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The proof of perfect square having $0,1,4,5,6,9$ as units digit

I need explanation for this solution for the proof. (Perfect square ends with 0,1,4,5,6,9)Prove that if n is a perfect square, then n+2 is not a perfect squareA perfect square and a perfect cube: is it also a perfect sixth power?A number is a perfect square if and only if it has odd number of positive divisorsPermutation: How many numbers of n digits are possible for which product of its digits is a perfect square.Prove Perfect square of the form 4k or 4k+1Finding the last digit of an integer $n$“Prove that 98765432 is not the square of an integer” vs “Deduce that 1234567 is not a perfect square”Help understanding proof verification: Prove if n is a perfect square, n+2 is not a perfect squareShowing that the $n$-th positive integer that is not a perfect square is $n+sqrtn$, where $$ is the “closest integer” function

0

$begingroup$

To prove this, we defined $n$ as an integer, then said that it must have the form: $n= 10k +b$ where $b=0,1,....,9$

This is the step that I didn't get, why this integer has to have this form to prove that a perfect square has $0,1,4,5,6,9$ as ones digit ?

discrete-mathematics

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edited Mar 11 at 10:56

Vinyl_cape_jawa

3,33211433

asked Sep 17 '15 at 8:06

user271573user271573

1

$endgroup$

add a comment | 

0

$begingroup$

To prove this, we defined $n$ as an integer, then said that it must have the form: $n= 10k +b$ where $b=0,1,....,9$

This is the step that I didn't get, why this integer has to have this form to prove that a perfect square has $0,1,4,5,6,9$ as ones digit ?

discrete-mathematics

share|cite|improve this question

edited Mar 11 at 10:56

Vinyl_cape_jawa

3,33211433

asked Sep 17 '15 at 8:06

user271573user271573

1

$endgroup$

add a comment | 

$begingroup$

To prove this, we defined $n$ as an integer, then said that it must have the form: $n= 10k +b$ where $b=0,1,....,9$

This is the step that I didn't get, why this integer has to have this form to prove that a perfect square has $0,1,4,5,6,9$ as ones digit ?

discrete-mathematics

share|cite|improve this question

edited Mar 11 at 10:56

Vinyl_cape_jawa

3,33211433

asked Sep 17 '15 at 8:06

user271573user271573

1

$endgroup$

share|cite|improve this question

edited Mar 11 at 10:56

Vinyl_cape_jawa

3,33211433

asked Sep 17 '15 at 8:06

user271573user271573

1

share|cite|improve this question

edited Mar 11 at 10:56

Vinyl_cape_jawa

3,33211433

asked Sep 17 '15 at 8:06

user271573user271573

1

edited Mar 11 at 10:56

Vinyl_cape_jawa

3,33211433

edited Mar 11 at 10:56

Vinyl_cape_jawa

3,33211433

asked Sep 17 '15 at 8:06

user271573user271573

1

asked Sep 17 '15 at 8:06

user271573user271573

1

2 Answers
2

active

oldest

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0

$begingroup$

We, human, have ten fingers, and the last digit of a perfect square is never 2, 3, 7, or 8. Indeed,

$$n^2 = (10k + b)^2 = 100k^2 + 20kb + b^2 equiv b^2 mod 10$$

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edited Sep 17 '15 at 8:23

answered Sep 17 '15 at 8:14

user160928

$endgroup$

add a comment | 

0

$begingroup$

This form is needed to be able to separate the ones digit from the rest of the number.

If $ninmathbbZ$ then $n$ can be written in the form $n=10k+b$ with $kinmathbbZ$ and $bin0,1,2,ldots, 9$.

