
  $endgroup$
  – Yanko
  Feb 3 at 12:20
  

  
  
  
  

1

1

$begingroup$
What are your thoughts? Have you tried several $n$ at least?
$endgroup$
– lulu
Feb 3 at 12:06

$begingroup$
What are your thoughts? Have you tried several $n$ at least?
$endgroup$
– lulu
Feb 3 at 12:06

1

1

$begingroup$
The last digit is $7$,maybe this helps you to come up with a proof.
$endgroup$
– Alexdanut
Feb 3 at 12:11

$begingroup$
The last digit is $7$,maybe this helps you to come up with a proof.
$endgroup$
– Alexdanut
Feb 3 at 12:11

$begingroup$
yes i have already figure it out that the last digit is seven, however i cannot prove it.
$endgroup$
– dburanszki
Feb 3 at 12:14

$begingroup$
yes i have already figure it out that the last digit is seven, however i cannot prove it.
$endgroup$
– dburanszki
Feb 3 at 12:14

2

2

$begingroup$
The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
$endgroup$
– Yanko
Feb 3 at 12:20

$begingroup$
The assumption that $n>2$ is unnecessary. You get $7$ even for $n=0,1$.
$endgroup$
– Yanko
Feb 3 at 12:20

add a comment | 

5 Answers
5

active

oldest

votes

2

$begingroup$

If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
$$n^5−5n^3+4n+7 equiv 7 mod{10}$$
$$n^5−5n^3+4n equiv 0 mod{10}$$
$$n(n^4−5n^2+4) equiv 0 mod{10}$$
$$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
$$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
$$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$

And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.

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answered Feb 3 at 12:13

Peter ForemanPeter Foreman

7,5361320

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add a comment | 

3

$begingroup$

Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$

share|cite|improve this answer

edited Feb 3 at 12:15

Yanko

8,4692830

answered Feb 3 at 12:13

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No! 25 seconds before me...
  $endgroup$
  – Peter Foreman
  Feb 3 at 12:14
  

  
  
  
  

add a comment | 

2

$begingroup$

Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).

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answered Feb 3 at 12:18

BonbonBonbon

45118

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really nice one.
  $endgroup$
  – Yanko
  Feb 3 at 12:19
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.

Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.

So you only need to check these $10$ cases.

Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.

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answered Feb 3 at 12:11

YankoYanko

8,4692830

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add a comment | 

1

$begingroup$

Another variant:

$$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$

so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
$$n^5-5n^3+4n+7equiv 7mod 10. $$

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edited Feb 3 at 12:47

answered Feb 3 at 12:41

BernardBernard

124k742117

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add a comment | 

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
$$n^5−5n^3+4n+7 equiv 7 mod{10}$$
$$n^5−5n^3+4n equiv 0 mod{10}$$
$$n(n^4−5n^2+4) equiv 0 mod{10}$$
$$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
$$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
$$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$

And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.

share|cite|improve this answer

answered Feb 3 at 12:13

Peter ForemanPeter Foreman

7,5361320

$endgroup$

add a comment | 

2

$begingroup$

If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
$$n^5−5n^3+4n+7 equiv 7 mod{10}$$
$$n^5−5n^3+4n equiv 0 mod{10}$$
$$n(n^4−5n^2+4) equiv 0 mod{10}$$
$$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
$$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
$$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$

And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.

share|cite|improve this answer

answered Feb 3 at 12:13

Peter ForemanPeter Foreman

7,5361320

$endgroup$

add a comment | 

2

2

2

$begingroup$

If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
$$n^5−5n^3+4n+7 equiv 7 mod{10}$$
$$n^5−5n^3+4n equiv 0 mod{10}$$
$$n(n^4−5n^2+4) equiv 0 mod{10}$$
$$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
$$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
$$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$

And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.