Now taking the square of $n$ gives

$$
n^2=(10k+b)^2=100k^2+20kb+b^2equiv_10b^2.
$$

So the last digit of $n^2$ is the last digit of the square of $b$ and since $b$ is a digit it can only be ten different things. So it is deffinitely doable by hand.

$$0^2equiv_100\1^2equiv_101\ 2^2equiv_104\ 3^2equiv_109\ 4^2equiv_106\ 5^2equiv_105\6^2equiv_106\ 7^2equiv_109\ 8^2equiv_104\ 9^2equiv_101 $$

share|cite|improve this answer

edited Mar 11 at 22:21

answered Mar 11 at 10:55

Vinyl_cape_jawaVinyl_cape_jawa

3,33211433

$endgroup$

add a comment | 

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0

$begingroup$

This form is needed to be able to separate the ones digit from the rest of the number.

If $ninmathbbZ$ then $n$ can be written in the form $n=10k+b$ with $kinmathbbZ$ and $bin0,1,2,ldots, 9$.

Now taking the square of $n$ gives

$$
n^2=(10k+b)^2=100k^2+20kb+b^2equiv_10b^2.
$$

So the last digit of $n^2$ is the last digit of the square of $b$ and since $b$ is a digit it can only be ten different things. So it is deffinitely doable by hand.

$$0^2equiv_100\1^2equiv_101\ 2^2equiv_104\ 3^2equiv_109\ 4^2equiv_106\ 5^2equiv_105\6^2equiv_106\ 7^2equiv_109\ 8^2equiv_104\ 9^2equiv_101 $$

share|cite|improve this answer

edited Mar 11 at 22:21

answered Mar 11 at 10:55

Vinyl_cape_jawaVinyl_cape_jawa

3,33211433

$endgroup$

add a comment | 

0

$begingroup$

This form is needed to be able to separate the ones digit from the rest of the number.

If $ninmathbbZ$ then $n$ can be written in the form $n=10k+b$ with $kinmathbbZ$ and $bin0,1,2,ldots, 9$.

Now taking the square of $n$ gives

$$
n^2=(10k+b)^2=100k^2+20kb+b^2equiv_10b^2.
$$

So the last digit of $n^2$ is the last digit of the square of $b$ and since $b$ is a digit it can only be ten different things. So it is deffinitely doable by hand.

$$0^2equiv_100\1^2equiv_101\ 2^2equiv_104\ 3^2equiv_109\ 4^2equiv_106\ 5^2equiv_105\6^2equiv_106\ 7^2equiv_109\ 8^2equiv_104\ 9^2equiv_101 $$

share|cite|improve this answer

edited Mar 11 at 22:21

answered Mar 11 at 10:55

Vinyl_cape_jawaVinyl_cape_jawa

3,33211433

$endgroup$

add a comment | 

$begingroup$

This form is needed to be able to separate the ones digit from the rest of the number.

If $ninmathbbZ$ then $n$ can be written in the form $n=10k+b$ with $kinmathbbZ$ and $bin0,1,2,ldots, 9$.

Now taking the square of $n$ gives

$$
n^2=(10k+b)^2=100k^2+20kb+b^2equiv_10b^2.
$$

So the last digit of $n^2$ is the last digit of the square of $b$ and since $b$ is a digit it can only be ten different things. So it is deffinitely doable by hand.

$$0^2equiv_100\1^2equiv_101\ 2^2equiv_104\ 3^2equiv_109\ 4^2equiv_106\ 5^2equiv_105\6^2equiv_106\ 7^2equiv_109\ 8^2equiv_104\ 9^2equiv_101 $$

share|cite|improve this answer

edited Mar 11 at 22:21

answered Mar 11 at 10:55

Vinyl_cape_jawaVinyl_cape_jawa

3,33211433

$endgroup$

share|cite|improve this answer

edited Mar 11 at 22:21

answered Mar 11 at 10:55

Vinyl_cape_jawaVinyl_cape_jawa

3,33211433

answered Mar 11 at 10:55

Vinyl_cape_jawaVinyl_cape_jawa

3,33211433

answered Mar 11 at 10:55

Vinyl_cape_jawaVinyl_cape_jawa

3,33211433