share|cite|improve this answer

answered Feb 3 at 12:13

Peter ForemanPeter Foreman

7,5361320

$endgroup$

If $n in mathbb{N}$ we can try several values to see that the digit of the expression seems to always be $7$. So we need to prove that:
$$n^5−5n^3+4n+7 equiv 7 mod{10}$$
$$n^5−5n^3+4n equiv 0 mod{10}$$
$$n(n^4−5n^2+4) equiv 0 mod{10}$$
$$n(n^2-1)(n^2-4) equiv 0 mod{10}$$
$$n(n+1)(n-1)(n+2)(n-2) equiv 0 mod{10}$$
$$(n-2)(n-1)n(n+1)(n+2) equiv 0 mod{10}$$

And the above is true for all $n in mathbb{N}$ as $5$ consecutive numbers must contain a number divisible by $2$ and one divisible by $5$. Hence $10|(n-2)(n-1)n(n+1)(n+2)$ for all $n in mathbb{N}$.

share|cite|improve this answer

answered Feb 3 at 12:13

Peter ForemanPeter Foreman

7,5361320

share|cite|improve this answer

share|cite|improve this answer

answered Feb 3 at 12:13

Peter ForemanPeter Foreman

7,5361320

answered Feb 3 at 12:13

Peter ForemanPeter Foreman

7,5361320

answered Feb 3 at 12:13

Peter ForemanPeter Foreman

7,5361320

7,5361320

add a comment | 

add a comment | 

3

$begingroup$

Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$

share|cite|improve this answer

edited Feb 3 at 12:15

Yanko

8,4692830

answered Feb 3 at 12:13

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No! 25 seconds before me...
  $endgroup$
  – Peter Foreman
  Feb 3 at 12:14
  

  
  
  
  

add a comment | 

3

$begingroup$

Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$

share|cite|improve this answer

edited Feb 3 at 12:15

Yanko

8,4692830

answered Feb 3 at 12:13

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No! 25 seconds before me...
  $endgroup$
  – Peter Foreman
  Feb 3 at 12:14
  

  
  
  
  

add a comment | 

3

3

3

$begingroup$

Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$

share|cite|improve this answer

edited Feb 3 at 12:15

Yanko

8,4692830

answered Feb 3 at 12:13

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

Hint: Your term can be written in the form $$7+(n-2)(n-1)n(n+1)(n+2)$$

share|cite|improve this answer

edited Feb 3 at 12:15

Yanko

8,4692830

answered Feb 3 at 12:13

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

share|cite|improve this answer

share|cite|improve this answer

edited Feb 3 at 12:15

Yanko

8,4692830

edited Feb 3 at 12:15

Yanko

8,4692830

edited Feb 3 at 12:15

Yanko

8,4692830

8,4692830

answered Feb 3 at 12:13

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

answered Feb 3 at 12:13

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

answered Feb 3 at 12:13

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

79k42867

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No! 25 seconds before me...
  $endgroup$
  – Peter Foreman
  Feb 3 at 12:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No! 25 seconds before me...
  $endgroup$
  – Peter Foreman
  Feb 3 at 12:14
  

  
  
  
  

$begingroup$
No! 25 seconds before me...
$endgroup$
– Peter Foreman
Feb 3 at 12:14

$begingroup$
No! 25 seconds before me...
$endgroup$
– Peter Foreman
Feb 3 at 12:14

add a comment | 

2

$begingroup$

Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).

share|cite|improve this answer

answered Feb 3 at 12:18

BonbonBonbon

45118

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really nice one.
  $endgroup$
  – Yanko
  Feb 3 at 12:19
  

  
  
  
  

add a comment | 

2

$begingroup$

Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).

share|cite|improve this answer

answered Feb 3 at 12:18

BonbonBonbon

45118

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really nice one.
  $endgroup$
  – Yanko
  Feb 3 at 12:19
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).

share|cite|improve this answer

answered Feb 3 at 12:18

BonbonBonbon

45118

$endgroup$

Need to compute $n^5-5n^3+4n+7 $ (mod $10$). By Fermat's little theorem, $n^5-5n^3+4n+7equiv n-5n^3+4n+7equiv 2 $ (mod $5$), and $n^5-5n^3+4n+7equiv n-5n+4n+7equiv 1 $ (mod $2$), hence $n^5-5n^3+4n+7equiv 7 $ (mod $10$).

share|cite|improve this answer

answered Feb 3 at 12:18

BonbonBonbon

45118

share|cite|improve this answer

share|cite|improve this answer

answered Feb 3 at 12:18

BonbonBonbon

45118

answered Feb 3 at 12:18

BonbonBonbon

45118

answered Feb 3 at 12:18

BonbonBonbon

45118

45118

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really nice one.
  $endgroup$
  – Yanko
  Feb 3 at 12:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really nice one.
  $endgroup$
  – Yanko
  Feb 3 at 12:19
  

  
  
  
  

$begingroup$
Really nice one.
$endgroup$
– Yanko
Feb 3 at 12:19

$begingroup$
Really nice one.
$endgroup$
– Yanko
Feb 3 at 12:19

add a comment | 

1

$begingroup$

Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.

Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.

So you only need to check these $10$ cases.

Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.

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answered Feb 3 at 12:11

YankoYanko

8,4692830

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1

$begingroup$

Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.

Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.

So you only need to check these $10$ cases.

Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.

share|cite|improve this answer

answered Feb 3 at 12:11

YankoYanko

8,4692830

$endgroup$

add a comment | 

1

1

1

$begingroup$

Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.

Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.

So you only need to check these $10$ cases.

Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.

share|cite|improve this answer

answered Feb 3 at 12:11

YankoYanko

8,4692830

$endgroup$

Let $n$ be a number whose last digit is $din{0,1,2,...,9}$.

Then the last digit of $n^5 - 5n^3+4n^2+7$ is the last digit of $d^5-5d^3+4d^2+7$.

So you only need to check these $10$ cases.

Another option: Note that $n^5-5n^3+4n^2 = (n-2)(n-1)n(n+1)(n+2)$ and so it has to be divisible by $10$.

share|cite|improve this answer

answered Feb 3 at 12:11

YankoYanko

8,4692830

share|cite|improve this answer

share|cite|improve this answer

answered Feb 3 at 12:11

YankoYanko

8,4692830

answered Feb 3 at 12:11

YankoYanko

8,4692830

answered Feb 3 at 12:11

YankoYanko

8,4692830

8,4692830

add a comment | 

add a comment | 

1

$begingroup$

Another variant:

$$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$

so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
$$n^5-5n^3+4n+7equiv 7mod 10. $$

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edited Feb 3 at 12:47

answered Feb 3 at 12:41

BernardBernard

124k742117

$endgroup$

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1

$begingroup$

Another variant:

$$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$

so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
$$n^5-5n^3+4n+7equiv 7mod 10. $$

share|cite|improve this answer

edited Feb 3 at 12:47

answered Feb 3 at 12:41

BernardBernard

124k742117

$endgroup$

add a comment | 

1

1

1

$begingroup$

Another variant:

$$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$

so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
$$n^5-5n^3+4n+7equiv 7mod 10. $$

share|cite|improve this answer

edited Feb 3 at 12:47

answered Feb 3 at 12:41

BernardBernard

124k742117

$endgroup$

Another variant:

$$begin{array}{l}n^5-5n^3+4nequiv n^5+4nequiv n-n=0mod5\
n^5-5n^3+4nequiv n^5-n^3equiv n-n=0mod 2
end{array}biggr} quadtext{by } textit{ lil' Fermat}, $$

so $;n^5-5n^3+4nequiv 0mod 10$, and ultimately
$$n^5-5n^3+4n+7equiv 7mod 10. $$

share|cite|improve this answer

edited Feb 3 at 12:47

answered Feb 3 at 12:41

BernardBernard

124k742117

share|cite|improve this answer

share|cite|improve this answer

edited Feb 3 at 12:47

edited Feb 3 at 12:47

edited Feb 3 at 12:47

answered Feb 3 at 12:41

BernardBernard

124k742117

answered Feb 3 at 12:41

BernardBernard

124k742117

answered Feb 3 at 12:41

BernardBernard

124k742117

124k742117

add a comment | 

add a comment | 

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